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Thermochemistry Past Papers
Solved past paper MCQs for Thermochemistry from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Official printable A4 booklet containing all 65 verified past paper MCQs with unbroken options, 5-column rapid answer key, and Propolis cognitive error autopsies.
Burning of coal is a highly energetic combustion reaction. Carbon reacts with oxygen to form highly stable carbon dioxide bonds, releasing massive amounts of heat and light. This is definitively an exothermic chemical reaction.
Note: Freezing and condensation are exothermic physical phase changes, but combustion is universally recognized as the primary exothermic chemical reaction in this context.
Why other options are incorrect:
Melting of ice is an endothermic phase change (absorbs heat). While freezing and condensation release heat, "burning of coal" is the textbook chemical reaction sought by the examiner.
Born Haber cycle is used to calculate the lattice energy of which compound?
A
\( PbS \)
B
\( RbBr \)
C
\( SO_2 \)
D
\( NH_3 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The Born-Haber cycle strictly applies to the formation of ionic lattices. It models the electrostatic attraction between discrete gaseous cations (Group 1/2 metals) and anions (Halogens/Oxygen).
Solution:
We must identify the most archetypal ionic compound among the choices.
RbBr (Rubidium Bromide) is formed from a highly electropositive Group 1 alkali metal and a highly electronegative Group 17 halogen. It forms a perfect, purely ionic lattice.
The cycle perfectly maps the ionization of Rb and electron affinity of Br to calculate its lattice energy.
Why other options are incorrect:
\( SO_2 \) and \( NH_3 \) are covalent molecules. \( PbS \) has significant covalent character and does not fit the ideal ionic Born-Haber model as cleanly as an alkali halide like RbBr.
(Note: The cycle describes a path from A to B directly, or indirectly via C and D: A → C → D → B. Where \( A
ightarrow C = -540 \text{ kJ} \), \( C
ightarrow D = +200 \text{ kJ} \), \( D
ightarrow B = -100 \text{ kJ} \))
A
\( -400 \text{ KJ/mol} \)
B
\( -600 \text{ KJ/mol} \)
C
\( -440 \text{ KJ/mol} \)
D
\( -640 \text{ KJ/mol} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Hess's Law states that the overall enthalpy change of a reaction is equal to the algebraic sum of the enthalpy changes of individual steps in a thermodynamic cycle, provided the initial and final states are identical.
Calculate the work done when 1 mole of an ideal gas expands from \( 15 \text{ dm}^3 \) to \( 20 \text{ dm}^3 \) against a constant external pressure of 2 atmospheres.
A
\( -5 \text{ atm.dm}^3 \)
B
\( 5 \text{ atm.dm}^3 \)
C
\( 10 \text{ atm.dm}^3 \)
D
\( -10 \text{ atm.dm}^3 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The mechanical work done during the expansion or compression of a gas is calculated using pressure-volume formulas. Depending on the convention (physics vs. chemistry), the absolute magnitude is identical.
Formula:
$$ W = P \times \Delta V $$
Solution:
The constant external pressure (\( P \)) = 2 atmospheres.
The change in volume (\( \Delta V \)) = \( V_{\text{final}} - V_{\text{initial}} = 20 - 15 = 5 \text{ dm}^3 \).
Note: In standard IUPAC chemistry convention, expansion work is done by the system and is technically negative (\( -10 \text{ atm.dm}^3 \)). However, the ETEA key and provided options utilize the absolute magnitude / physics convention yielding a positive \( 10 \text{ atm.dm}^3 \).
Why other options are incorrect:
Values of 5 or -5 arise if the pressure is completely ignored or misapplied.
When 1 mole of ice melts at \( 0^\circ\text{C} \) and constant pressure of 1 atmosphere, 6025 J of heat is absorbed by the system. The molar volume of ice and water are 0.020 and \( 0.018 \text{ dm}^3 \), respectively. Calculate \( \Delta E \). (\( 1 \text{ dm}^3\text{.atm} = 101.33 \text{ J} \))
A
\( 6010.20 \text{ J} \)
B
\( 6020.20 \text{ J} \)
C
\( 6015.20 \text{ J} \)
D
\( 6025.20 \text{ J} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The first law of thermodynamics requires accounting for the work done when a system changes volume, even during phase changes like melting.
Formula:
$$ \Delta E = q - P\Delta V $$
Solution:
Heat absorbed (\( q \)) = \( +6025 \text{ J} \).
Change in volume (\( \Delta V \)) = \( V_{\text{water(liquid)}} - V_{\text{ice(solid)}} \). \( \Delta V = 0.018 - 0.020 = -0.002 \text{ dm}^3 \) (Water uniquely shrinks when melting).
Calculate Work (\( W = P\Delta V \)) in Joules: \( W = 1 \text{ atm} \times -0.002 \text{ dm}^3 = -0.002 \text{ atm.dm}^3 \). Convert to Joules: \( -0.002 \times 101.33 = -0.2026 \text{ J} \).
One slice of bread with a tablespoon of peanut butter on it contains 20g carbohydrate, 10g protein, and 9g fat. Calculate total energy consumed in this intake:
A
173 kcal
B
218 kcal
C
158 kcal
D
201 kcal
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
In nutritional thermochemistry, the total metabolic energy of food is calculated using standard physiological caloric values: Carbohydrates = \( 4 \text{ kcal/g} \), Proteins = \( 4 \text{ kcal/g} \), and Fats = \( 9 \text{ kcal/g} \).
The heat of sublimation of potassium is \( 98 \text{ kJ/mol} \), the heat of dissociation of bromine gas is \( 192.5 \text{ kJ/mol} \). The ionization energy of K is \( 414 \text{ kJ/mol} \). The electron affinity of Br is \( -334.7 \text{ kJ/mol} \) and the heat of formation of KBr is \( -405.8 \text{ kJ/mol} \). Calculate the lattice energy of KBr.
A
\( -669.5 \)
B
\( 679.3 \)
C
\( -679.3 \)
D
\( 669.5 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The Born-Haber cycle uses Hess's law to map out the entire energy profile of an ionic solid. We must sum all steps to equal the total heat of formation.
Formula:
$$ \Delta H_f = \Delta H_{\text{sub}} + IE + \frac{1}{2}\Delta H_{\text{diss}} + EA + LE $$
Solution:
Because KBr contains only one Br atom, we only need half a mole of \( Br_2 \) gas. Thus, dissociation energy = \( 192.5 / 2 = 96.25 \text{ kJ/mol} \).
Set up the equation: \( -405.8 = 98 \text{ (sub)} + 414 \text{ (IE)} + 96.25 \text{ (diss)} + (-334.7) \text{ (EA)} + LE \)
Sum the indirect step energies: \( 98 + 414 + 96.25 - 334.7 = 273.55 \text{ kJ/mol} \).
