Question

The free energy of formation of NO is $78kJmo{l}^{-1}$ at the temperature of $1000K$. What is the 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

The correct option is A $8.4\times {10}^{-5}$

The relation between Gibbs free energy and K is given by
$-\Delta G^{\circ }=2.303RTlogK$
simplifying and taking antilog on both sides we get,
$K=\text{anti log}\left[\frac{-\Delta G^{\circ }}{2.303RT}\right]\text{Putting the values we get,}\left[\frac{-78\times 1000}{2.303\times 8.314\times 1000}\right]=8.4\times {10}^{-5}$