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Question: The free energy of formation of \( {\text{'NO'}} \) is \( {\text{78kJ/mol}} \) at the temperature of...

The free energy of formation of ’NO’{\text{'NO'}} is 78kJ/mol{\text{78kJ/mol}} at the temperature of an automobile engine (1000K)\left( {{\text{1000K}}} \right) . What is the KC equilibrium constant for this reaction at 1000K{\text{1000K}} ?
12N2(g)+12O2(g)⇌NO(g)\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) 8.4×10−58.4 \times {10^{ - 5}}
(B) 7.1×10−97.1 \times {10^{ - 9}}
(C) 4.2×10−104.2 \times {10^{ - 10}}
(D) 1.7×10−191.7 \times {10^{ - 19}}

Explanation

Solution

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:
ΔG0=−RTlnKC\Delta G^{0}= -RTlnK_{C}
Here, ΔG0\Delta G^{0} is the free energy of formation for the reaction, KC{{\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 ’NO’{\text{'NO'}} is 78kJ/mol{\text{78kJ/mol}} .
Also given, the temperature of an automobile engine is equal to 1000K{\text{1000K}} .
We need to find out the value of the equilibrium constant (KC)\left( {{{\text{K}}_{\text{C}}}} \right) for the given reaction which is
12N2(g)+12O2(g)⇌NO(g)\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:
ΔG0=−2.303RTlogKC\Delta G^{0}= -2.303RTlogK_{C}
Now, according to the question, the free energy of formation is equal to 78kJ/mol{\text{78kJ/mol}} and so ΔG0=78kJ/mol\Delta G^{0} = 78kJ/mol . Convert the value of ΔG0\Delta G^{0} from kJ/mol{\text{kJ/mol}} to J/mol{\text{J/mol}} . So now we will have ΔG0=78000J/mol\Delta G^{0} = 78000J/mol . Since the temperature of the reaction is given to be equal to 1000K{\text{1000K}} , so T = 1000K{\text{T = 1000K}} . Also since R is the gas constant, so the value of R will be R = 8.314J/K/mol{\text{R = 8}}{\text{.314J/K/mol}} .
Now, substitute the values of ΔG0=78000J/mol\Delta G^{0} = 78000J/mol , R = 8.314J/K/mol{\text{R = 8}}{\text{.314J/K/mol}} and T = 1000K{\text{T = 1000K}} in the expression mentioned above.
Therefore, we will now have, 78000Jmol−1=−2.303×8.314J/K/mol×1000K×logKC{78000Jmol^{-1}= -2.303\times 8.314J/K/mol\times 1000K\times logK_{C}}
All the units will be cancelled and we will have,
logKC=78000−2.303×8.314×1000⇒logKC=−4.07 logK_{C}= \dfrac{78000}{-2.303\times 8.314\times 1000} \Rightarrow logK_{C}=-4.07
Now, take antilog on both sides:
KC=antilog(−4.07)⇒KC=8.4×10−5 K_{C}= antilog\left ( -4.07 \right ) \Rightarrow K_{C}=8.4\times 10^{-5}
Thus, the value of the equilibrium constant (KC)\left( {{{\text{K}}_{\text{C}}}} \right) for the given reaction is found to be 8.4×10−58.4 \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 J/K/mol{\text{J/K/mol}} . For spontaneous processes, free energy change is negative and in equilibrium, it is equal to zero.