Solve for Lattice Energy (LE): \( LE = -405.8 - 273.55 = -679.35 \text{ kJ/mol} \).
The closest matched option is exactly \( -679.3 \).
Why other options are incorrect:
Option A (-669.5) results from mathematical errors. Positive values (B and D) imply the formation of the lattice absorbs energy, which violates the fundamental stability laws of ionic compounds.
Concept: Volume (V) is an extensive state function whose value depends solely on the current thermodynamic state of the system, independent of the path taken. Work (w) and heat (q) are path functions.
Photosynthesis is the process by which green plants convert carbon dioxide and water into glucose.
Because this reaction is utterly dependent on absorbing incoming solar photon energy to drive the non-spontaneous uphill synthesis of sugar, it is heavily endothermic.
Why other options are incorrect:
Nuclear fission releases world-ending amounts of energy (highly exothermic). Synthesis of ammonia (Haber process) is exothermic (\( -92 \text{ kJ} \)). Oxidation reactions (combustion-like) are universally exothermic.
\( NaOH \) is a strong base, and \( HNO_3 \) is a strong nitric acid.
They undergo a classic double displacement reaction where the \( H^+ \) and \( OH^- \) ions combine to form water.
The specific enthalpy change (\( -57.4 \text{ kJ/mol} \)) associated with the formation of 1 mole of water via an acid and base is formally defined as Heat of neutralization.
Why other options are incorrect:
It is not combustion (requires \( O_2 \)). It is not formation (which requires elemental reactants like solid Na and gaseous \( N_2/O_2 \)). It is not decomposition (which requires a single reactant breaking down).
This conceptual question tests the ability to distinguish between exclusive, specific thermodynamic definitions. "Enthalpy of reaction" is the broad, generalized term used when a reaction does not cleanly fit the strict definitions of formation, combustion, or atomization.
Solution:
Option A is the enthalpy of Formation of water (and also Combustion of hydrogen).
Option D is the enthalpy of Combustion of carbon monoxide.
Option B has sulfur in a gaseous state (\( S_{(g)} \)), which violates standard state rules (Sulfur is a solid at standard conditions), disqualifying it from standard terminology.
Option C is a classic thermite displacement reaction. It is neither a formation from elements, nor a standard combustion with oxygen gas. Therefore, its energy change is exclusively classified by the generalized term: standard enthalpy of reaction (\( \Delta H_{rxn} \)).
Why other options are incorrect:
As noted, options A and D hold more highly specific thermochemical names. The question seeks the equation where the broad term is its only applicable label.
Expansion takes place when a gas is evolved during a chemical reaction between marble chips and dilute HCl. Predict the energy change and work done:
A
w is positive, q is negative
B
w is negative, q is negative
C
q is positive, no work done
D
w is positive, q is positive
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The reaction between Calcium Carbonate (marble) and Hydrochloric Acid produces \( CaCl_2 \), water, and Carbon Dioxide gas. This is a classic exothermic reaction resulting in a large volume expansion.
Note on Exam Fidelity: This past paper question is fundamentally flawed. No single option contains the physically correct pair of signs under strict IUPAC conventions.
Heat (q): The reaction releases heat (exothermic). Therefore, \( q \) should be negative.
Work (w): A gas (\( CO_2 \)) is evolved, causing the system to expand. The system does work on the surroundings. In IUPAC chemistry conventions, expansion work is negative (\( w = -P\Delta V \)).
Since the correct answer is (\( w \) is negative, \( q \) is negative), all given options are false. We map to B to maintain alignment with historical attempts to grade this impossible question, but the foundational physics is clear.
Why other options are incorrect:
Work cannot be positive during an expansion (rules out A and D). Heat cannot be positive during an exothermic acid-carbonate reaction (rules out B, C, D).
\( \Delta H^\circ \) for the sublimation of one mole of iodine from the following equations will be: $$ H_{2(g)} + I_{2(s)} \rightarrow 2HI_{(g)} \quad \Delta H^\circ = +51.8 \text{ kJ/mol} $$ $$ H_{2(g)} + I_{2(g)} \rightarrow 2HI_{(g)} \quad \Delta H^\circ = -10.5 \text{ kJ/mol} $$ (Note: Original paper contained state typos; corrected here for thermodynamic mathematical validity)
A
\( 41.3 \text{ kJ/mol} \)
B
\( 53.5 \text{ kJ/mol} \)
C
\( 62.3 \text{ kJ/mol} \)
D
\( 36.5 \text{ kJ/mol} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Sublimation is the direct phase transition from a solid to a gas: \( I_{2(s)} \rightarrow I_{2(g)} \). We can determine its enthalpy using Hess's Law by algebraically manipulating given reactions.
41.3 results from blindly adding the numbers without flipping the necessary equation (\( 51.8 - 10.5 \)). The other numbers are mathematically irrelevant.
What is the reason of energy changes in chemical reactions?
[USH (2023)]
A
Bond formation
B
Bond breakage
C
Ionic bonds
D
Bond formation and breakage
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Chemical reactions are fundamentally the rearrangement of atoms. This structural rearrangement necessitates the destruction of existing molecular architectures and the creation of new ones.
Reactant molecules must first have their constituent bonds broken apart. This step strictly requires an input of energy.
The isolated atoms then rearrange and form new bonds to create products. This step strictly releases energy.
The net energy change (exothermic or endothermic) is merely the difference between these two concurrent processes.
Therefore, energy changes in a reaction are due to both bond formation and breakage.
Why other options are incorrect:
Selecting only bond formation or only bond breakage represents just half of the chemical process. Ionic bonds are a specific type of bond, not a universal reason for all reaction energy changes.
Heat capacity (C) of a body or system is defined as the quantity of heat energy required to change its temperature by exactly 1 Kelvin (or 1 degree Celsius).
Formula:
$$ C = \frac{q}{\Delta T} $$
Solution:
The numerator, heat (q), is measured in Joules (J) or kiloJoules (kJ).
The denominator, change in temperature (\( \Delta T \)), is measured in Kelvin (K).
Therefore, the unit becomes \( \text{J/K} \) or, scaled up, \( \text{kJ/K} \).
Note: The past paper key accepts \( \text{kJ/K} \) (Option C in original print) as the specific answer here, though \( \text{JK}^{-1} \) is also physically valid. We faithfully preserve the source's verified correct answer.
Why other options are incorrect:
\( \text{kJmol}^{-1} \) is the unit for molar enthalpy, not heat capacity (it lacks a temperature dimension). \( \text{kJmol} \) is dimensional nonsense.
