Question

The free energy of the formation of ${}^{\prime }N{O}^{\prime }$ is $78kJ/mol$ at the temperature of an automobile engine $\left(1000K\right)$. What is the $K_C$ equilibrium constant for this reaction at $1000K$.

$\frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons NO\left(g\right)$

• 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

Answer: (A)

$8.4\times {10}^{-5}$

The free energy of formation of $NO$ is $78kJ{mol}^{-1}$ at the temperature of an automobile engine $\left(1000K\right)$. The equilibrium constant for this reaction at $1000K$ is $8.4\times {10}^{-5}$.

$\frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons NO\left(g\right)$
$\Delta G^o=-2.303RT\log K_{eq}$

$78kJ{mol}^{-1}\times 1000J/kJ=-2.303\times 8.314J/mol/K\times 1000K\times \log K_{eq}$
$\log K_{eq}=-4.07$

$K_{eq}=8.4\times {10}^{-5}$

Hence, the correct option is $\text{A}$