[UHS (2024)]
The principle that states if a stress is applied to a system at equilibrium the system nullify the effect of stress as far as possible is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This is the literal textbook definition of the fundamental principle of chemical equilibrium dynamics.
Solution:- In 1884, French chemist Henri Le Chatelier postulated that "If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium."
- The "stress" refers to changes in concentration, pressure, volume, or temperature.
- This is universally known as Le-Chatelier's Principle.
Why other options are incorrect:Haber developed a specific industrial process for ammonia. Boyle's and Charles's laws relate strictly to the physical properties of ideal gases (Volume vs. Pressure/Temperature), not chemical equilibrium.
[UHS (2024)]
Identify the correct option required for the maximum yield of ammonia in Haber's process:
A
High pressure low temperature continual removal of ammonia
B
Low pressure low temperature continual removal of ammonia
C
High pressure high temperature continual removal of ammonia
D
High pressure low temperature continual addition of ammonia
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Maximizing yield in an industrial reversible reaction requires manipulating pressure, temperature, and concentrations according to Le Chatelier's Principle.
Formula:$$ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \quad \Delta H < 0 $$
Solution:- High Pressure: Shifts equilibrium right because there are fewer moles of gas on the product side (4 moles vs 2 moles).
- Low Temperature: Shifts equilibrium right because the forward reaction is exothermic (releases heat).
- Continual Removal: Dropping the concentration of \( \text{NH}_3 \) forces the system to constantly try and replace it, preventing equilibrium from ever halting the reaction.
- Combining these three factors yields the absolute maximum amount of product.
Why other options are incorrect:Low pressure favors reactants. High temperature favors reactants. Continual addition of ammonia would violently drive the reaction backwards.
[UHS (2024)]
Consider the following reaction in the equilibrium and tell addition of which will turn the cloudy solution into clear solution?
$$ \text{BiCl}_3 + \text{H}_2\text{O} \rightleftharpoons \text{BiOCl} + 2\text{HCl} $$
B
\( \text{H}_2\text{O} \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Bismuth oxychloride (\( \text{BiOCl} \)) is highly insoluble in water and forms a thick, cloudy white precipitate. To clear the solution, the equilibrium must be forced backward to dissolve the solid.
Formula:$$ \text{BiCl}_{3(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{BiOCl}_{(s)} \text{ [Cloudy]} + 2\text{HCl}_{(aq)} $$
Solution:- The cloudiness is entirely due to the presence of the product, solid \( \text{BiOCl} \).
- To make the solution clear, we must push the equilibrium to the left (towards the fully soluble \( \text{BiCl}_3 \) reactants).
- According to Le Chatelier's Principle, adding an excess of a product forces the reaction backward.
- Therefore, adding concentrated \( \text{HCl} \) increases product concentration, forcing the solid \( \text{BiOCl} \) to react and dissolve back into clear \( \text{BiCl}_3 \).
Why other options are incorrect:Adding \( \text{BiCl}_3 \) or \( \text{H}_2\text{O} \) (reactants) shifts the reaction forward, making it even cloudier. Adding \( \text{BiOCl} \) directly adds more cloudy precipitate.
[UHS (2024)]
In endothermic reaction, the heat content of the
A
Reactants and products is equal
B
Reactants is more than that of products
C
Products is more than that of reactants
D
Reactants & products will not change
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:An endothermic reaction fundamentally involves the absorption of thermal energy from the surroundings, converting it into internal chemical potential energy (heat content or enthalpy, \( H \)).
Formula:$$ \Delta H = H_{\text{products}} - H_{\text{reactants}} > 0 $$
Solution:- Because energy is absorbed during the reaction, this energy is stored in the chemical bonds of the newly formed products.
- Consequently, the total internal enthalpy (heat content) of the system increases.
- This means the final state (Products) possesses a higher heat content than the initial state (Reactants).
- Therefore, the heat content of the Products is more than that of reactants.
Why other options are incorrect:If they were equal, \( \Delta H = 0 \). If reactants had more, energy would be released (an exothermic reaction).
[SZABMU (2024)]
Chemical equilibrium given below will shift to backward direction by
$$ 2\text{NO} + \text{O}_2 \rightleftharpoons 2\text{NO}_2 + \text{Heat} $$
A
Decreasing pressure and increasing temperature
B
Decreasing the temperature
C
Increasing the temperature
D
Increasing the pressure
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:To shift a reaction in the backward (reactant) direction, we must manipulate conditions against the thermodynamic and stoichiometric flow of the forward reaction.
Solution:- Temperature: The forward reaction is exothermic (produces heat). Adding heat (increasing temperature) creates an excess on the product side, forcing the equilibrium to shift backward to consume it.
- Pressure: There are 3 moles of gaseous reactants (\( 2\text{NO} + 1\text{O}_2 \)) and only 2 moles of gaseous products (\( 2\text{NO}_2 \)). Decreasing pressure shifts equilibrium toward the side with more moles of gas to restore pressure. This side is the reactant side (left).
- Therefore, combining both Decreasing pressure and increasing temperature effectively forces the reaction backward.
Why other options are incorrect:Decreasing temperature (Options B/C) favors the forward exothermic reaction. Increasing pressure (Option D) favors the forward reaction (fewer moles).
[SZABMU (2024)]
Which of the following mixture will constitute the buffer solution?
A
Acetic acid & sodium acetate
C
Acetic acid and ammonium acetate
D
Ammonia & ammonium acetate
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:A classical chemical buffer solution must consist of a weak acid and its conjugate base (supplied via a highly soluble salt), or a weak base and its conjugate acid.
Formula:$$ \text{Weak Acid (HA)} + \text{Salt of Conjugate Base (A}^- \text{)} $$
Solution:- Acetic acid (\( \text{CH}_3\text{COOH} \)) is a classic weak acid.
- Sodium acetate (\( \text{CH}_3\text{COONa} \)) is a highly soluble salt that fully dissociates to provide the acetate ion (\( \text{CH}_3\text{COO}^- \)), which is the direct conjugate base of acetic acid.
- Mixing these two creates a solution containing both an acid to neutralize added \( \text{OH}^- \) and a base to neutralize added \( \text{H}^+ \). This is the definition of an acidic buffer.
Why other options are incorrect:Acetic acid and ammonia are an acid and a base that will just neutralize each other. Ammonium acetate does not provide the proper counter-ion metal (like Na or K) to act solely as a robust conjugate base salt for acetic acid in a simple binary buffer.
[SZABMU (2024)]
According to law of mass action, \( K_p > K_c \) when reaction occurs with ____.
A
Decrease in volume on product side
B
Increase in volume on product side
C
Increase in volume on reactant side
D
Simultaneous increase and decrease
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The mathematical relationship between \( K_p \) and \( K_c \) relies on \( \Delta n \) (change in gaseous moles). According to Avogadro's law, volume is directly proportional to moles of gas at constant T and P.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- For \( K_p \) to be mathematically greater than \( K_c \), the term \( \Delta n \) must be positive (\( \Delta n > 0 \)).
- \( \Delta n = (\text{moles of products}) - (\text{moles of reactants}) \).
- A positive \( \Delta n \) means there are more moles of gas on the product side than the reactant side.
- Because more moles occupy more volume, this translates directly to an increase in volume on the product side.
Why other options are incorrect:A decrease in volume on the product side implies \( \Delta n < 0 \), which would make \( K_p < K_c \).
[SZABMU (2024)]
(Out of syllabus)
What will be the molarity of HCl solution with pH=4?
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:pH is the negative logarithm (base 10) of the hydrogen ion concentration. HCl is a strong acid, meaning its molarity is directly equal to the hydrogen ion concentration.
Formula:$$ [\text{H}^+] = 10^{-\text{pH}} $$
Solution:- The given pH is 4.
- Plug this into the inverse logarithm formula: \( [\text{H}^+] = 10^{-4} \text{ M} \).
- Convert scientific notation to decimal form: \( 10^{-4} = 0.0001 \text{ M} \).
- Since HCl fully dissociates (\( \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \)), the initial molarity of the HCl solution is exactly 0.0001 M.
Why other options are incorrect:Option D (4) is the pH value itself, not the concentration. Options B and C introduce a '4' into the significant figures, which is a common math error confusing the exponent with the coefficient.
[SZABMU (2024)]
(Out of syllabus)
If weak acid is diluted with water, then \( \text{H}^+ \) ions concentration will ____.
B
Gradually decreases then increase
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Dilution affects both the volume of the solution and the degree of ionization (Ostwald's Dilution Law), but the overwhelming effect of the added volume dominates the absolute concentration metric (Molarity).
Formula:$$ C_1V_1 = C_2V_2 \quad \text{and} \quad \alpha = \sqrt{\frac{K_a}{C}} $$
Solution:- Adding water dramatically increases the total volume (\( V \)) of the solution.
- According to Ostwald's law, dilution does cause a weak acid to ionize more (\( \alpha \) increases), producing slightly more total moles of \( \text{H}^+ \).
- However, the massive increase in the denominator (Volume) far outpaces the small increase in the numerator (moles of \( \text{H}^+ \)).
- Therefore, the overall molar concentration (moles/Liter) of \( \text{H}^+ \) ions must definitively decrease.
Why other options are incorrect:While total moles of \( \text{H}^+ \) might increase, the
concentration (which is what pH and this question measure) always drops upon dilution.
#10 of 92
SZABMU-RC (2024)
[SZABMU-RC (2024)]
The pH of human blood is
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Human blood operates as a natural buffer system to maintain strict physiological homeostasis, protecting enzymes from denaturing.
Solution:- The optimal blood pH for enzyme function, cellular respiration, and oxygen transport is tightly controlled between 7.35 and 7.45.
- This is a slightly alkaline (basic) state.
- It is maintained primarily by the carbonic acid-bicarbonate buffer system in the plasma.
Why other options are incorrect:Deviations outside this extremely narrow range rapidly lead to acidosis (below 7.35) or alkalosis (above 7.45), both of which cause death.
#11 of 92
SZABMU-RC (2024)
[SZABMU-RC (2024)]
Maximum yield of ammonia can be obtained by
A
Low temperature, high pressure
B
Low temperature, low pressure
C
High temperature, high pressure
D
High temperature, low pressure
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The Haber process for synthesizing ammonia is a classic application of Le Chatelier's Principle regarding temperature and pressure manipulation.
Formula:$$ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \quad \Delta H = -92.4 \text{ kJ/mol} $$
Solution:- Temperature: The forward reaction is exothermic. Lowering the temperature removes heat, forcing the system to shift forward to produce more heat (and thus more ammonia).
- Pressure: The reaction goes from 4 total moles of reactant gas down to 2 moles of product gas. Increasing the pressure forces the system to shift towards the side with fewer moles to relieve the stress.
- Combining these, Low temperature and high pressure mathematically guarantee the highest theoretical yield.
Why other options are incorrect:High temperature shifts the reaction backwards. Low pressure shifts the reaction backwards.
[ETEA (2024)]
Which of the following factor will disturb chemical equilibrium in a given reaction?
$$ 2\text{HI}_{(g)} \rightleftharpoons \text{H}_{2(g)} + \text{I}_{2(g)} \quad \Delta H^{\circ} = +ive $$
1. Change in concentration
2. Change in temperature
3. Change in pressure
4. Change in volume
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Le Chatelier's Principle identifies which specific stresses will cause a shift in the equilibrium position, heavily depending on the stoichiometry of the gases involved.