For exothermic reaction \( \Delta H^\circ \) is written as:
A
Zero
B
Positive
C
Constant
D
Negative
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The sign convention for the change in enthalpy (\( \Delta H \)) is rooted in the system's perspective. Loss of energy to the surroundings is a negative change for the system.
Formula:
$$ \Delta H = H_{\text{products}} - H_{\text{reactants}} $$
Solution:
In an exothermic reaction, the chemical system releases energy into the surroundings as heat.
This means the final internal enthalpy of the products is lower than the initial internal enthalpy of the reactants (\( H_P < H_R \)).
Subtracting a larger number from a smaller number results in a mathematical deficit, so \( \Delta H \) must strictly be written as Negative.
Why other options are incorrect:
A positive \( \Delta H \) means energy was absorbed (endothermic). Zero means no net heat exchange occurred (isothermic). Constant implies no change, which contradicts a chemical reaction occurring.
Processes involving solids and liquids have their:
A
\( \Delta H = \Delta E + P\Delta V \)
B
\( \Delta H \approx \Delta E \)
C
\( \Delta H = P\Delta V \)
D
\( \Delta H = 0 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Enthalpy (\( H \)) is defined as the internal energy (\( E \)) plus the product of pressure and volume (\( PV \)). For chemical processes, the change in enthalpy relates to the change in internal energy and volume.
Formula:
$$ \Delta H = \Delta E + P\Delta V $$
Solution:
When a reaction involves strictly solids and liquids, the change in volume (\( \Delta V \)) is extremely negligible because condensed phases are virtually incompressible.
Therefore, the term \( P\Delta V \) approaches zero.
Substituting this into the equation yields: \( \Delta H = \Delta E + 0 \).
Thus, for solids and liquids, the enthalpy change is approximately equal to the internal energy change: \( \Delta H \approx \Delta E \).
Why other options are incorrect:
Option A is the general formula for all states (including gases). Options C and D are mathematically incorrect simplifications that ignore the massive internal energy component.
What is the mathematical statement of the first law of thermodynamics?
A
\( \Delta E = q - w \)
B
\( \Delta E = -q - w \)
C
\( \Delta E = -q + w \)
D
\( \Delta E = q + w \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The first law of thermodynamics is the law of conservation of energy. It states that the internal energy of an isolated system is constant, and any change in internal energy (\( \Delta E \)) must equal the sum of heat and work exchanged.
Formula:
$$ \Delta E = q + w $$
Solution:
In the standard IUPAC chemical convention, heat added to the system is positive (\( +q \)).
Work done on the system is also positive (\( +w \)).
Therefore, the total accumulation of internal energy is strictly represented as \( \Delta E = q + w \).
Why other options are incorrect:
Equations with negative signs represent non-standard sign conventions (e.g., work done by the system) or violate the conservation of energy if applied improperly.
What is \( \Delta H \) equal to —— according to the first law of thermodynamics?
A
\( \Delta E + P\Delta V \)
B
\( \Delta E - P\Delta V \)
C
\( -\Delta E - P\Delta V \)
D
\( -\Delta E + P\Delta V \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Enthalpy (\( H \)) is a state function defined fundamentally as internal energy plus pressure-volume energy.
Formula:
$$ H = E + PV $$
Solution:
If a thermodynamic process occurs at constant pressure, the change in enthalpy is given by taking the delta of the definition:
$$ \Delta H = \Delta E + P\Delta V $$
This directly aligns with the First Law when substituting heat at constant pressure (\( q_p = \Delta H \)) and expansion work (\( w = -P\Delta V \)): \( \Delta E = \Delta H - P\Delta V \), which rearranges to the same equation.
Why other options are incorrect:
Option B uses a negative sign which contradicts the definition of enthalpy. Options C and D improperly apply negative signs to the internal energy term.
State functions are macroscopic properties that depend only on the current state of a system. Path functions depend on the specific sequence of steps taken to transition between states.
Solution:
Pressure, Temperature, and Volume define the exact thermodynamic coordinate of a system. They are classic state functions.
Work (W) is energy transferred due to an applied force over a distance (like a piston expanding). The amount of work done drastically changes depending on whether the expansion was done reversibly (maximum work) or irreversibly.
Therefore, Work is a path function, not a state function.
Why other options are incorrect:
P, T, and V are universally recognized as fundamental thermodynamic state functions.
The equation shows isolated gaseous ions (\( Mg^{2+} \) and \( O^{2-} \)) coming together to form a solid crystal lattice (\( MgO_{(s)} \)).
The massive amount of energy released during this highly specific phase and structural transition is universally defined as the Lattice energy (or Lattice enthalpy of formation).
Why other options are incorrect:
Atomization forms gaseous atoms from a solid, not a solid from gaseous ions. Neutralization involves \( H^+ \) and \( OH^- \) forming liquid water. Solution involves dissolving a solid into aqueous ions.
Born-Haber's cycle is used to determine the lattice energy of ionic compounds. It is the application of:
A
Le-Chatelier's principle
B
Henry's law
C
Hess's law
D
Common ion effects
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The Born-Haber cycle breaks down the complex formation of an ionic solid into a hypothetical sequence of discrete, measurable steps (sublimation, ionization, etc.).
The foundation of the Born-Haber cycle relies entirely on the principle that the total enthalpy change of a reaction is path-independent.
Because we equate the direct formation path (\( \Delta H_f \)) to the sum of the multi-step indirect path, we are directly applying Hess's law of constant heat summation.
Why other options are incorrect:
Le-Chatelier's principle deals with dynamic equilibrium shifts. Henry's law concerns gas solubility in liquids. Common ion effect involves shifting solubility product equilibria.
The correct equation for the first law of thermodynamics is:
A
\( \Delta E = \Delta q + p \Delta v \)
B
\( \Delta E = w - q \)
C
\( \Delta E = q + w \)
D
\( \Delta E = \Delta q - p \Delta v \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The first law of thermodynamics defines the conservation of internal energy (\( \Delta E \)) relative to heat exchange (\( q \)) and work (\( w \)).
Formula:
$$ \Delta E = q + w $$
Solution:
The universally standard expression is that the change in internal energy equals the heat added to the system plus the work done on the system.
Hence, the correct mathematical equation is \( \Delta E = q + w \).
Note: In some physics conventions, work done by the system is used, leading to \( \Delta E = q - w \) or \( \Delta E = q - P\Delta V \), but the primary IUPAC chemical convention is \( q + w \).
Why other options are incorrect:
Option B uses minus q, which misaligns with conservation rules. Option D uses minus \( p\Delta v \) but paired with a delta q (heat is not a state function, so \( \Delta q \) is improper notation).
The thermal energy at constant pressure is called:
A
Heat capacity
B
Internal energy
C
Enthalpy
D
Work done
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
When heat is transferred to a system, the conditions under which it occurs (constant volume vs. constant pressure) dictate which thermodynamic property changes.
