Question:

For the reaction,
\(\mathrm{2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)}\), \(\Delta H=-57.2\ \mathrm{kJ\,mol^{-1}}\) and \(K_C=1.7\times10^{16}\)
Which of the following statement is incorrect ?

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A catalyst affects the rate, not the equilibrium constant; a large K says nothing about the speed.
Updated On: Nov 5, 2026
  • The equilibrium will shift in forward direction as the pressure increase.
  • The addition of inert gas at constant volume will be not affect the equilibrium constant.
  • The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.
  • The equilibrium constant decreases as the temperature increase.
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The Correct Option is C

Solution and Explanation

(1) \(\Delta n_g=2-3=-1\); increasing pressure shifts equilibrium to fewer moles, i.e. forward. Correct.
(2) Adding an inert gas at constant volume changes no concentrations, and \(K\) depends only on temperature. Correct.
(3) A large \(K_C\) shows the extent of the reaction (thermodynamics), not its rate. The reaction can still be very slow, which is why a catalyst (\(V_2O_5\)) is used. Incorrect.
(4) The reaction is exothermic (\(\Delta H<0\)), so \(K\) decreases when temperature rises. Correct.
The incorrect statement is (3).
Answer: (3) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required.
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