Formula:$$ \Delta n = \text{Moles}_{\text{products}} - \text{Moles}_{\text{reactants}} $$
Solution:- Calculate \( \Delta n \): Reactants = 2 moles. Products = 1 + 1 = 2 moles. \( \Delta n = 2 - 2 = 0 \).
- Because there is no change in the number of gaseous moles, changes in Pressure (3) or Volume (4) will have absolutely zero effect on the equilibrium position.
- However, changing Concentration (1) of any species will always shift the equilibrium to consume the excess.
- Changing Temperature (2) will also shift the equilibrium (forward in this case, as it is endothermic / +ive).
- Therefore, only factors 1 and 2 disturb this specific equilibrium.
Why other options are incorrect:Any option including 3 or 4 is incorrect because pressure/volume changes cannot stress a gaseous system where \( \Delta n = 0 \).
[ETEA (2024)]
The \( \text{pK}_a \), of n-propyl amine is
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Note: The question text uses the term "\( \text{pK}_a \)" but chemically provides the literature value for the \( \text{pK}_b \) of n-propylamine. Amines are bases.Solution:- Amines are basic organic compounds. The strength of their basicity is measured by their base dissociation constant, \( K_b \), or its negative log, \( \text{pK}_b \).
- n-propylamine (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \)) is a primary aliphatic amine.
- Historical textbook data and chemical tables list its \( \text{pK}_b \) value as exactly 3.32.
- (For strict accuracy: its actual \( \text{pK}_a \) as a conjugate acid is ~10.68. The exam question contained a typo but 3.32 is the universally expected answer key for this specific rote-memorization metric.)
Why other options are incorrect:These are simply incorrect numerical values for this specific constant.
[ETEA (2024)]
Forward reaction is the one that
A
Is very slow at the beginning of the reaction
B
Reacts to form reactants
C
Speeds up gradually and at equilibrium its rate becomes constant
D
Takes place from left to right as given in chemical equation
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:This is a fundamental definition of standard chemical equation notation.
Solution:- By universal IUPAC convention, chemical equations are written with original Reactants on the left side and newly formed Products on the right side.
- The Forward reaction is defined as the process of Reactants converting into Products.
- Therefore, it geometrically takes place from left to right as written in the equation.
Why other options are incorrect:The forward reaction is actually at its
fastest at the very beginning (eliminating A and C). Reacting to form reactants (Option B) is the definition of the
backward reaction.
[ETEA (2024)]
In the production of \( \text{SO}_3 \), from \( \text{SO}_2 \) and Oxygen, the yield of \( \text{SO}_3 \), is increased by
B
Adding more \( \text{SO}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This is an application of Le Chatelier's Principle regarding concentration changes in the Contact Process.
Formula:$$ 2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)} \quad \Delta H = \text{-ive} $$
Solution:- According to Le Chatelier, adding more of a reactant creates an imbalance, forcing the system to consume the excess.
- By continuously adding more \( \text{SO}_2 \) (a reactant), the equilibrium is constantly shifted forward (to the right).
- This forward shift directly increases the production and yield of the \( \text{SO}_3 \) product.
Why other options are incorrect:Catalysts do not increase yield, only rate. Increasing temperature decreases yield (it's exothermic). Removing oxygen would shift the reaction backward, ruining yield.
[ETEA (2024)]
Consider \( \text{N}_2 + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)} \quad \Delta H = -92.46 \text{ kJ/mol} \)
The optimum temperature, (\( ^{\circ}\text{C} \)) to produce ammonia is
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The Haber-Bosch process requires an "optimum" compromise temperature to balance thermodynamic yield against kinetic speed.
Solution:- Thermodynamically, because the reaction is exothermic (negative \( \Delta H \)), low temperatures maximize the theoretical yield of ammonia.
- Kinetically, low temperatures make the reaction far too slow to be industrially viable because the activation energy required to break the \( \text{N}\equiv\text{N} \) triple bond is massive.
- Industrial chemists use a compromise temperature of exactly 450\( ^{\circ}\text{C} \) (with an iron catalyst). This provides enough kinetic energy for a fast reaction rate while maintaining an acceptable equilibrium yield.
Why other options are incorrect:0\( ^{\circ}\text{C} \) is kinetically dead (reaction won't happen). 5000\( ^{\circ}\text{C} \) would shift the equilibrium totally backward, destroying the ammonia. "Constant" is not a numerical value.
[ETEA (2024)]
The unit of \( K_c \) for the system \( \text{PCl}_5 \rightarrow \text{PCl}_3 + \text{Cl}_2 \) is
A
\( \text{dm}^3 / \text{mol} \)
B
\( \text{mol} / \text{dm}^3 \)
C
\( \text{mol} / \text{dm}^2 \)
D
\( \text{mol} / \text{dm}^6 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The overall unit for the equilibrium constant \( K_c \) is derived from the difference in the stoichiometric coefficients of gaseous products and reactants.
Formula:$$ \text{Units} = (\text{mol/dm}^3)^{\Delta n} $$
Solution:- First, calculate \( \Delta n = \text{Moles of Products} - \text{Moles of Reactants} \).
- The balanced equation is \( \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \).
- Moles of Reactants = 1 (from \( \text{PCl}_5 \)).
- Moles of Products = 1 + 1 = 2 (from \( \text{PCl}_3 \) and \( \text{Cl}_2 \)).
- \( \Delta n = 2 - 1 = 1 \).
- Therefore, the unit is \( (\text{mol/dm}^3)^1 = \text{mol/dm}^3 \).
Why other options are incorrect:Option A is the unit if \( \Delta n = -1 \). Options C and D do not follow correct dimensional analysis for standard molarity powers.
[DUHS (2024)]
The equilibrium of this reaction would not be affected by an increase in pressure
A
\( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \)
B
\( \text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO} \)
C
\( 2\text{NO} + \text{Cl}_2 \rightleftharpoons 2\text{NOCl} \)
D
\( \text{PCl}_5 \rightleftharpoons \text{PCl}_3 + \text{Cl}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Pressure changes only affect the equilibrium position of gaseous reactions if there is a difference in the number of gaseous moles between products and reactants.
Formula:$$ \text{If } \Delta n = 0, \text{ Pressure has no effect} $$
Solution:- Analyze \( \Delta n \) for each option.
- Option A: \( \Delta n = 2 - 3 = -1 \). (Affected)
- Option C: \( \Delta n = 2 - 3 = -1 \). (Affected)
- Option D: \( \Delta n = 2 - 1 = +1 \). (Affected)
- Option B: \( \text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO} \). Reactant moles = \( 1 + 1 = 2 \). Product moles = 2.
- \( \Delta n = 2 - 2 = 0 \). Because the volume footprint of both sides is identical, pressure changes do not stress the system.
Why other options are incorrect:All other options have a non-zero \( \Delta n \), meaning Le Chatelier's principle would dictate a shift to compensate for pressure changes.
[DUHS (2024)]
The pH value of human blood is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Human blood is a heavily regulated buffer system that maintains strict pH homeostasis to prevent enzyme denaturation.
Solution:- The normal physiological pH range for human arterial blood is strictly between 7.35 and 7.45.
- This represents a very slightly basic (alkaline) environment.
- Option B (7.35) falls perfectly on the lower boundary of this standard medical range.
Why other options are incorrect:6.62 and 7.33 represent clinical acidosis (fatal if untreated). 7.8 represents severe clinical alkalosis.
[DUHS (2024)]
The term "active mass" used in the Law of Mass Action means:
A
Number of moles per \( \text{dm}^3 \)
B
Number of moles per litter
C
Gram per \( \text{dm}^3 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Guldberg and Waage formulated the Law of Mass Action, standardizing how chemical quantities are expressed in rate and equilibrium equations.
Formula:$$ \text{Active Mass} = [ ] = \frac{n}{V} = \text{Molarity} $$
Solution:- "Active mass" mathematically refers to the molar concentration of a reacting substance.
- Concentration is defined as the number of moles (\( n \)) divided by the volume in cubic decimeters (\( \text{dm}^3 \)).
- Therefore, it means Number of moles per \( \text{dm}^3 \).
- Note: While Option B (moles per liter) is technically identical since \( 1 \text{ dm}^3 = 1 \text{ L} \), SI unit conventions in this curriculum strictly favor \( \text{dm}^3 \), making A the technically 'more correct' textbook answer.
Why other options are incorrect:Grams per \( \text{dm}^3 \) is mass concentration (density), not molar concentration. Simply "moles" ignores the volume requirement entirely.
[DUHS (2024)]
Conjugated acid \( \text{NH}_3 \) is
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:According to the Brønsted-Lowry theory, a conjugate acid is formed when a base accepts a proton (\( \text{H}^+ \)).
Formula:$$ \text{Base} + \text{H}^+ \rightleftharpoons \text{Conjugate Acid} $$
Solution:- Ammonia (\( \text{NH}_3 \)) acts as a weak base due to the lone pair of electrons on the nitrogen atom.
- It accepts a hydrogen ion (\( \text{H}^+ \)) from an acid (like water or HCl).
- Adding one Hydrogen atom and one positive charge yields the ammonium ion: \( \text{NH}_3 + \text{H}^+ = \text{NH}_4^+ \).
Why other options are incorrect:\( \text{NH}_2^- \) is the conjugate
base of ammonia (formed by losing a proton). The other options are chemically unstable radical or invalid formulas.
[DUHS (2024)]
(Out of syllabus)
According to Raoult's law, the relative lowering of vapor pressure is equal to:
A
Mole fraction of solute
B
Mole fraction of solvent
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Raoult's Law relates the physical properties of a solution to the concentration of a non-volatile solute dissolved within it, establishing colligative properties.
Formula:$$ \frac{\Delta P}{P^{\circ}} = X_{\text{solute}} $$
Solution:- \( \Delta P \) is the lowering of vapor pressure (\( P^{\circ} - P \)).
- Dividing by the pure solvent pressure (\( P^{\circ} \)) gives the "relative" lowering.
- Raoult experimentally proved that this fractional drop is exactly equal to the Mole fraction of the solute (\( X_{\text{solute}} \)).
- This implies that vapor pressure drops linearly as more solute particles take up space at the liquid surface, preventing solvent evaporation.
Why other options are incorrect:The mole fraction of the solvent defines the
remaining vapor pressure (\( P = P^{\circ} X_{\text{solvent}} \)), not the
lowering. Molarity and Molality are different concentration units entirely.
[DUHS (2024)]
(Out of syllabus)
A colloidal solution of liquid into liquid is known as:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Colloids are classified based on the physical state of the dispersed phase (what is being mixed in) and the dispersion medium (what it is mixed into).
Solution:- When tiny droplets of one liquid (dispersed phase) are scattered throughout another immiscible liquid (dispersion medium), the resulting mixture is highly specific.
- This specific liquid-in-liquid colloid is scientifically termed an Emulsion.
- Common everyday examples include milk (liquid fat dispersed in water) and mayonnaise.
Why other options are incorrect:Foam is gas-in-liquid. Solid foam is gas-in-solid. Fog is liquid-in-gas (aerosol).