Formula:
$$ q_p = \Delta H $$
Solution:
When heat is added at a constant volume (\( q_v \)), all energy goes into changing the internal energy (\( \Delta E \)) because no \( P\Delta V \) work can be done.
When heat is added at a constant pressure (\( q_p \)), the system is allowed to expand and do work. The total heat added accounts for both the internal energy change and the work done.
This specific quantity is defined as the change in Enthalpy (H).
Why other options are incorrect:
Internal energy equals heat at constant volume. Heat capacity is the ratio of heat to temperature change. Work done is a separate component (\( P\Delta V \)).
A system that can exchange or transfer both matter and energy with the surroundings is:
A
Open system
B
Adiabatic system
C
Isolated system
D
Closed system
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Thermodynamic systems are classified based on the permeability of their boundaries to matter and energy (heat/work).
Categories:
Open: Exchanges both Energy and Matter.
Closed: Exchanges Energy only.
Isolated: Exchanges Neither.
Solution:
A beaker of boiling water without a lid is a classic example. Heat can enter or leave (energy transfer), and water vapor can escape (matter transfer).
By definition, a system permitting both exchanges is an open system.
Why other options are incorrect:
A closed system restricts matter transfer. An isolated system restricts both matter and energy transfer. An adiabatic system strictly prevents heat exchange.
The sum of all the energies of all the molecules or atoms of a substance is called its:
A
Latent heat
B
Internal energy
C
Specific heat
D
Heat capacity
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A macroscopic chemical substance contains countless microscopic particles, all possessing kinetic and potential energies.
Formula:
$$ U (\text{or } E) = E_{\text{kinetic}} + E_{\text{potential}} $$
Solution:
Atoms and molecules possess translational, vibrational, and rotational kinetic energies, as well as electronic and nuclear potential energies.
The absolute sum of all these microscopic kinetic and potential energies within a defined system is known as the system's Internal energy (denoted as E or U).
Why other options are incorrect:
Specific heat and heat capacity are properties measuring how temperature changes relative to added heat. Latent heat is energy associated with phase changes. None of these represent the total energy pool of the substance.
The equation of the first law of thermodynamics \( q = \Delta E + W \) will reduce at constant volume to:
A
\( q_v = \Delta E + P\Delta V \)
B
\( q_v = \Delta E + W \)
C
\( q_v = \Delta E \)
D
\( \Delta E + W \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The work done (W) by an expanding gas against an external pressure is given by \( P\Delta V \). If the volume is artificially kept constant, no physical expansion work can occur.
Formula:
$$ W = P\Delta V $$ $$ q = \Delta E + P\Delta V $$
Solution:
At constant volume, the change in volume (\( \Delta V \)) is strictly zero.
Therefore, the expansion work \( W = P \times 0 = 0 \).
Substituting this back into the first law equation (\( q = \Delta E + W \)) yields: \( q_v = \Delta E + 0 \).
The equation cleanly reduces to \( q_v = \Delta E \).
Why other options are incorrect:
Option A and B still include the work term (\( P\Delta V \) or \( W \)), which violates the condition of constant volume. Option D lacks the equality to \( q_v \).
Reaction taking place in one step or in several have same \( \Delta H \) is statement of:
A
Hess's law
B
Charle's law
C
Lussac's Law of combining volume
D
Boyle's law
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Because enthalpy is a state function, the overall change in enthalpy depends solely on the initial reactants and final products, completely independent of the route or number of intermediate steps taken.
This fundamental principle of thermochemistry is formally defined as Hess's Law of Constant Heat Summation.
It states that if a chemical change occurs by multiple routes, the overall enthalpy change is identical.
Why other options are incorrect:
Boyle's and Charles's laws govern the pressure, volume, and temperature relationships of ideal gases. Lussac's law governs the volume ratios of reacting gases. None of these relate to reaction enthalpy.
The conversion of a neutral atom into a positively charged ion (cation) involves the removal of an electron. This is quantified by the Ionization Energy.
Electrons are held in orbit by the electrostatic attractive pull of the positive nucleus.
To overcome this attraction and remove the valence electron from a sodium atom, a strict input of energy is required.
Because energy must be absorbed from the surroundings to drive the process, it is strictly an endothermic process (\( \Delta H = +\text{ve} \)).
Why other options are incorrect:
It is never exothermic because stripping an electron from an atom always requires energy. Sublimation refers to solid converting to gas, not ionization.
The lattice energy of an ionic compound cannot be measured directly through a single experiment. It must be deduced indirectly using Hess's Law via a structured cycle of measurable steps.
The Born-Haber cycle was specifically developed by Max Born and Fritz Haber to relate the standard enthalpy of formation of an ionic solid to its component ionic formation steps.
By substituting the known experimental values into the cycle, one can algebraically solve for the unknown Lattice energy of ionic compounds.
Why other options are incorrect:
Covalent compounds do not form electrostatic lattices of ions. Enthalpy of combustion is calculated via bomb calorimetry. Ionization enthalpies are direct inputs into the cycle, not the final calculated objective.
Specific heat capacity (c) is the amount of heat energy required to raise the temperature of exactly one gram of a substance by one degree Celsius (or one Kelvin).
Water has an exceptionally high specific heat capacity due to the extensive network of strong intermolecular hydrogen bonds holding the molecules together.
Experimentally, it takes exactly 1 calorie of energy to raise 1g of water by 1°C.
Since 1 calorie = 4.18 Joules, the specific heat of water is \( 4.18 \text{ J/g/}^\circ\text{C} \).
Why other options are incorrect:
Options A, C, and D are random numerical values that do not correspond to the known physical properties of liquid water.
Enthalpy change of combustion is exothermic because:
A
Energy absorbed in bond breaking is greater
B
Energy absorbed in bond making is greater
C
Energy released in bond breaking is greater
D
Energy released in bond making is greater
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
A chemical reaction involves two distinct thermochemical steps: breaking old bonds (which always requires absorbing energy) and forming new bonds (which always releases energy). The overall enthalpy (\( \Delta H \)) is the net sum.
Formula:
$$ \Delta H = \Sigma H_{\text{bond breaking (absorbed)}} - \Sigma H_{\text{bond making (released)}} $$
Solution:
In a combustion reaction, fuel molecules and oxygen molecules are broken apart. This requires an initial input of energy (endothermic).
However, the newly formed products (usually \( CO_2 \) and \( H_2O \)) have exceedingly strong, stable covalent bonds.
The formation of these highly stable bonds releases a massive amount of energy.
Because the energy released in bond making is much greater than the energy required for bond breaking, the net reaction is highly exothermic.
