Chemistry Equilibrium MDCAT 2019
PMDC Verified Question 83 of 102
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
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: The concentrations of \( \text{SO}_2 \) and \( \text{O}_2 \) increase and concentration of \( \text{SO}_3 \) decreases as the temperature increases
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.

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