[NUMS (2024)]
The solubility of \( \text{PbS} \) at \( 25^{\circ}\text{C} \) is \( 4.0 \times 10^{-28} \). Then ionic concentration will be:
A
\( 4 \times 10^{-14} \)
B
\( 2 \times 10^{-14} \)
C
\( 1 \times 10^{-14} \)
D
\( 3.0 \times 10^{-14} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Note: The question wording is slightly flawed. Based on the numerical options, it implies that \( 4.0 \times 10^{-28} \) is actually the Solubility Product (\( K_{sp} \)), and it is asking for the molar solubility (\( s \)), which equals the ionic concentration.Formula:$$ \text{PbS}_{(s)} \rightleftharpoons \text{Pb}^{2+} + \text{S}^{2-} \implies K_{sp} = (s)(s) = s^2 $$
Solution:- Assuming \( K_{sp} = 4.0 \times 10^{-28} \).
- Set up the equation: \( s^2 = 4.0 \times 10^{-28} \).
- Take the square root of both sides to find solubility (\( s \)): \( s = \sqrt{4.0 \times 10^{-28}} \).
- The square root of 4.0 is 2. The square root of \( 10^{-28} \) (divide exponent by 2) is \( 10^{-14} \).
- Therefore, the molar solubility (and thus individual ionic concentration) is \( 2 \times 10^{-14} \text{ M} \).
Why other options are incorrect:Option A forgot to square root the base number 4. Options C and D are mathematically incorrect square roots.
[NUMS (2024)]
What conditions should be applied to minimize the leftover reactants in ammonia synthesis?
A
200 atm, \( 500^{\circ}\text{C} \)
B
400 atm, \( 200^{\circ}\text{C} \)
C
100 atm, \( 400^{\circ}\text{C} \)
D
200 atm, \( 400^{\circ}\text{C} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:To minimize "leftover reactants," we must maximize the theoretical equilibrium yield (shift as far right as possible) using Le Chatelier's Principle.
Formula:$$ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \quad \Delta H = -92.4 \text{ kJ/mol} $$
Solution:- Pressure: The forward reaction goes from 4 gaseous moles to 2. To force the equilibrium forward (reducing reactants), we need the highest possible pressure. Comparing options: 400 atm is the highest.
- Temperature: The forward reaction is exothermic. To force the equilibrium forward, we need to continually remove heat, meaning we need the lowest possible temperature. Comparing options: \( 200^{\circ}\text{C} \) is the lowest.
- Combining these two thermodynamic extremes, 400 atm and \( 200^{\circ}\text{C} \) provides the absolute maximum theoretical yield, minimizing leftover reactants (ignoring kinetic speed constraints).
Why other options are incorrect:Higher temperatures (400, 500) shift the reaction backward, creating more leftover reactants. Lower pressures (100, 200) fail to push the reaction as far forward as 400 atm.
[NUMS (2024)]
Select the buffer solution having highest pH:
A
0.1M \( \text{CH}_3\text{COOH} \), 0.01M \( \text{CH}_3\text{COO}^- \)
B
0.1M \( \text{CH}_3\text{COOH} \), 0.05M \( \text{CH}_3\text{COO}^- \)
C
0.1M \( \text{CH}_3\text{COOH} \), 0.10M \( \text{CH}_3\text{COO}^- \)
D
0.1M \( \text{CH}_3\text{COOH} \), 0.15M \( \text{CH}_3\text{COO}^- \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The pH of an acidic buffer is dictated by the Henderson-Hasselbalch equation, which relies on the logarithmic ratio of the conjugate base (salt) to the weak acid.
Formula:$$ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) $$
Solution:- The weak acid concentration (\( \text{CH}_3\text{COOH} \)) is held perfectly constant at 0.1M across all four options.
- Therefore, the pH will increase strictly as the concentration of the conjugate base (\( [\text{Salt}] \)) increases, because the mathematical fraction \( \frac{[\text{Salt}]}{[\text{Acid}]} \) becomes larger.
- Let's analyze the ratios:
- Option A: \( \log(0.01 / 0.1) = \log(0.1) = -1 \)
- Option B: \( \log(0.05 / 0.1) = \log(0.5) = -0.3 \)
- Option C: \( \log(0.10 / 0.1) = \log(1) = 0 \)
- Option D: \( \log(0.15 / 0.1) = \log(1.5) = +0.176 \)
- Option D adds a positive value to the \( \text{pK}_a \), resulting in the mathematically highest pH. This conceptually makes sense: having more of the basic component makes the buffer more basic (higher pH).
Why other options are incorrect:Options A, B, and C have lower concentrations of the conjugate base, leading to lower, more acidic pH values.
[UHS (2023)]
What is the ultimate fate of reversible reaction?
B
Complete consumption of reactants
C
Complete consumption of products
D
A state when there is no net concentration change
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Reversible reactions do not go to completion; instead, they eventually reach a state of dynamic chemical equilibrium.
Solution:- In a closed system, a reversible reaction will proceed until the forward reaction rate equals the backward reaction rate.
- Once these rates are locked in balance, the macroscopic concentrations of all reactants and products stop changing over time.
- Therefore, the ultimate fate is reaching a state where there is no net concentration change.
Why other options are incorrect:Reversible reactions never reach "completion" (where reactants are fully consumed, eliminating A and B). Products are also never fully consumed (eliminating C).
[UHS (2023)]
In reversible reaction, when product is removed, the equilibrium shift towards the:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Le Chatelier's Principle states that a system at equilibrium will shift to counteract any applied stress.
Formula:$$ \text{Reactants} \rightleftharpoons \text{Products} $$
Solution:- Removing a product from the system creates a concentration deficit on the right side of the chemical equation.
- To counteract this loss and restore the equilibrium constant (\( K_c \)), the system speeds up the forward reaction.
- The reactants are consumed to replace the lost product, meaning the equilibrium shifts towards the Product side.
Why other options are incorrect:Shifting toward the reactant side would only happen if reactants were removed or products were added.
[UHS (2023)]
One can estimate the direction in which equilibrium will shift with the help of:
A
Le Chatelier's principle
D
Law of heat of formation
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Predicting the shift of a dynamic equilibrium when disturbed (by concentration, temperature, or pressure) is the central purpose of Le Chatelier's Principle.
Solution:- Henri Le Chatelier postulated that a system at equilibrium will dynamically shift to counteract any applied stress.
- This principle acts as a predictive tool for chemists to manipulate reaction yields and control the direction of the shift.
Why other options are incorrect:The Law of Mass Action provides the mathematical equation for \( K_c \), but doesn't predict shifts. Hess's Law determines total enthalpy. The Law of heat of formation deals with thermodynamics, not equilibrium shifts.
[SZABMU (2023)]
$$ \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \quad (\Delta H = -92.5) $$
C
Continuous removal \( \text{NH}_3 \)
D
Continuous addition of \( \text{NH}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Note: Historically, this past paper question was printed with a missing stem. Based on the official answer key (D), the implied question was: "Which of the following conditions shifts the equilibrium in the backward (reactant) direction?"Formula:$$ \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \quad \Delta H < 0 $$
Solution:- According to Le Chatelier's Principle, adding a product to an equilibrium mixture forces the system to consume the excess.
- Continuous addition of \( \text{NH}_3 \) creates an excess on the product side.
- The system shifts in the reverse (backward) direction to consume the added \( \text{NH}_3 \), turning it back into \( \text{N}_2 \) and \( \text{H}_2 \).
Why other options are incorrect:Low temperature and continuous removal of \( \text{NH}_3 \) (Options A and C) would shift the reaction forward to maximize yield, which are the standard industrial conditions (Haber Process).
[SZABMU (2023)]
What is not true about the Le-Chatelier's principle?
A
Allow to predict the change in concentration on system at equilibrium
B
Allow to predict the change in reaction rate on system at equilibrium
C
Allow to predict the change in pressure on system at equilibrium
D
Allow to predict the change in temperature on system at equilibrium
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Le Chatelier's Principle is strictly a thermodynamic and equilibrium concept. It does not provide any mathematical information about chemical kinetics (rates of reaction).
Solution:- The principle explains how changes in concentration, pressure, and temperature shift the position of an equilibrium (predicting yields and directions).
- However, predicting how fast a reaction gets there (the reaction rate) is the domain of Chemical Kinetics (Rate Laws, Arrhenius equation).
- Therefore, predicting changes in reaction rates is NOT a function of Le Chatelier's Principle.
Why other options are incorrect:Options A, C, and D are the three fundamental stresses (concentration, pressure, temperature) that Le Chatelier's Principle is explicitly designed to address.
[SZABMU (2023)]
For a gaseous phase reaction, when number of moles of reactant and product are equal:
A
The values of \( K_p \) and \( K_c \) are different
B
The value of \( K_p \) is greater than \( K_c \)
C
The value of \( K_c \) is greater than \( K_p \)
D
The values of \( K_p \) and \( K_c \) are the same
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:When the stoichiometry of gaseous reactants perfectly matches the products, the change in moles (\( \Delta n \)) is zero.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- Since gaseous moles are equal on both sides, \( \Delta n = 0 \).
- Substitute this into the equation: \( K_p = K_c(RT)^0 \).
- Any non-zero mathematical expression raised to the power of zero equals 1.
- Therefore, \( K_p = K_c \times 1 \), meaning their numerical values are completely identical.
Why other options are incorrect:Any discrepancy between \( K_p \) and \( K_c \) fundamentally requires a non-zero change in the number of gaseous moles.
[ETEA (2023)]
Haber's process is used for the synthesis of ammonia. The optimum temperature for the Haber process is:
A
\( 35 - 50^{\circ}\text{C} \)
B
\( 130 - 150^{\circ}\text{C} \)
C
\( 400 - 450^{\circ}\text{C} \)
D
\( 500 - 600^{\circ}\text{C} \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The Haber-Bosch process requires a compromise temperature. While a lower temperature favors the exothermic equilibrium yield, a moderate-to-high temperature is required so the reaction proceeds at an economically viable rate.
Solution:- The synthesis of ammonia (\( \Delta H = -92.4 \text{ kJ/mol} \)) is exothermic, so Le Chatelier's principle suggests low temperatures maximize yield.
- However, at very low temperatures, the kinetic energy is too low to break the strong \( \text{N}\equiv\text{N} \) triple bonds, making the reaction agonizingly slow.
- Industrial chemists use an "optimum" compromise temperature of roughly \( 400 - 450^{\circ}\text{C} \), paired with an iron catalyst, to get a reasonable yield in a short amount of time.
Why other options are incorrect:Temperatures below 150°C are too slow kinetically. Temperatures above 500°C shift the equilibrium too far backwards, drastically ruining the yield.
[ETEA (2023)]
If ionic product is less than \( K_{sp} \) then:
A
Solution will be saturated
B
Precipitation will occur
C
Solution will be super saturated
D
No precipitation will occur
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The relationship between the ionic product (\( Q_{sp} \)) and the solubility product (\( K_{sp} \)) dictates whether a solution can hold more dissolved ions or if it will precipitate.
Formula:$$ \text{If } Q_{sp} < K_{sp} \implies \text{Unsaturated Solution} $$
Solution:- \( K_{sp} \) is the maximum equilibrium threshold for dissolved ions.