Why other options are incorrect:
Bond breaking always absorbs energy, it never releases it (disproving option C). Bond making releases energy, it never absorbs it (disproving option B). If energy absorbed in bond breaking were greater (Option A), the reaction would be endothermic.
The calorie (cal) is defined historically based on the thermal properties of water. It is a fundamental conversion factor in thermodynamics to transition to the SI unit of energy (Joule).
Formula:
$$ 1 \text{ cal} = 4.184 \text{ J} $$
Solution:
By strict international definition, one standard thermochemical calorie is exactly \( 4.184 \text{ Joules} \).
This is derived from the specific heat capacity of water.
Why other options are incorrect:
Option A has a decimal error. Option C uses Kilojoules (which would equal 4184 Joules). Option D is \( 0.4184 \text{ KJ} \), which equals 418.4 Joules, not 4.184.
Energy, whether it is heat, work, or internal energy, can be measured using multiple metric and non-metric unit systems depending on the scientific context.
Which one of the following is not an example of state function?
A
Volume(V)
B
Heat(q)
C
Temperature(T)
D
Enthalpy(H)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Thermodynamic properties are divided into state functions (path-independent) and path functions (path-dependent).
Formula:
$$ \Delta E = q + w $$
Solution:
Temperature (T), Volume (V), and Enthalpy (H) are strictly determined by the current macroscopic state of a system. You can calculate their change purely by knowing the initial and final states. Thus, they are state functions.
Heat (q) represents the transfer of thermal energy. The amount of heat transferred depends entirely on how the process is executed (the path taken between states).
Therefore, Heat is a path function, not a state function.
Why other options are incorrect:
Volume, Temperature, and Enthalpy are all classic examples of state functions.
Thermochemical sign conventions dictate the flow of energy relative to the system. An exothermic reaction is a system that releases thermal energy to its surroundings.
Formula:
$$ \Delta H = H_{\text{products}} - H_{\text{reactants}} $$
Solution:
Because heat leaves the system, the internal enthalpy of the system decreases.
This means the final state (products) has less energy than the initial state (reactants).
Subtracting a larger initial value from a smaller final value yields a mathematically Negative result (\( \Delta H < 0 \)).
Why other options are incorrect:
Positive \( \Delta H \) indicates an endothermic process. "More than 1" and "Neutral" do not describe standard thermodynamic sign conventions.
The First Law of Thermodynamics is the law of conservation of energy applied to chemical systems. It states that energy cannot be created or destroyed, only transferred as heat or work.
Formula:
$$ \Delta E = q + w $$
Solution:
The mathematical statement of the 1st law dictates that the total change in a system's internal energy (\( \Delta E \) or \( \Delta U \)) is equal to the heat added to the system (\( q \)) plus the work done on the system (\( w \)).
Therefore, \( \Delta E = q + w \) is the universally recognized correct equation.
Why other options are incorrect:
Option B is Boyle's Law for gases. Option C is the de Broglie wavelength equation from quantum mechanics. Option D is Einstein's mass-energy equivalence equation.
The thermal energy at constant pressure is called:
A
Work done
B
Internal energy
C
Heat capacity
D
Enthalpy
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
According to the first law of thermodynamics, heat transferred to a system can either change its internal energy or perform pressure-volume work.
Formula:
$$ q_p = \Delta E + P\Delta V = \Delta H $$
Solution:
When heat (q) is supplied to a system at constant pressure, the system may expand, doing work (\( P\Delta V \)).
The total heat absorbed accounts for both the change in internal energy (\( \Delta E \)) and the expansion work.
This specific quantity (\( q_p \)) is defined thermodynamically as the change in Enthalpy (\( \Delta H \)).
Why other options are incorrect:
Internal energy (\( \Delta E \)) equals the thermal energy at constant volume (\( q_v \)), not pressure. Heat capacity is the ratio of heat added to temperature change, not the thermal energy itself.
Born-Haber cycle is used to determine the lattice energies of:
A
Covalent solids
B
Metallic solids
C
Ionic solids
D
Molecular solids
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The Born-Haber cycle is an application of Hess's Law designed specifically to analyze the thermochemical steps involved in forming crystal lattices from charged particles.
The cycle relies on measuring ionization energies (to form cations) and electron affinities (to form anions).
These discrete charged gaseous ions then condense to form a solid lattice.
This exact mechanism of cation-anion electrostatic attraction strictly defines the formation of Ionic solids (e.g., NaCl, KBr).
Why other options are incorrect:
Covalent, metallic, and molecular solids are held together by covalent bonding, metallic "sea of electrons", and intermolecular forces respectively, which do not fit the cation-anion formation model of the Born-Haber cycle.
The given diagram shows the enthalpy changes during a chemical reaction:
This diagram represents:
A
An exothermic reaction
B
An endothermic reaction
C
A non-spontaneous process
D
An isothermic process
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
An enthalpy profile diagram plots the internal energy of substances against the progress of a reaction. The relative vertical positions of reactants and products indicate the nature of the heat exchange.
Formula:
$$ \Delta H = H_{\text{products}} - H_{\text{reactants}} $$
Solution:
In the provided diagram, the energy level of the Reactants is higher than the energy level of the Products (\( E_R > E_P \)).
To drop to this lower energy state, the system must shed the excess energy to the surroundings in the form of heat.
Because heat is released (\( \Delta H = -\text{ve} \)), this specifically represents an exothermic reaction.
Why other options are incorrect:
If it were an endothermic reaction, the products would sit at a higher energy level than the reactants. Isothermic means no temperature change occurs, which contradicts a standard enthalpy drop.
Heat of formation of \( MgO \) is given below. $$ Mg + \frac{1}{2}O_{2(g)} \rightarrow MgO_{(s)} \quad \Delta H = -692 \text{ kJmol}^{-1} $$ This equation shows that:
A
The product is very stable
B
The product is highly unstable
C
The reactants are very stable
D
The reaction is endothermic
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In thermodynamics, the stability of a compound is directly tied to its standard enthalpy of formation (\( \Delta H_f^\circ \)). The more negative the enthalpy of formation, the more energy was lost, and the deeper the compound sits in a potential energy well.
Formula:
$$ \Delta H = -692 \text{ kJmol}^{-1} $$
Solution:
The formation of \( MgO \) releases a massive amount of energy (\( 692 \text{ kJ} \) for every mole).
Because so much internal energy has been lost to the surroundings, the resulting solid \( MgO \) lattice is in a much lower energy state than the starting elements.
Lower energy directly correlates with higher thermodynamic stability. Hence, the product is very stable.
Why other options are incorrect:
The reaction is clearly exothermic (negative sign), ruling out endothermic. Because the reactants spontaneously lost so much energy to become \( MgO \), they were relatively less stable compared to the product.
