Chemistry Equilibrium MDCAT 2012
PMDC Verified Question 99 of 102
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
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: More product (\( \text{NH}_3 \)) will be formed
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.

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