- When the actual ionic product (\( Q_{sp} \)) is less than \( K_{sp} \), the solution has not yet reached its maximum capacity.
- This describes an unsaturated solution. Because there is room for more solute to dissolve, no precipitation will occur.
Why other options are incorrect:Saturated means \( Q_{sp} = K_{sp} \). Supersaturated (leading to precipitation) means \( Q_{sp} > K_{sp} \).
[ETEA (2023)]
For \( \Delta n = 0 \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The mathematical relationship between pressure-based (\( K_p \)) and concentration-based (\( K_c \)) equilibrium constants relies entirely on the change in gaseous moles.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- The variable \( \Delta n \) represents the moles of gaseous products minus moles of gaseous reactants.
- When \( \Delta n = 0 \), the equation becomes \( K_p = K_c(RT)^0 \).
- Since any non-zero value to the zeroth power is 1, the \( RT \) term vanishes (becomes 1).
- This leaves the strict equality: \( K_p = K_c \).
Why other options are incorrect:Any inequality requires \( \Delta n \) to be either a positive or negative integer, not zero.
[DUHS (2023)]
Which of the following reactions has same value of \( K_c \) & \( K_p \)?
A
\( \text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3 \)
B
\( \text{PCl}_5 \rightarrow \text{PCl}_3 + \text{Cl}_2 \)
C
\( \text{H}_2 + \text{I}_2 \rightarrow 2\text{HI} \)
D
\( 2\text{SO}_2 + \text{O}_2 \rightarrow 2\text{SO}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The equilibrium constants \( K_p \) and \( K_c \) are mathematically identical only when there is no net change in the total number of gaseous moles during the reaction.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- We must find the reaction where \( \Delta n = (\text{Moles of Products}) - (\text{Moles of Reactants}) = 0 \).
- For Option A: \( \Delta n = 2 - 4 = -2 \).
- For Option B: \( \Delta n = (1+1) - 1 = +1 \).
- For Option C: \( \text{H}_2 + \text{I}_2 \rightarrow 2\text{HI} \). Reactants = 1 + 1 = 2. Products = 2. Therefore, \( \Delta n = 2 - 2 = 0 \).
- Because \( \Delta n = 0 \), \( (RT)^0 = 1 \), making \( K_p = K_c \).
Why other options are incorrect:Options A, B, and D all feature a change in the total moles of gas, meaning the \( (RT)^{\Delta n} \) factor will alter the value between \( K_p \) and \( K_c \).
[DUHS (2023)]
Which of the following conditions required for maximum yield of ammonia through Haber's process?
B
Decreasing concentration of reactant
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:According to Le Chatelier's Principle, the yield of an exothermic reaction is maximized by removing heat to shift the equilibrium forward.
Formula:$$ \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \quad \Delta H = -92.4 \text{ kJ/mol} $$
Solution:- The synthesis of ammonia releases heat (exothermic).
- Heat can be conceptually treated as a product on the right side of the equation.
- By decreasing temperature (removing heat), the system experiences a stress. It responds by shifting forward (to the right) to replace the lost heat.
- This forward shift drastically increases the concentration of the actual product, Ammonia.
Why other options are incorrect:Increasing temperature shifts the reaction backwards. Decreasing reactants shifts it backwards. Decreasing pressure shifts it to the side with more moles (reactants, backwards).
[DUHS (2023)]
Which of the following is true for reversible reaction at equilibrium?
A
The rate of forward reaction is greater than backward reaction
B
The rate of forward reaction is lesser than backward reaction
C
The rate of backward reaction is equal to forward reaction
D
The rate of backward reaction is greater than forward reaction
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Dynamic equilibrium is defined kinetically by the perfect balance of opposing reaction rates.
Formula:$$ Rate_f = Rate_r $$
Solution:- In a closed system, as reactants are consumed, the forward rate slows down.
- Simultaneously, as products accumulate, the backward rate speeds up.
- Equilibrium is the exact moment these two rates intersect.
- At this point, the rate of the backward reaction is perfectly equal to the rate of the forward reaction, meaning there is no longer any net change in macroscopic concentrations.
Why other options are incorrect:If one rate is greater than the other (Options A, B, D), the system is actively shifting and, by definition, has not yet reached equilibrium.
[BUMHS (2023)]
The solubility of \( \text{KClO}_3 \) is decreased by adding
C
\( \text{H}_2\text{O} \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The Common Ion Effect suppresses the solubility of a sparingly soluble salt when a highly soluble salt sharing one of the same ions is added to the solution.
Formula:$$ \text{KClO}_{3(s)} \rightleftharpoons \text{K}^+_{(aq)} + \text{ClO}^-_{3(aq)} $$
Solution:- When \( \text{KCl} \) is added, it completely dissociates into \( \text{K}^+ \) and \( \text{Cl}^- \).
- This floods the solution with \( \text{K}^+ \) ions.
- Because \( \text{K}^+ \) is an ion common to the \( \text{KClO}_3 \) equilibrium, Le Chatelier's Principle states the system will shift left to consume the excess \( \text{K}^+ \).
- This backward shift forces dissolved ions back into solid \( \text{KClO}_3 \), thereby decreasing its overall solubility.
Why other options are incorrect:Adding more solid \( \text{KClO}_3 \) to a saturated solution has no effect. Adding water (\( \text{H}_2\text{O} \)) dilutes the solution, allowing
more salt to dissolve (increasing absolute solubility).
[BUMHS (2023)]
The equilibrium stage that is not affected by temperature is called?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The fundamental
nature of dynamic equilibrium (the fact that opposing microscopic processes continue simultaneously) remains true regardless of temperature changes.
Solution:- While changing the temperature definitely shifts the position (the \( K_c \) value) of an equilibrium, it does not stop the microscopic motion.
- The defining characteristic of dynamic equilibrium is that the forward and backward reactions never cease. They just re-balance at a new rate.
- Therefore, the core "dynamic" nature of the stage is unaffected by temperature, unlike static equilibrium where macroscopic forces simply sum to zero and motion stops.
Why other options are incorrect:Static equilibrium applies to physics (forces on resting objects), not typical chemical reactions. Natural and Unstable are not standard classifications for chemical equilibrium states in this context.
[BUMHS (2023)]
Temperature increase in an exothermic reversible reaction, shift the equilibrium to:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:In an exothermic reaction, heat is released as a product. According to Le Chatelier's Principle, adding heat to the system forces it to shift backwards.
Formula:$$ \text{Reactants} \rightleftharpoons \text{Products} + \text{Heat} $$
Solution:- By increasing the temperature, you are artificially adding "Heat" to the product side of the equilibrium.
- To relieve this applied stress, the system must consume the excess heat.
- It does this by shifting the reaction to the left (the endothermic direction).
- This breaks down products and forms more reactants, thus shifting to the Reactant side.
Why other options are incorrect:Shifting to the product side would happen if temperature was decreased (removing heat). The equilibrium definitely changes, ruling out C and D.
[NUMS (2023)]
The high pressure of 200 atm in Haber's process is used for:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Le Chatelier's principle dictates that increasing pressure on a gaseous system favors the direction that produces fewer moles of gas.
Formula:$$ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} $$
Solution:- In the Haber process, 4 total moles of reactant gases form only 2 moles of product gas.
- By applying a massive pressure of 200 atm, the system is highly stressed.
- To reduce this pressure, the system shifts forcefully to the right, where the gas occupies less volume (fewer moles).
- This deliberate engineering choice results in a massively better yield of Ammonia.
Why other options are incorrect:Lowering yield makes no industrial sense. High pressure increases the reaction rate, not lowers it. High pressure requires extremely expensive, thick-walled steel pipes, meaning costs drastically increase, not decrease.
[NUMS (2023)]
By which of the following factors equilibrium state is attained earlier?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A catalyst provides an alternative reaction pathway with a lower activation energy, which accelerates both the forward and reverse reaction rates proportionally.
Formula:$$ \text{Rate} \propto e^{-E_a / RT} $$
Solution:- When activation energy (\( E_a \)) is lowered by a Catalyst, a much larger fraction of molecules possess enough kinetic energy to react.
- Both the forward and backward rates spike simultaneously.
- Because both opposing rates are faster, they intersect and balance each other out in a fraction of the time.
- Therefore, the state of equilibrium is attained much earlier, even though the final yield (position) remains completely unchanged.
Why other options are incorrect:While temperature and concentration changes can affect rates, they also shift the position of the equilibrium. A catalyst is the only factor whose sole equilibrium purpose is speeding up the attainment of balance.
[UHS (2022)]
The decrease in solubility of the salt in a solution that already contains an ion common to that salt is known as:
A
Le-Chatelier's principle
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:This is the standard definition of a specific phenomenon that arises as a consequence of Le Chatelier's Principle in solubility scenarios.
Solution:- When an external ion identical to one in the salt is added, the overall concentration of that specific ion increases.
- The equilibrium shifts left to precipitate the solid salt and lower the ionic concentration.
- The specific term for this solubility decrease is the Common Ion Effect.
Why other options are incorrect:While Le Chatelier's Principle is the underlying rule, "Common ion effect" is the specific term for this exact manifestation. Solubility product is a mathematical constant.
[UHS (2022)]
The precipitation occurs if the ionic concentration is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Precipitation is the physical manifestation of a solution pushing excess dissolved ions out of the liquid phase to restore equilibrium.
Formula:$$ \text{Ionic Product } (Q_{sp}) > K_{sp} $$
Solution:- \( K_{sp} \) defines the absolute maximum limit of ions a solution can stably hold at equilibrium.
- If the product of the ionic concentrations in the solution exceeds this limit (\( Q_{sp} > K_{sp} \)), the solution is unstable (supersaturated).
- The excess ions will clump together and form a solid precipitate until the concentration drops back to the \( K_{sp} \) limit.
Why other options are incorrect:Less than \( K_{sp} \) means more can dissolve. Equal to \( K_{sp} \) means it is perfectly saturated and stable without precipitation.
[UHS (2022)]
One can estimate the direction in which equilibrium will shift with the help of:
A
Le-Chatelier's principle
D
Law of heat of formation
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Predicting the shift of a dynamic equilibrium when disturbed is the central purpose of Le Chatelier's Principle.
Solution:- Henri Le Chatelier postulated that a system at equilibrium will dynamically shift to counteract any applied stress (changes in concentration, pressure, or temperature).
- This principle acts as a predictive tool for chemists to manipulate reaction yields.
Why other options are incorrect:The Law of Mass Action provides the mathematical equation, but doesn't predict shifts. Hess's Law determines total enthalpy. The Law of heat of formation deals with thermodynamics, not equilibrium shifts.
[SZABMU (2022)]
For the chemical reaction;
\( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} + \text{Heat} \)
We can maximize the yield of \( \text{NH}_3 \):
A
By increasing the temperature
B
By decreasing the pressure
C
By increasing the volume of the reaction vessel
D
By continuous withdrawal of ammonia
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:According to Le Chatelier's Principle, continuously removing a product prevents the system from ever reaching equilibrium, forcing the forward reaction to proceed indefinitely.
Formula:$$ Q_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} $$
Solution:- If \( \text{NH}_3 \) is continuously withdrawn from the vessel, its concentration drops to near zero.
- This makes the reaction quotient \( Q_c \) practically zero, which is far less than \( K_c \).