Which enthalpy change is relevant in the following process? \( Na_{(s)} \rightarrow Na_{(g)} \quad \Delta H = +108 \text{ kJmol}^{-1} \)
A
Enthalpy of fusion
B
Enthalpy of formation
C
Enthalpy of vaporization
D
Enthalpy of atomization
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Identifying the correct thermodynamic term requires matching the physical states and stoichiometry of the provided reaction equation to established definitions.
The equation shows solid sodium (\( Na_{(s)} \)), which is its standard state, transforming directly into isolated gaseous sodium atoms (\( Na_{(g)} \)).
Exactly 1 mole of gaseous atoms is produced.
This precise transformation is the definition of the Enthalpy of atomization (\( \Delta H_{at}^\circ \)). Note: For a monoatomic solid like Na, atomization is numerically equal to the enthalpy of sublimation.
Why other options are incorrect:
Enthalpy of fusion refers to Solid \( \rightarrow \) Liquid. Enthalpy of vaporization refers to Liquid \( \rightarrow \) Gas. Neither applies because this equation skips the liquid phase and starts from a solid to form gaseous atoms.
When two moles of \( H_2 \) and one mole of \( O_2 \) react to form \( H_2O \) 484KJ heat is evolved. What is \( \Delta H_f \) for one mole of \( H_2O \)?
A
\( +242 \text{ KJmol}^{-1} \)
B
\( -121 \text{ KJmol}^{-1} \)
C
\( -242 \text{ KJmol}^{-1} \)
D
\( -484 \text{ KJmol}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The standard enthalpy of formation (\( \Delta H_f \)) is strictly defined as the heat evolved or absorbed when exactly one mole of a compound is formed. Extensive properties like enthalpy scale linearly with molar quantities.
The negative sign must be retained because the word "evolved" indicates an exothermic process.
Why other options are incorrect:
Option D is the enthalpy for two moles, not one. Option A has the wrong sign (implies endothermic). Option B results from incorrectly dividing by 4 instead of 2.
For a given reaction \( CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O \) the change in enthalpy under standard conditions is called?
A
Standard enthalpy of neutralization
B
Standard enthalpy change of solution
C
Standard enthalpy change of formation
D
Standard enthalpy change of hydration
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A reaction between an acid and a base that produces a salt and water is known as a neutralization reaction. The heat released during this process is classified accordingly.
\( CH_3COOH \) is acetic acid (a weak acid) and \( NaOH \) is sodium hydroxide (a strong base).
They undergo a double displacement reaction to form sodium acetate (a salt) and one mole of water.
The enthalpy change associated with forming 1 mole of water from the reaction of \( H^+ \) and \( OH^- \) ions is defined precisely as the Standard enthalpy of neutralization.
Why other options are incorrect:
It is not a formation reaction (which requires elemental reactants). It is not a solution or hydration process because those involve dissolving a solute in a solvent, not a chemical acid-base reaction.
Which of the equations shows the same "twice" the enthalpy changes of neutralization as the following equation? \( HCl + NaOH \rightarrow NaCl + H_2O \)
The standard enthalpy of neutralization (\( \Delta H_n \)) is the energy released (approx. \( -57.4 \text{ kJ} \)) for the formation of exactly 1 mole of water from \( H^+ \) and \( OH^- \). To double this energy, a reaction must produce exactly 2 moles of water from strong acid-base components.
To yield "twice" the enthalpy change (\( 2 \times 57.4 = 114.8 \text{ kJ} \)), the reaction must yield 2 moles of water.
Look at Option C: \( H_2SO_4 \) provides 2 moles of \( H^+ \), and \( Mg(OH)_2 \) provides 2 moles of \( OH^- \), generating exactly \( 2H_2O \). Therefore, the enthalpy change will be double.
Why other options are incorrect:
Options A and D only produce 1 mole of water. Option B is a reaction with a carbonate, which involves the formation of \( CO_2 \) and not a simple strong acid-base double neutralization, so the thermochemistry differs.
The given diagram is a Born-Haber cycle for the formation of KBr:
Using the given data, the lattice energy of potassium bromide is calculated to be:
A
\( -672 \text{ kJmol}^{-1} \)
B
\( -672 \text{ KCalmol}^{-1} \)
C
\( -787 \text{ kJmol}^{-1} \)
D
\( +672 \text{ Jmol}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The Born-Haber cycle equates the standard enthalpy of formation of an ionic solid to the sum of the energies of sublimation, ionization, dissociation, electron affinity, and lattice energy.
Option B uses the wrong units (KCal instead of kJ). Option D uses the wrong units (Joules) and incorrect sign. Option C is the lattice energy for \( NaCl \), acting as a distractor.
Which one of the following is not a state function?
A
Work
B
Enthalpy
C
Internal energy
D
Pressure
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A state function is a thermodynamic property whose value depends strictly on the current state of the system (e.g., its temperature, pressure, and volume) and is entirely independent of the path or history used to reach that state.
Formula:
$$ \Delta U = q + w $$
Solution:
Enthalpy (H), Internal energy (U or E), and Pressure (P) are all state functions; their changes (\( \Delta H, \Delta U, \Delta P \)) only depend on final minus initial states.
Work (w) and Heat (q) are "path functions". The amount of work done or heat transferred depends entirely on how the process was carried out (e.g., reversible vs irreversible expansion).
Therefore, Work is not a state function.
Why other options are incorrect:
Pressure, Internal energy, and Enthalpy all have fixed values at a specific thermodynamic state regardless of the system's history.
Reaction of water with quick lime result in the rise in the temperature of the system. Using the temperature change, indicate the nature of the reaction?
A
Third order reaction
B
Exothermic reaction
C
Non spontaneous reaction
D
Endothermic reaction
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When a chemical reaction releases heat energy, that energy is transferred to the immediate surroundings (the system's medium), resulting in an observable rise in temperature.
Quick lime (\( CaO \)) reacts vigorously with water to form slaked lime (\( Ca(OH)_2 \)).
The question states there is a "rise in the temperature of the system", which is the hallmark macroscopic observation of heat being evolved.
Reactions that evolve heat are exothermic (\( \Delta H < 0 \)).
Why other options are incorrect:
Endothermic reactions absorb heat and cause a drop in temperature. Reaction order (Third order) describes kinetics, not thermodynamics. The reaction is highly spontaneous, ruling out non-spontaneous.
Which of the following enthalpy change is always exothermic?
A
Enthalpy of formation
B
Enthalpy of atomization
C
Enthalpy of solution
D
Enthalpy of combustion
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Different types of standard enthalpy changes can be positive (endothermic) or negative (exothermic), but certain specific thermochemical processes exclusively release energy by definition.
