Question

The free energy of formation of $ {\text{'NO'}} $ is $ {\text{78kJ/mol}} $ at the temperature of an automobile engine $ \left( {{\text{1000K}}} \right) $ . What is the KC equilibrium constant for this reaction at $ {\text{1000K}} $ ?
 $ \dfrac{{\text{1}}}{{\text{2}}}{{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + \dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \rightleftharpoons {\text{NO}}\left( {\text{g}} \right) $
(A) $ 8.4 \times {10^{ - 5}} $
(B) $ 7.1 \times {10^{ - 9}} $
(C) $ 4.2 \times {10^{ - 10}} $
(D) $ 1.7 \times {10^{ - 19}} $

Answer 1

Hint :For non-spontaneous processes, free energy is positive and equilibrium constant is less than 1. The relationship between free energy of formation and the equilibrium constant of a reaction is given by the following expression:
 $ \Delta G^{0}= -RTlnK_{C} $
Here, $ \Delta G^{0} $ is the free energy of formation for the reaction, $ {{\text{K}}_{\text{C}}} $ is the equilibrium constant of the reaction, T is the temperature of the reaction and R is the gas constant.

Complete Step By Step Answer:
Given that the free energy of formation of $ {\text{'NO'}} $ is $ {\text{78kJ/mol}} $ .
 Also given, the temperature of an automobile engine is equal to $ {\text{1000K}} $ .
We need to find out the value of the equilibrium constant $ \left( {{{\text{K}}_{\text{C}}}} \right) $ for the given reaction which is
 $ \dfrac{{\text{1}}}{{\text{2}}}{{\text{N}}_{\text{2}}}\left( {\text{g}} \right) + \dfrac{{\text{1}}}{{\text{2}}}{{\text{O}}_{\text{2}}}\left( {\text{g}} \right) \rightleftharpoons {\text{NO}}\left( {\text{g}} \right) $
The relation between equilibrium constant and free energy can also be written by converting ln to log:
 $ \Delta G^{0}= -2.303RTlogK_{C} $
Now, according to the question, the free energy of formation is equal to $ {\text{78kJ/mol}} $ and so $ \Delta G^{0} = 78kJ/mol $ . Convert the value of $ \Delta G^{0} $ from $ {\text{kJ/mol}} $ to $ {\text{J/mol}} $ . So now we will have $ \Delta G^{0} = 78000J/mol $ . Since the temperature of the reaction is given to be equal to $ {\text{1000K}} $ , so $ {\text{T = 1000K}} $ . Also since R is the gas constant, so the value of R will be $ {\text{R = 8}}{\text{.314J/K/mol}} $ .
Now, substitute the values of $ \Delta G^{0} = 78000J/mol $ , $ {\text{R = 8}}{\text{.314J/K/mol}} $ and $ {\text{T = 1000K}} $ in the expression mentioned above.
Therefore, we will now have, $ {78000Jmol^{-1}= -2.303\times 8.314J/K/mol\times 1000K\times logK_{C}} $
All the units will be cancelled and we will have,
 $
  logK_{C}= \dfrac{78000}{-2.303\times 8.314\times 1000}
\Rightarrow logK_{C}=-4.07
  $
Now, take antilog on both sides:
 $
  K_{C}= antilog\left ( -4.07 \right )
\Rightarrow K_{C}=8.4\times 10^{-5}
  $
Thus, the value of the equilibrium constant $ \left( {{{\text{K}}_{\text{C}}}} \right) $ for the given reaction is found to be $ 8.4 \times {10^{ - 5}} $ .
So the correct option is A.

Note :
While solving these kind of problems, always convert the energy from kilojoule per mol to Joule per mol so that these get cancelled by the unit of the gas constant R when we take its value in the unit of $ {\text{J/K/mol}} $ . For spontaneous processes, free energy change is negative and in equilibrium, it is equal to zero.