- To try and restore \( K_c \), the reactants will continuously convert into more product.
Why other options are incorrect:Increasing temperature shifts this exothermic reaction backwards. Decreasing pressure (or increasing volume) favors the side with more moles (reactants).
[SZABMU (2022)]
The high pressure of 200 atm in Haber process is used for:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:In reactions where gases combine to form fewer moles of gas, high pressure thermodynamically favors the product side.
Formula:$$ 4 \text{ moles of reactant gas} \rightleftharpoons 2 \text{ moles of product gas} $$
Solution:- Le Chatelier's Principle states that increasing pressure shifts equilibrium towards the side with fewer gas molecules.
- In the Haber process, 4 moles of reactants (\( \text{N}_2 + 3\text{H}_2 \)) form 2 moles of product (\( 2\text{NH}_3 \)).
- Applying 200 atm of pressure heavily forces the reaction forward, drastically increasing the yield of Ammonia.
Why other options are incorrect:It certainly doesn't lower yield or rate. High pressure actually increases costs due to the need for thick, heavy-duty industrial piping, so it is strictly done for yield, not cost reduction.
[SZABMU (2022)]
By which of the following factors equilibrium state is attained earlier?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A catalyst provides an alternative reaction pathway with a lower activation energy, accelerating both forward and backward rates equally.
Formula:$$ \text{Rate} = k[\text{Reactants}]^x \quad (k \text{ increases via lower } E_a) $$
Solution:- Because the activation energy barrier is lowered from both sides, particles require less kinetic energy to react successfully.
- Both the forward and reverse reaction rates spike proportionally.
- As a result, the time required to balance these two rates (attain equilibrium) is significantly reduced.
Why other options are incorrect:While temperature increases rate, it also shifts the equilibrium position. A catalyst uniquely accelerates the attainment of equilibrium without shifting its final position.
[SZABMU (2022)]
Which of the following is not the use of the Buffer solution?
A
Used for the calibration of pH meters
B
Used to preserve biological specimen
C
Maintain the pH of the human blood
D
Predict the concentration of a substance
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:A buffer's sole function is to resist changes in pH upon the addition of acids or bases, making it vital for environmental control, not analytical prediction.
Solution:- Buffers are standard references for pH meters (Option A).
- They maintain stable environments for tissue preservation (Option B).
- They are critical in physiology, like blood pH regulation (Option C).
- However, predicting unknown concentrations is done via analytical techniques like titration, spectroscopy, or utilizing equilibrium constants, not by using a buffer.
Why other options are incorrect:Options A, B, and C describe completely standard and vital real-world applications of buffer solutions.
[ETEA (2022)]
Buffer capacity is maximum when both components have:
D
High and equal concentration
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Buffer capacity refers to the amount of acid or base a buffer can absorb before a significant change in pH occurs. It is maximized when the ratio of salt to acid is 1:1.
Formula:$$ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) $$
Solution:- When the concentration of the weak acid and its conjugate base are perfectly equal, the ratio is 1, and \( \log(1) = 0 \).
- At this exact point (\( \text{pH} = \text{pK}_a \)), the buffer has an equal ability to neutralize added \( \text{H}^+ \) or \( \text{OH}^- \).
- Therefore, it operates at its maximum theoretical buffer capacity relative to its components.
- Note: While high concentration (Option D) technically yields a larger absolute capacity, standard chemical theory teaches that the optimal "state" for maximum capacity is defined by equal concentrations (ratio=1). This is why B is often the accepted key in these specific exams.
Why other options are incorrect:Low concentration lowers absolute capacity. Unequal concentrations mean the buffer will fail quickly in one direction.
[ETEA (2022)]
If solubility product (\( K_{sp} \)) value is large the salt in water is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The \( K_{sp} \) value mathematically represents the extent to which a solid salt dissociates into its aqueous ions.
Formula:$$ K_{sp} = [\text{Products}] $$
Solution:- Because \( K_{sp} \) is the product of the ion concentrations, a larger \( K_{sp} \) directly correlates with higher concentrations of ions in solution at equilibrium.
- More ions in solution mean that more of the solid salt was able to dissolve.
- Therefore, a large \( K_{sp} \) indicates high (more) solubility.
Why other options are incorrect:A small \( K_{sp} \) (e.g., \( 10^{-20} \)) indicates less solubility. A large \( K_{sp} \) is the opposite.
[ETEA (2022)]
The value of solubility products depends only on ____
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The solubility product (\( K_{sp} \)) is a specific type of equilibrium constant. Like all equilibrium constants, it is only fundamentally altered by temperature.
Solution:- Dissolution of a salt is either endothermic or exothermic.
- Changing the temperature changes the kinetic energy of the system and shifts the equilibrium either towards dissolution or precipitation.
- Because the fundamental ratio of equilibrium shifts, the actual constant \( K_{sp} \) changes.
- Concentrations, catalysts, and pressure do not change the constant itself.
Why other options are incorrect:Solvent changes the absolute solubility, but \( K_{sp} \) is defined for a specific solvent (usually water). Pressure rarely affects solids/liquids. Catalysts never change equilibrium constants.
[DUHS (2022)]
In exothermic reaction, by decreasing the temperature, equilibrium constant:
B
Reaction moves backward
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:For exothermic reactions, heat is considered a product. Removing heat (decreasing temperature) shifts the equilibrium forward, resulting in more products and less reactants.
Formula:$$ K_c = \frac{[\text{Products}]}{[\text{Reactants}]} $$
Solution:- When temperature is decreased, Le Chatelier's Principle dictates a shift to the right to generate more heat.
- This forward shift drastically increases the concentration of Products in the numerator.
- Simultaneously, Reactants in the denominator are consumed and decrease.
- A larger numerator divided by a smaller denominator mathematically results in a larger overall ratio.
- Therefore, the equilibrium constant \( K_c \) inherently increases.
Why other options are incorrect:Constants only remain the same if temperature is unchanged. Decreasing would happen in an endothermic reaction.
[DUHS (2022)]
At equilibrium in a reversible reaction:
A
The rate of backward reaction > rate of forward reaction
B
The rate of forward reaction > rate of backward reaction
C
The concentration of reactant & products becomes constant
D
The concentration of reactants & products becomes constant & the rate of forward reaction is same as the rate of backward reaction.
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Dynamic equilibrium is defined by two simultaneous macroscopic conditions occurring in a reversible system.
Solution:- Kinetic condition: The speed at which reactants form products exactly equals the speed at which products revert into reactants (Rate forward = Rate backward).
- Thermodynamic condition: Because the rates are locked in balance, the net amount of reactants and products stops changing over time, meaning their macroscopic concentrations remain perfectly constant.
- Option D perfectly encapsulates both halves of this definition.
Why other options are incorrect:Rates being unequal implies the system is still shifting and has not reached equilibrium. Option C is true but incomplete compared to D.
[DUHS (2022)]
The solubility product \( K_{sp} \) predicts whether:
C
Precipitation will take place or not
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The primary analytical use of the \( K_{sp} \) value in chemistry is determining the saturation threshold of a solution to predict precipitate formation.
Formula:$$ \text{If } Q_{sp} > K_{sp} \implies \text{Precipitate forms} $$
Solution:- By calculating the ionic product (\( Q_{sp} \)) of a mixture and comparing it to the literature \( K_{sp} \) value, chemists can perfectly predict outcomes.
- If the calculated \( Q_{sp} \) exceeds the \( K_{sp} \) threshold, the solid will drop out of solution as a precipitate.
- Therefore, predicting precipitation is its primary function.
Why other options are incorrect:While it relates to solubility, "predicting precipitation" is the specific active analytical application. It has nothing to do with boiling point or diffusion rates.
[NUMS (2022)]
The pH of human blood is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Human blood operates as a natural buffer system to maintain physiological homeostasis, strictly regulating its acid-base balance.
Formula:$$ \text{Normal Blood pH} \approx 7.4 $$
Solution:- The optimal blood pH for enzyme function and oxygen transport is tightly controlled between 7.35 and 7.45.
- This slightly alkaline state is maintained primarily by the carbonic acid-bicarbonate (\( \text{H}_2\text{CO}_3 / \text{HCO}_3^- \)) buffer system.
Why other options are incorrect:A pH below 7.35 causes acidosis, and above 7.45 causes alkalosis, both of which are extremely dangerous. The other ranges are physically fatal for human survival.
[NUMS (2022)]
Which term describes a solution in which dissolved solute is in equilibrium with undissolved solute?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:A saturated solution represents a state of dynamic equilibrium between the solid phase and the aqueous phase of a solute.
Formula:$$ \text{Solute}_{(s)} \rightleftharpoons \text{Solute}_{(aq)} $$
Solution:- In a saturated solution, the maximum amount of solute has dissolved at a given temperature.
- Any additional solute added will simply sink to the bottom.
- At this exact point, the rate of dissolution (solid turning to aqueous) perfectly equals the rate of crystallization (aqueous turning back to solid).
- This defines a state of equilibrium.
Why other options are incorrect:Unsaturated solutions have not reached equilibrium yet (more can dissolve). Supersaturated solutions are highly unstable and technically past the equilibrium point until precipitation occurs.
[PMC (2021)]
\( \text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI} \). Relation between \( K_p \) and \( K_c \) for this reaction is:
D
\( K_p = K_c(RT)^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The relationship depends on \( \Delta n \), the difference in moles between gaseous products and reactants.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- For the reaction \( \text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI} \), count the gaseous moles.
- Moles of products = 2. Moles of reactants = \( 1 + 1 = 2 \).
- \( \Delta n = 2 - 2 = 0 \).
- Therefore, \( K_p = K_c(RT)^0 = K_c \).
Why other options are incorrect:These options introduce \( RT \) factors mathematically which only apply if \( \Delta n \) is not zero.
[PMC (2021)]
According to Le-Chatelier's principle, exothermic reactions are favored by:
C
Decrease in temperature
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Exothermic reactions release heat as a product. Le Chatelier's principle states that removing a product shifts the equilibrium forward.
Formula:$$ \text{Reactants} \rightleftharpoons \text{Products} + \text{Heat} $$
Solution:- Since heat is essentially a product, lowering the temperature (removing heat) creates a deficit on the right side of the equation.
- The system shifts forward to produce more heat, thus favoring the exothermic reaction.
Why other options are incorrect:Pressure and volume changes depend strictly on the change in gas moles (\( \Delta n \)), NOT on whether the reaction is exothermic or endothermic.
[PMC (2021)]
The equilibrium constant is always written as a ratio of:
A
Reactants over products
C
Product times reactants
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:By universal convention in chemistry, governed by the Law of Mass Action, the equilibrium constant compares product concentration to reactant concentration.
Formula:$$ K = \frac{[\text{Products}]^p}{[\text{Reactants}]^r} $$
Solution:- The formula explicitly places the equilibrium concentrations of the products in the numerator.
- The equilibrium concentrations of the reactants are placed in the denominator.
Why other options are incorrect:Reactants over products is the inverse of the equilibrium constant. Multiplying them has no standard physical meaning in equilibrium.
[PMC (2021)]
Manufacturing of Ammonia by Haber's process is an:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The Haber-Bosch process involves the formation of strong nitrogen-hydrogen bonds, which releases energy.