Formula:
$$ \Delta H_c^\circ < 0 \text{ (Always)} $$
Solution:
Enthalpy of combustion (\( \Delta H_c \)) is the heat released when one mole of a substance is completely burned in excess oxygen.
Combustion is a highly energetic oxidation process that universally evolves heat due to the formation of stable oxides (like \( CO_2 \) and \( H_2O \)).
Therefore, it is always an exothermic process (\( \Delta H = -\text{ve} \)).
Why other options are incorrect:
Enthalpy of formation and solution can be either positive or negative depending on the specific substance. Enthalpy of atomization requires breaking bonds to form gaseous atoms, so it is always endothermic.
Determinate the value of Enthalpy of formation of \( NH_4Cl \):
A
\( -314.55 \text{ kJmol}^{-1} \)
B
\( -788 \text{ kJmol}^{-1} \)
C
None of these
D
\( -692 \text{ kJmol}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The enthalpy of formation (\( \Delta H_f^\circ \)) is a specific thermochemical constant for a given compound, indicating the energy change when 1 mole of it forms from its elemental constituents.
The formation of solid ammonium chloride (\( NH_4Cl \)) from gaseous nitrogen, hydrogen, and chlorine is highly exothermic.
The historically accepted and experimentally determined value for the standard enthalpy of formation of \( NH_4Cl \) is exactly \( -314.5 \text{ kJ/mol} \).
Why other options are incorrect:
\( -788 \text{ kJ/mol} \) is roughly the lattice energy of \( NaCl \). \( -692 \text{ kJ/mol} \) is the enthalpy of formation for \( MgO \). These are incorrect values for ammonium chloride.
To compare enthalpy changes across different reactions universally, scientists define "standard conditions". When properties are measured under these conditions, they are referred to as standard enthalpy changes (\( \Delta H^\circ \)).
In chemical thermodynamics, standard conditions specify that the pressure must be 1 atmosphere (or 100 kPa in modern IUPAC definitions, but historically 1 atm).
The designated standard temperature is strictly \( 25^\circ \text{C} \), which converts to 298 K.
Thus, standard enthalpy is measured at 298 K and 1 atm.
Why other options are incorrect:
300 K (\( 27^\circ \text{C} \)) and 295 K (\( 22^\circ \text{C} \)) are not accepted IUPAC standard thermodynamic temperatures. 2 atm is non-standard pressure.
Calculate the lattice energy of sodium chloride on the basis of Born-Haber cycle when: \( \Delta H_f [NaCl] = -411 \text{ KJmol}^{-1}, \Delta H_{at} [Na] = +107 \text{ KJmol}^{-1}, \Delta H_{at} [Cl] = +122 \text{ KJmol}^{-1}, \) \( \Delta H_{i} [Na] = +496 \text{ KJmol}^{-1}, \Delta H_{ea} [Cl] = -349 \text{ KJmol}^{-1} \)
A
\( 376 \text{ kJ/mole} \)
B
\( +787 \text{ kJ/mole} \)
C
\( -376 \text{ kJ/mole} \)
D
\( -787 \text{ kJ/mole} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The Born-Haber cycle utilizes Hess's law to equate the direct enthalpy of formation (\( \Delta H_f \)) to the sum of all individual step enthalpies: atomization, ionization, electron affinity, and lattice energy.
\( +787 \) is the magnitude but reversed sign, representing lattice dissociation instead of formation. \( 376 \) and \( -376 \) are merely the sum of the non-lattice steps in the cycle, omitting the crucial \( \Delta H_f \) term.
The equation shows gaseous molecular hydrogen (\( H_2 \)), which is its standard state, being converted into purely gaseous atomic hydrogen (\( H \)).
Furthermore, exactly 1 mole of the atomic product is being formed.
This perfectly fits the definition of standard enthalpy of atomization (\( \Delta H_{at}^\circ \)).
Why other options are incorrect:
It is not formation because formation applies to compounds formed from elements. While it involves bond dissociation, 'Enthalpy of dissociation' typically refers to breaking 1 full mole of bonds (yielding 2 moles of atoms), whereas atomization specifically scales to yielding 1 mole of atoms.
\( Mg + \frac{1}{2} O_2 \rightarrow MgO_{(s)} \quad \Delta H = -692 \text{ kJmol}^{-1} \) at STP. Enthalpy of the above reaction will be called:
A
\( \Delta H^\circ_{at} \)
B
\( \Delta H^\circ_n \)
C
\( \Delta H^\circ_f \)
D
\( \Delta H^\circ_{sol} \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The standard enthalpy of formation (\( \Delta H_f^\circ \)) is the enthalpy change associated with the creation of exactly 1 mole of a compound directly from its constituent elements in their standard states at STP.
Magnesium (\( Mg \)) and Oxygen (\( O_2 \)) are in their standard elemental states.
They combine to form exactly 1 mole of the solid compound Magnesium Oxide (\( MgO \)).
Because it represents the formation of a compound from its raw elements, the heat change (\( -692 \text{ kJ/mol} \)) is defined as the standard enthalpy of formation, \( \Delta H_f^\circ \).
Why other options are incorrect:
It is not \( \Delta H_{at} \) (atomization) because a solid compound is formed. It is not \( \Delta H_n \) (neutralization) because there is no acid/base reaction producing water. It is not \( \Delta H_{sol} \) (solution) because no solute is dissolving in a solvent.
The standard enthalpy of atomization (\( \Delta H_{at}^\circ \)) is specifically defined as the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state.
Hydrogen's standard state is the diatomic gas, \( H_{2(g)} \).
To form exactly 1 mole of \( H_{(g)} \) atoms, we must break half a mole of \( H-H \) bonds. The balanced equation is: \( \frac{1}{2} H_{2(g)} \rightarrow H_{(g)} \).
Because bond breaking always requires energy, the process is endothermic and the sign must be positive (\( +218 \text{ kJmol}^{-1} \)).
Why other options are incorrect:
Option D has the right equation but a negative sign (implies atomization releases energy, which is false). Options B and C describe the decomposition of water, not the atomization of elemental hydrogen.
\( 2H_2 + O_2 \rightarrow 2H_2O \quad \Delta H = 205.5 \text{ kJmol}^{-1} \). What will be the enthalpy change in the above reaction?
A
\( -205.5 \text{ kJ mol}^{-1} \)
B
\( 205.5 \text{ kJ mol}^{-1} \)
C
\( 1 \text{ kJ mol}^{-1} \)
D
Zero \( \text{kJ mol}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Thermochemical equations must reflect the correct sign convention indicating whether heat is released or absorbed. The formation of water from hydrogen and oxygen gases is an exothermic process.
The question text provides a magnitude of \( 205.5 \text{ kJ/mol} \) without a sign.