Formula:$$ \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \quad \Delta H = -92.4 \text{ kJ/mol} $$
Solution:- The reaction has a negative enthalpy change (\( \Delta H = -92.4 \text{ kJ/mol} \)).
- A negative enthalpy change definitively classifies the reaction as exothermic, meaning it releases heat to the surroundings.
Why other options are incorrect:It is not endothermic (it releases heat, doesn't absorb it). It is highly reversible (hence the need for specific high pressure/low temp constraints). With an iron catalyst, the industrial rate is optimized, not inherently slow.
[NMDCAT (2020)]
Which of the following reaction has greater \( K_p \) than \( K_c \) (\( K_p > K_c \))?
A
\( 2\text{NO} + \text{Cl}_2 \rightleftharpoons 2\text{NOCl} \)
B
\( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \)
C
\( 2\text{NOCl} \rightleftharpoons 2\text{NO} + \text{Cl}_2 \)
D
\( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The condition \( K_p > K_c \) occurs when the reaction produces a net increase in the number of gaseous moles (\( \Delta n > 0 \)).
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- Analyze the change in moles (\( \Delta n \)) for each option.
- A: \( \Delta n = 2 - 3 = -1 \). (Here \( K_p < K_c \))
- B: \( \Delta n = 2 - 3 = -1 \). (Here \( K_p < K_c \))
- C: \( \Delta n = (2+1) - 2 = +1 \). Because \( \Delta n > 0 \), \( (RT)^1 \) is greater than 1, making \( K_p > K_c \).
- D: \( \Delta n = 2 - 4 = -2 \). (Here \( K_p < K_c \))
Why other options are incorrect:Options A, B, and D all result in fewer gaseous moles, leading to \( K_p < K_c \).
[NMDCAT (2020)]
The equation \( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \) represents:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Different industrial chemical processes are identified by their core chemical reactions.
Solution:- The synthesis of ammonia (\( \text{NH}_3 \)) from nitrogen and hydrogen gases under high pressure and temperature with an iron catalyst is globally known as the Haber-Bosch process.
Why other options are incorrect:The Contact process manufactures Sulfuric acid (\( \text{H}_2\text{SO}_4 \)). The Solvay process manufactures Sodium carbonate (\( \text{Na}_2\text{CO}_3 \)). Avogadro's law relates gas volume to moles.
[NMDCAT (2020)]
For a gaseous phase reaction, when number of moles of reactants and products are equal:
A
The values of \( K_p \) and \( K_c \) are different
B
The value of \( K_p \) is greater than \( K_c \)
C
The value of \( K_c \) is greater than \( K_p \)
D
The value of \( K_p \) and \( K_c \) are the same
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:When the stoichiometry of gaseous reactants perfectly matches the products, the change in moles (\( \Delta n \)) is zero.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- Since moles are equal, \( \Delta n = 0 \).
- Substitute this into the equation: \( K_p = K_c(RT)^0 \).
- Any non-zero number to the power of zero equals 1.
- Therefore, \( K_p = K_c \times 1 \), meaning they are identical.
Why other options are incorrect:Any discrepancy between \( K_p \) and \( K_c \) fundamentally requires a non-zero \( \Delta n \).
[NMDCAT (2020)]
Purification of table salt (\( \text{NaCl} \)) by passing \( \text{HCl} \) gas through its saturated aqueous solution is an example of
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The purification of table salt relies on disrupting the solubility equilibrium by adding an external source of a shared ion.
Formula:$$ \text{NaCl}_{(s)} \rightleftharpoons \text{Na}^+_{(aq)} + \text{Cl}^-_{(aq)} $$
Solution:- When \( \text{HCl} \) gas is passed through the solution, it heavily dissociates, providing a massive influx of \( \text{Cl}^- \) ions.
- Because \( \text{Cl}^- \) is "common" to both \( \text{NaCl} \) and \( \text{HCl} \), Le Chatelier's principle dictates the equilibrium shifts left to consume the excess \( \text{Cl}^- \).
- This forces pure \( \text{NaCl} \) to crystallize and precipitate, leaving impurities in the solution. This is a classic application of the Common Ion Effect.
Why other options are incorrect:Hess's law relates to enthalpy changes. Henry's law relates to gas solubility. The Law of mass action defines the equilibrium expression but doesn't specifically name this precipitation phenomenon.
[NUMS (2019)]
The \( K_c \) Unit for the reaction \( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightarrow 2\text{NH}_{3(g)} \) is:
A
\( \text{mole}^{-1}\text{dm}^{+6} \)
B
\( \text{mole}^{-2}\text{dm}^{+3} \)
C
\( \text{mole}^{-2}\text{dm}^{+6} \)
D
\( \text{mole}^{-1}\text{dm}^{+3} \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The unit of the equilibrium constant \( K_c \) is derived from the difference in the number of gaseous moles (\( \Delta n \)).
Formula:$$ \text{Unit} = (\text{mol dm}^{-3})^{\Delta n} $$
Solution:- Calculate \( \Delta n \) for \( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \).
- \( \Delta n = n_{\text{products}} - n_{\text{reactants}} = 2 - 4 = -2 \).
- Substitute into the formula: \( (\text{mol dm}^{-3})^{-2} \).
- Distribute the exponent: \( \text{mol}^{-2} \cdot (\text{dm}^{-3})^{-2} = \text{mol}^{-2} \text{dm}^{+6} \).
Why other options are incorrect:They do not match the correct mathematical derivation of the exponent \( -2 \) applied to molarity.
[NUMS (2019)]
Which one of the following bases has highest \( K_b \) value?
A
\( \text{NH}_4\text{OH} \)
D
\( \text{CH}_3\text{NH}_2 \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The base dissociation constant (\( K_b \)) is a measure of base strength. Strong bases completely dissociate in water and have exceptionally high \( K_b \) values.
Formula:$$ K_b = \frac{[B^+][\text{OH}^-]}{[B\text{OH}]} $$
Solution:- \( \text{NaOH} \) is an alkali metal hydroxide, which makes it a very strong base that dissociates \( \approx 100\% \) in aqueous solutions.
- Because the denominator \( [B\text{OH}] \) approaches zero, its \( K_b \) value is astronomically large compared to weak bases.
- \( \text{NH}_4\text{OH} \) and \( \text{CH}_3\text{NH}_2 \) are weak bases. \( \text{Ca(OH)}_2 \) is a strong base but has lower solubility than NaOH.
Why other options are incorrect:Weak bases have \( K_b \) values typically in the range of \( 10^{-4} \) to \( 10^{-10} \), which are infinitely smaller than a strong base like \( \text{NaOH} \).
[NUMS (2019)]
The \( \text{pKa} \) values of \( \text{CH}_3\text{COOH} \) is 4.74, the pH of equimolar solution of acetic acid and sodium acetate is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The pH of a buffer solution depends on the ratio of the salt (conjugate base) to the weak acid according to the Henderson-Hasselbalch equation.
Formula:$$ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) $$
Solution:- The solution is equimolar, meaning \( [\text{Salt}] = [\text{Acid}] \).
- Therefore, the ratio \( \frac{[\text{Salt}]}{[\text{Acid}]} = 1 \).
- The logarithm of 1 is zero: \( \log(1) = 0 \).
- The equation simplifies to \( \text{pH} = \text{pK}_a + 0 \).
- Since \( \text{pKa} = 4.74 \), the pH is also exactly 4.74.
Why other options are incorrect:Any deviation from 4.74 would require unequal concentrations of the acid and its salt.
[NUMS (2019)]
Precipitation occurs when the product of ionic concentration is?
A
Greater than \( K_{sp} \)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:The relationship between the ionic product (\( Q_{sp} \)) and the solubility product (\( K_{sp} \)) determines whether a solution is unsaturated, saturated, or supersaturated.
Formula:$$ \text{If } Q_{sp} > K_{sp} \implies \text{Precipitation} $$
Solution:- \( Q_{sp} \) represents the actual product of ion concentrations in a solution at any given moment.
- When \( Q_{sp} > K_{sp} \), the solution contains more dissolved ions than it can stably hold at equilibrium (supersaturated).
- To restore equilibrium, the excess ions precipitate out as a solid until \( Q_{sp} \) drops down to equal \( K_{sp} \).
Why other options are incorrect:Less than \( K_{sp} \) means the solution is unsaturated (no precipitation). Equal to \( K_{sp} \) means it is perfectly saturated (equilibrium, no net precipitation).
[ETEA (2019)]
Consider the reversible reaction. \( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} + \text{Heat} \)
The yield of \( \text{NH}_3 \) will be maximum at:
A
High temperature and low pressure
B
High temperature and high pressure
C
Low temperature and low pressure
D
Low temperature and high pressure
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Le Chatelier's Principle dictates optimal conditions for the Haber process based on thermodynamics and stoichiometry.
Formula:$$ \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \quad \Delta H < 0, \Delta n = -2 $$
Solution:- The reaction produces heat (exothermic). Decreasing the temperature shifts the equilibrium forward to replace lost heat, yielding more \( \text{NH}_3 \).
- The reaction goes from 4 moles of gas to 2 moles of gas. Increasing pressure favors the side with fewer moles of gas to reduce pressure. Thus, high pressure shifts it forward.
- Therefore, maximum yield requires Low Temperature and High Pressure.
Why other options are incorrect:High temperature would shift the reaction backwards. Low pressure would favor the side with more moles (reactants).
[ETEA (2019)]
Ice and water is in equilibrium with each other. By increasing the pressure the equilibrium will shift in
C
To all system at equilibrium
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:Water has the unique property where its solid phase (ice) is less dense (occupies more volume) than its liquid phase.
Formula:$$ \text{Ice}_{(s, \text{higher vol})} \rightleftharpoons \text{Water}_{(l, \text{lower vol})} $$
Solution:- According to Le Chatelier's Principle, increasing pressure on a system shifts the equilibrium towards the state that occupies less volume.
- Since liquid water occupies about 9% less space than solid ice, increasing pressure forces the ice to melt.
- This represents a shift to the right, which is the forward direction.
Why other options are incorrect:Shifting in reverse would form more ice, which occupies more volume and would further increase the stress of pressure.
[MDCAT (2019)]
For an equilibrium reaction;
\( 2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)} \)
The forward reaction is exothermic, increase in temperature shifts the equilibrium position towards left because:
A
The concentrations of \( \text{SO}_3 \), \( \text{SO}_2 \) and \( \text{O}_2 \) increase as the temperature increases
B
The concentrations of \( \text{SO}_2 \) and \( \text{O}_2 \) increase and concentration of \( \text{SO}_3 \) decreases as the temperature increases
C
The concentrations of \( \text{SO}_2 \) and \( \text{O}_2 \) decrease and concentration of \( \text{SO}_3 \) increases as the temperature increases
D
The concentrations of \( \text{SO}_2 \) and \( \text{O}_2 \) increase and concentration of \( \text{SO}_3 \) stays same as the temperature increases
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:Increasing the temperature of an exothermic reaction supplies excess heat, which the system counteracts by shifting in the endothermic (reverse) direction.
Formula:$$ 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 + \text{Heat} $$
Solution:- Because heat is a product, adding heat pushes the reaction to the left (backwards).
- As the reaction shifts left, the product \( \text{SO}_3 \) is consumed and broken down.