Because the combustion of hydrogen to form water releases energy, the enthalpy change (\( \Delta H \)) must mathematically carry a negative sign to denote an exothermic process.
Therefore, the correct representation of the enthalpy change for the reaction as written is \( -205.5 \text{ kJ mol}^{-1} \). (Note: While the universally accepted true value for \( \Delta H \) of this reaction is \( -571.6 \text{ kJ} \) or \( -285.8 \text{ kJ/mol} \) for formation, we must adapt to the magnitude given in the past paper).
Why other options are incorrect:
A positive value (205.5) would falsely imply the formation of water is endothermic. Zero and 1 are mathematically arbitrary distractors.
The total energy or enthalpy of a chemical system dictates whether a reaction will release or absorb heat. This is visualized on a potential energy diagram.
Formula:
$$ \Delta H = E_{\text{products}} - E_{\text{reactants}} $$
Solution:
If the reactants possess higher internal energy than the products (\( E_R > E_P \)), the excess energy must be released into the surroundings as heat.
A reaction that releases heat to the surroundings is strictly classified as an exothermic reaction.
The result is a more stable product and a negative \( \Delta H \).
Why other options are incorrect:
In endothermic and photochemical reactions, energy is actively absorbed, meaning the products end up with a higher energy than the reactants (\( E_P > E_R \)).
Heat of formation (\( \Delta H_f^\circ \)) for \( CO_2 \) is:
A
\( -390 \text{ kJ/mole} \)
B
\( +394 \text{ kJ/mole} \)
C
\( -294 \text{ kJ/mole} \)
D
\( -394 \text{ kJ/mole} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The standard enthalpy of formation is the heat change when one mole of a compound is synthesized from its constituent elements in their standard physical states under standard conditions (298 K, 1 atm).
The formation of carbon dioxide from solid graphite and gaseous oxygen is highly exothermic because it creates very stable covalent double bonds.
The experimentally determined value for this complete oxidation (which is also the enthalpy of combustion of graphite) releases exactly \( -394 \text{ kJ/mol} \).
Why other options are incorrect:
Option B is positive (endothermic), which is impossible for this combustion process. Options A and C are incorrect numerical values.
The change in enthalpy (\( \Delta H \)) represents the heat absorbed or released by a system at constant pressure. The sign convention is determined by the direction of heat flow.
Formula:
$$ \Delta H = H_{\text{products}} - H_{\text{reactants}} $$
Solution:
In an exothermic reaction, the system loses heat to the surroundings.
Because the energy of the products is lower than the energy of the reactants, the final enthalpy is less than the initial enthalpy.
Mathematically, this results in a negative value: \( \Delta H = -ve \).
Why other options are incorrect:
Endothermic, dissociation, and decomposition reactions generally require an input of energy to break chemical bonds, meaning they absorb heat from the surroundings and have a positive \( \Delta H \).
Combustion of graphite to form \( CO_2 \) can be done by two ways. Reactions are given as follow: \( C + O_2 \rightarrow CO_2 \quad \Delta H = -393.7 \text{ kJmol}^{-1} \) \( C + \frac{1}{2}O_2 \rightarrow CO \quad \Delta H_1 = ? \) \( CO + \frac{1}{2}O_2 \rightarrow CO_2 \quad \Delta H_2 = -283 \text{ kJmol}^{-1} \)
A
\( -110 \text{ kJ mol}^{-1} \)
B
\( +110 \text{ kJ mol}^{-1} \)
C
\( -676 \text{ kJ mol}^{-1} \)
D
\( 676 \text{ kJ mol}^{-1} \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Hess's Law of Constant Heat Summation states that if a reaction can take place by more than one route, the total enthalpy change is identical regardless of the route taken, provided initial and final states are the same.
Rounding to the nearest whole number given in the options yields \( -110 \text{ kJ mol}^{-1} \).
Why other options are incorrect:
Option B has a positive sign, incorrectly implying the partial combustion of carbon is endothermic. Options C and D result from incorrectly adding the two values (\( -393.7 - 283 \)) instead of substituting them properly into the algebraic sum.
In standard enthalpy of atomization heat of surrounding:
A
Increases
B
Increases then decreases
C
Decreases
D
Remains same
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The standard enthalpy of atomization (\( \Delta H_{at}^\circ \)) is the enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.
To turn a solid (like Na) or a diatomic gas (like \( Cl_2 \)) into separated gaseous atoms, chemical bonds or strong intermolecular forces must be broken.
Bond breaking strictly requires an input of energy, making atomization an endothermic process (\( \Delta H = +\text{ve} \)).
In an endothermic process, the reacting system actively absorbs heat from the immediate surroundings, which causes the heat of the surroundings to decrease.
Why other options are incorrect:
The heat of the surroundings would only increase if the reaction were exothermic (releasing heat). It does not remain the same because energy transfer is required to break the bonds.
Lattice energy of an ionic crystal is the enthalpy of:
A
Dissolution
B
Dissociation
C
Formation
D
Combustion
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Lattice energy can be defined from two opposing perspectives: the energy required to break apart an ionic lattice, or the energy released when forming it. Standard IUPAC definitions often align it with a specific type of formation.
By convention, Lattice Energy is often defined as the enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions.
Because gas-phase ions are coming together to form a highly stable, tightly packed solid matrix, a tremendous amount of energy is released.
Therefore, in this context, it represents the enthalpy of Formation of the lattice.
Why other options are incorrect:
Combustion involves reaction with oxygen. Dissolution is dissolving in a solvent. While dissociation is the reverse of lattice formation (breaking the lattice apart), the test context explicitly maps lattice energy conceptually to the structural formation step in Born-Haber cycles.
A spontaneous process is one that occurs on its own without external intervention. These processes have a natural directional flow and cannot be reversed without applying external work.
Formula:
$$ \Delta G < 0 \text{ (for spontaneous processes)} $$
Solution:
Spontaneous processes proceed in a specific direction, making them unidirectional.
Because they occur naturally in real-world conditions with an overall increase in entropy, they are irreversible and real processes.
Therefore, all the given characteristics accurately describe a spontaneous process.
Why other options are incorrect:
Options A, B, and C are incomplete individually because a spontaneous reaction encompasses all three listed characteristics. Hence, 'All of above' is the only fully correct choice.
When one mole of gaseous hydrogen ions are dissolved in water to form infinitely dilute solution, amount of heat liberated is:
A
\( -1562 \text{ kJ/mol} \)
B
\( -499 \text{ kJ/mol} \)
C
\( -1891 \text{ kJ/mol} \)
D
\( -1075 \text{ kJ/mol} \)
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The standard enthalpy of hydration (\( \Delta H_{\text{hyd}} \)) is the enthalpy change when one mole of gaseous ions dissolves in sufficient water to give an infinitely dilute solution.
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