- Simultaneously, the reactants \( \text{SO}_2 \) and \( \text{O}_2 \) are generated.
- Thus, \( \text{SO}_2 \) and \( \text{O}_2 \) increase while \( \text{SO}_3 \) decreases.
Why other options are incorrect:Option A violates the conservation of mass (everything cannot increase simultaneously). Option C describes a forward shift. Option D is impossible because \( \text{SO}_3 \) must be consumed to form the reactants.
[MDCAT (2018)]
The product of the concentrations of each ion in saturated solution of a sparingly soluble salt at 298K raised to the power of their relative concentrations is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
Concept:This is the fundamental textbook definition of the Solubility Product Constant.
Formula:$$ K_{sp} = [M^{+y}]^x [A^{-x}]^y $$
Solution:- For a sparingly soluble salt \( M_xA_y \rightleftharpoons xM^{+y} + yA^{-x} \), the equilibrium constant is purely based on aqueous ions.
- This specific constant is termed \( K_{sp} \) (Solubility Product).
Why other options are incorrect:\( K_a \) and \( K_b \) represent acid and base dissociation respectively. \( K_w \) is the autoionization constant of water.
[MDCAT (2017)]
For which of the following equilibrium reaction, \( K_c \) has no units?
A
\( \text{N}_{2(g)} + 3\text{H}_{2(g)} \rightleftharpoons 2\text{NH}_{3(g)} \)
B
\( \text{CO}_{(g)} + \text{H}_2\text{O}_{(g)} \rightleftharpoons \text{CO}_{2(g)} + \text{H}_{2(g)} \)
C
\( 2\text{SO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{SO}_{3(g)} \)
D
\( 2\text{NO}_{2(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{NO}_{(g)} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The unit of \( K_c \) is given by \( (\text{mol dm}^{-3})^{\Delta n} \). It will be unitless only if \( \Delta n = 0 \).
Formula:$$ \Delta n = \Sigma \text{Moles of gaseous products} - \Sigma \text{Moles of gaseous reactants} $$
Solution:- For option B: \( \text{CO}_{(g)} + \text{H}_2\text{O}_{(g)} \rightleftharpoons \text{CO}_{2(g)} + \text{H}_{2(g)} \)
- Reactant moles = \( 1 + 1 = 2 \). Product moles = \( 1 + 1 = 2 \).
- \( \Delta n = 2 - 2 = 0 \). Therefore, units cancel out.
Why other options are incorrect:Option A has \( \Delta n = 2 - 4 = -2 \). Option C has \( \Delta n = 2 - 3 = -1 \). Option D is unbalanced as written but represents \( \Delta n \neq 0 \).
[MDCAT (2017)]
Consider the following reversible reaction;
$$ \text{CH}_3\text{CH}_2\text{OH}_{(l)} + \text{CH}_3\text{COOH}_{(l)} \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_{3(l)} + \text{H}_2\text{O}_{(l)} $$
Initial concentration:
\( 1\text{ mol} \quad 1\text{ mol} \quad 0\text{ mol} \quad 0\text{ mol} \)
Equilibrium concentration:
\( 0.333\text{ mol} \quad 0.333\text{ mol} \quad 0.666\text{ mol} \quad 0.666\text{ mol} \)
\( K_c = 4 \) at \( 100^{\circ}\text{C} \).
What are new equilibrium concentrations of all species if 1 mole of each of \( \text{CH}_3\text{CH}_2\text{OH} \) and \( \text{CH}_3\text{COOH} \) are added to this equilibrium mixture? (Apply Le-Chatelier's Principle) (Temperature remained same)
A
\( (\text{CH}_3\text{COOH})=0.333\text{mol} \), \( (\text{C}_2\text{H}_5\text{OH})=1.333\text{mol} \), \( (\text{CH}_3\text{COOC}_2\text{H}_5)=1.666\text{mol} \), \( (\text{H}_2\text{O})=0.666\text{mol} \)
B
\( (\text{CH}_3\text{COOH})=1.333\text{mol} \), \( (\text{C}_2\text{H}_5\text{OH})=0.333\text{mol} \), \( (\text{CH}_3\text{COOC}_2\text{H}_5)=0.666\text{mol} \), \( (\text{H}_2\text{O})=1.666\text{mol} \)
C
\( (\text{CH}_3\text{COOH})=0.666\text{mol} \), \( (\text{C}_2\text{H}_5\text{OH})=0.666\text{mol} \), \( (\text{CH}_3\text{COOC}_2\text{H}_5)=1.333\text{mol} \), \( (\text{H}_2\text{O})=1.333\text{mol} \)
D
\( (\text{CH}_3\text{COOH})=0.333\text{mol} \), \( (\text{C}_2\text{H}_5\text{OH})=0.333\text{mol} \), \( (\text{CH}_3\text{COOC}_2\text{H}_5)=1.333\text{mol} \), \( (\text{H}_2\text{O})=1.333\text{mol} \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The value of \( K_c \) is a constant at a given temperature. If initial concentrations are doubled proportionally, the final equilibrium concentrations will also double to maintain the ratio of \( K_c \).
Formula:$$ K_c = \frac{[\text{Ester}][\text{Water}]}{[\text{Acid}][\text{Alcohol}]} = 4 $$
Solution:- Initial experiment started with 1 mol of each reactant and produced 0.666 mol of products.
- By adding another 1 mol of each reactant to the equilibrium mixture, the total moles put into the system is exactly 2 moles of each reactant.
- Since the system volume is unchanged and \( \Delta n = 0 \), this is identical to starting a fresh reaction with double the initial concentration (2 mol each).
- Because all stoichiometry is 1:1:1:1, doubling initial reactants exactly doubles the final equilibrium concentrations to maintain \( K_c = 4 \).
- New Reactants = \( 0.333 \times 2 = 0.666 \text{ mol} \).
- New Products = \( 0.666 \times 2 = 1.333 \text{ mol} \).
Why other options are incorrect:Other options represent asymmetrical shifts which violate the 1:1 stoichiometry of the balanced chemical equation.
[MDCAT (2017)]
\( \text{Ca(OH)}_2 \) is sparingly soluble having solubility product value \( 6.5 \times 10^{-6} \). What will be its solubility?
A
\( 2.75 \times 10^{-2} \)
B
\( 1.17 \times 10^{-2} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The solubility product (\( K_{sp} \)) relates to the molar solubility (\( s \)) based on the stoichiometry of the dissolving salt.
Formula:$$ \text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^- $$
$$ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3 $$
Solution:- Given \( K_{sp} = 6.5 \times 10^{-6} \).
- Set up the equation: \( 4s^3 = 6.5 \times 10^{-6} \).
- Divide by 4: \( s^3 = 1.625 \times 10^{-6} \).
- Take the cube root: \( s = \sqrt[3]{1.625 \times 10^{-6}} = \sqrt[3]{1.625} \times 10^{-2} \).
- Since \( 1^3 = 1 \) and \( 1.2^3 = 1.728 \), \( \sqrt[3]{1.625} \) is approximately 1.17.
- Therefore, \( s \approx 1.17 \times 10^{-2} \text{ M} \).
Why other options are incorrect:Option A is incorrect algebra. Options C and D have positive exponents, representing impossibly high solubility for a sparingly soluble salt.
[MDCAT (2016)]
Human blood maintains its pH between:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Human blood operates as a natural buffer system to maintain physiological homeostasis.
Formula:$$ \text{Normal Blood pH} = 7.35 \text{ to } 7.40 $$
Solution:- The optimal blood pH for enzyme function and oxygen transport is strictly regulated between 7.35 and 7.40.
- This slightly alkaline state is maintained primarily by the carbonic acid-bicarbonate buffer system.
Why other options are incorrect:A pH below 7.35 causes acidosis, and above 7.45 causes alkalosis, both of which can be fatal. The other ranges are physically dangerous for human survival.
[MDCAT (2016)]
Value of \( K_{sp} \) for \( \text{PbSO}_4 \) system at \( 25^{\circ}\text{C} \) is equal to:
A
\( 1.6 \times 10^{-5} \text{ mol}^2 \text{dm}^{-6} \)
B
\( 1.6 \times 10^{-8} \text{ mol}^2 \text{dm}^{-6} \)
C
\( 1.6 \times 10^{-6} \text{ mol}^2 \text{dm}^{-6} \)
D
\( 1.6 \times 10^{-7} \text{ mol}^2 \text{dm}^{-6} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:This is a memory-based fundamental constant for the sparingly soluble salt Lead(II) sulfate at standard room temperature.
Formula:$$ K_{sp}(\text{PbSO}_4) = 1.6 \times 10^{-8} \text{ at } 25^{\circ}\text{C} $$
Solution:- The dissociation is \( \text{PbSO}_{4(s)} \rightleftharpoons \text{Pb}^{2+} + \text{SO}_4^{2-} \).
- Experimental data establishes its solubility product exactly at \( 1.6 \times 10^{-8} \text{ mol}^2 \text{dm}^{-6} \).
Why other options are incorrect:These are incorrect magnitude variations of the standard accepted literature value.
[ETEA (2016)]
\( K_p = K_c(RT) \) in the equation if \( \Delta n < 0 \) then:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The mathematical relationship between \( K_p \) and \( K_c \) depends heavily on the sign of \( \Delta n \) (change in gas moles).
Formula:$$ K_p = K_c (RT)^{\Delta n} $$
Solution:- If \( \Delta n < 0 \) (negative), the term \( (RT)^{\Delta n} \) becomes a fraction \( \frac{1}{(RT)^{|\Delta n|}} \).
- Assuming \( RT > 1 \) (which is true for standard thermodynamic temperatures in Kelvins), multiplying \( K_c \) by a fraction less than 1 yields a smaller number.
- Therefore, \( K_p \) will be less than \( K_c \).
Why other options are incorrect:\( K_p = K_c \) requires \( \Delta n = 0 \). \( K_p > K_c \) requires \( \Delta n > 0 \). Equilibrium constants cannot be negative (eliminating D).
[ETEA (2016)]
\( \text{pK}_a \) values of some acids are given below; choose the weaker acid?
A
\( \text{HClO}_4 \) (\( -10 \))
B
\( \text{HBr} \) (\( -9 \))
C
\( \text{H}_2\text{SO}_4 \) (\( -3 \))
D
\( \text{HCl} \) (\( -7 \))
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The \( \text{pKa} \) is inversely proportional to acid strength. A lower (more negative) \( \text{pKa} \) indicates a stronger acid, while a higher (less negative/positive) \( \text{pKa} \) indicates a weaker acid.
Formula:$$ \text{pKa} = -\log(K_a) $$
Solution:- Compare the given values mathematically: \( -10, -9, -7, -3 \).
- The largest numerical value (closest to zero) is \( -3 \).
- Because \( -3 > -10 \), \( \text{H}_2\text{SO}_4 \) has the highest \( \text{pKa} \) among the listed options.
- Therefore, it is relatively the weakest acid in this specific grouping.
Why other options are incorrect:\( \text{HClO}_4 \) with \( -10 \) is the strongest. The others fall in between.
[MDCAT (2015)]
During the manufacture of nitric acid, nitric oxide is oxidized to nitrogen dioxide. This reaction is given as:
$$ 2\text{NO}_{(g)} + \text{O}_{2(g)} \rightleftharpoons 2\text{NO}_{2(g)} \quad \Delta H = -114 \text{ kJ/mol} $$
According to Le Chatelier's Principle:
A
Reaction must not be temperature dependent
B
Reaction must be carried out at room temperature
C
Reaction must be carried out at low temperature
D
Reaction must be carried out at high temperature
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:Le Chatelier's Principle predicts how an equilibrium shifts in response to temperature changes, depending on whether the reaction is endothermic or exothermic.
Formula:$$ \Delta H < 0 \implies \text{Exothermic Reaction} $$
Solution:- The negative sign in \( \Delta H = -114 \text{ kJ/mol} \) indicates the forward reaction releases heat (exothermic).
- To maximize the yield of \( \text{NO}_2 \), the equilibrium must be shifted forward.
- Decreasing the temperature (removing heat) shifts an exothermic reaction forward to produce more heat.
- Therefore, it must be carried out at a logically low temperature.
Why other options are incorrect:High temperature would favor the reverse endothermic reaction, breaking down \( \text{NO}_2 \). It is fundamentally dependent on temperature.
[MDCAT (2015)]
What is the correct relation between pH and pKa?
A
\( \text{pH} = \text{pKa} + \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \)
B
\( \text{pH} = \text{pKa} - \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \)
C
\( \text{pH} = \text{pKa} - \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \)
D
\( \text{pKa} = \text{pH} + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The Henderson-Hasselbalch equation defines the relationship between the pH of a buffer, the \( \text{pKa} \) of the weak acid, and the concentrations of the acid and its conjugate base.
Formula:$$ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) $$
Solution:- The standard form of the equation has a positive logarithm of Base over Acid.
- By logarithmic properties, \( +\log(x/y) = -\log(y/x) \).
- Applying this: \( +\log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right) = -\log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \).
- Substituting this back gives \( \text{pH} = \text{pK}_a - \log \left( \frac{[\text{Acid}]}{[\text{Base}]} \right) \).
Why other options are incorrect:Option A has a positive sign but inverted ratio. Option B applies a negative sign incorrectly to the standard ratio. Option D algebraically fails when rearranging the standard formula.
[MDCAT (2015)]
Which one of the following is the correct representation for \( K_{sp} \)?
$$ \text{AgCl} \rightleftharpoons \text{Ag}^+ + \text{Cl}^- $$
A
\( K_{sp} = \frac{[\text{AgCl}]}{[\text{Ag}^+][\text{Cl}^-]} \)
B
\( K_{sp} = \frac{[\text{Ag}^+][\text{Cl}^-]}{[\text{AgCl}]} \)
C
\( K_{sp} = [\text{Ag}^+][\text{Cl}^-] \)
D
\( K_{sp} = [\text{AgCl}] \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The solubility product constant (\( K_{sp} \)) is the equilibrium constant for a solid substance dissolving in an aqueous solution, representing the product of dissolved ion concentrations.
Formula:$$ K_{sp} = [\text{Cation}]^x [\text{Anion}]^y $$
Solution:- For the dissolution of \( \text{AgCl} \), the equilibrium expression is initially \( K_c = \frac{[\text{Ag}^+][\text{Cl}^-]}{[\text{AgCl}]} \).
- Because \( \text{AgCl} \) is a pure solid, its concentration is constant and incorporated into the equilibrium constant.
- This creates the new constant \( K_{sp} = [\text{Ag}^+][\text{Cl}^-] \).
Why other options are incorrect:Solid reactants are never included in the denominator of a \( K_{sp} \) expression, eliminating A and B. Option D ignores the ions completely.
[MDCAT (2014)]
The value of equilibrium constant \( K_c \) for the reaction \( 2\text{HF}_{(g)} \rightleftharpoons \text{H}_{2(g)} + \text{F}_{2(g)} \) is \( 10^{-13} \) at \( 2000^{\circ}\text{C} \). Calculate the value of \( K_p \) for this reaction
A
\( 2 \times 10^{-13} \)
B
\( 186 \times 10^{-13} \)
D
\( 3.48 \times 10^{-9} \)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The relationship between \( K_p \) and \( K_c \) is governed by the change in gaseous moles (\( \Delta n \)) during the reaction.
Formula:$$ K_p = K_c(RT)^{\Delta n} $$
Solution:- Calculate \( \Delta n \) for \( 2\text{HF}_{(g)} \rightleftharpoons \text{H}_{2(g)} + \text{F}_{2(g)} \).
- Moles of gaseous products = \( 1 + 1 = 2 \).
- Moles of gaseous reactants = 2.
- \( \Delta n = 2 - 2 = 0 \).
- Therefore, \( K_p = K_c(RT)^0 = K_c(1) = K_c \).
- Since \( K_c = 10^{-13} \), \( K_p \) must also be \( 10^{-13} \).
Why other options are incorrect:Other values would only be correct if \( \Delta n \neq 0 \) and the term \( (RT)^{\Delta n} \) altered the value of \( K_c \).
[MDCAT (2014)]
What will be the pH of a solution of \( \text{NaOH} \) with a concentration of \( 10^{-3} \) M?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:\( \text{NaOH} \) is a strong base that completely dissociates in water. The relationship between pH and pOH must be used.
Formula:$$ \text{pOH} = -\log[\text{OH}^-] $$
$$ \text{pH} + \text{pOH} = 14 $$
Solution:- Since \( \text{NaOH} \) is a strong base, \( [\text{OH}^-] = [\text{NaOH}] = 10^{-3} \text{ M} \).
- Calculate pOH: \( \text{pOH} = -\log(10^{-3}) = 3 \).
- Calculate pH: \( \text{pH} = 14 - \text{pOH} \).
- \( \text{pH} = 14 - 3 = 11 \).
Why other options are incorrect:Option A (3) is the pOH, not the pH, representing an acidic solution which is impossible for NaOH. Options 7 and 14 are standard pH reference points but mathematically incorrect for this concentration.
[MDCAT (2013)]
The chemical substance, when dissolved in water, gives "\( \text{H}^+ \)" is called:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:The classical Arrhenius definition classifies compounds based on the ions they yield in an aqueous solution.
Formula:$$ \text{HA}_{(aq)} \rightarrow \text{H}^+_{(aq)} + \text{A}^-_{(aq)} $$
Solution:- According to Arrhenius theory, an acid is a substance that ionizes in water to yield hydrogen ions (\( \text{H}^+ \)).
- For example, \( \text{HCl} \) dissolves in water to form \( \text{H}^+ \) and \( \text{Cl}^- \).
Why other options are incorrect:A base yields hydroxide ions (\( \text{OH}^- \)) in water. Neutral substances yield neither in excess. Amphoteric substances can act as both acids and bases depending on the environment.
[MDCAT (2013)]
The 'pH' of our blood is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:Human blood operates as a natural physiological buffer system, maintaining a very narrow and specific pH range essential for life and enzyme function.
Formula:$$ \text{pH}_{\text{blood}} \approx 7.4 $$
Solution:- The carbonic acid-bicarbonate buffer system (\( \text{H}_2\text{CO}_3 / \text{HCO}_3^- \)) is the primary buffer in human blood.
- This system tightly regulates blood pH between 7.35 and 7.40 (slightly alkaline).
- Deviations outside this range lead to conditions like acidosis or alkalosis.
Why other options are incorrect:6.7 is severely acidic, and 7.9 is severely alkaline for human blood, both being fatal. 7.5 is outside the strict normal physiological range.
[MDCAT (2012)]
Formation of \( \text{NH}_3 \) is reversible and exothermic process, what will happen on cooling?
A
More reactant will form
B
More \( \text{H}_2 \) will be formed
C
More \( \text{N}_2 \) will be formed
D
More product (\( \text{NH}_3 \)) will be formed
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:According to Le Chatelier's Principle, changing the temperature of a system at equilibrium will shift the reaction to counteract the imposed change.
Formula:$$ \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 + \text{Heat} $$
Solution:- The forward reaction is exothermic, meaning it produces heat.
- Cooling the system removes heat.
- To restore equilibrium, the system will shift in the direction that generates more heat.
- Therefore, it shifts forward, yielding more product (\( \text{NH}_3 \)).
Why other options are incorrect:Forming more reactants (\( \text{N}_2 \) or \( \text{H}_2 \)) would require shifting backwards, which is an endothermic process and would be favored by heating, not cooling.
[MDCAT (2012)]
A buffer solution is that which resists/minimizes the change in
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
Concept:A buffer is defined by its ability to maintain a relatively stable hydrogen ion concentration upon the addition of small amounts of acid or base.
Formula:$$ \text{pH} = \text{pK}_a + \log \frac{[\text{Salt}]}{[\text{Acid}]} $$
Solution:- Buffers contain a weak acid and its conjugate base (or a weak base and its conjugate acid).
- These components neutralize small added amounts of \( \text{H}^+ \) or \( \text{OH}^- \).
- As a direct result, the pH of the solution is kept stable.
Why other options are incorrect:While resisting pH implicitly resists pOH, the standard chemical definition specifically highlights the resistance to changes in pH. \( \text{pK}_a \) and \( \text{pK}_b \) are constants that do not change regardless of the buffer action.
[MDCAT (2011)]
If in \( \text{AgCl} \) solution, some salt of \( \text{NaCl} \) is added, \( \text{AgCl} \) will be precipitated due to:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
Concept:The addition of an ion that is already present in an equilibrium mixture of a sparingly soluble salt shifts the equilibrium toward solid precipitation.
Formula:$$ \text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} $$
Solution:- \( \text{NaCl} \) is a strong electrolyte and dissociates completely into \( \text{Na}^+ \) and \( \text{Cl}^- \) ions.
- This significantly increases the concentration of \( \text{Cl}^- \) ions in the solution.
- According to Le Chatelier's Principle, the equilibrium shifts backwards to consume the excess \( \text{Cl}^- \).
- This results in the precipitation of more solid \( \text{AgCl} \), a phenomenon known as the common ion effect.
Why other options are incorrect:Solubility generally describes the ability to dissolve, not precipitate. The term 'unsaturation' would imply more could dissolve, opposite to precipitation.
[MDCAT (2010)]
Units of \( K_c \) for the following reaction is:
$$ \text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI} $$
A
\( \text{mol}^2 \text{dm}^{-6} \)
C
\( \text{mol dm}^{-3} \)
D
\( \text{mol}^{-2} \text{dm}^{6} \)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
Concept:The unit of the equilibrium constant \( K_c \) is entirely dependent on the change in the number of moles of gas (\( \Delta n \)) between products and reactants.
Formula:$$ \text{Units of } K_c = (\text{mol dm}^{-3})^{\Delta n} $$
Solution:- First, calculate \( \Delta n = n_p - n_r \).
- For \( \text{H}_2 + \text{I}_2 \rightleftharpoons 2\text{HI} \), moles of product \( n_p = 2 \).
- Moles of reactant \( n_r = 1 + 1 = 2 \).
- \( \Delta n = 2 - 2 = 0 \).
- Thus, \( (\text{mol dm}^{-3})^0 = 1 \), leaving the constant with no unit.
Why other options are incorrect:Other options represent units for reactions where there is a net change in gaseous moles (\( \Delta n \neq 0 \)).
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