• # question_answer 54) The free energy formation of NO is $78\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}$at the temperature of an automobile engine (1000 K). What is the equilibrium constant for this reaction at 1000 K? $\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}^{2}}(g)\Leftrightarrow NO(g)$ A) $8.4\times {{10}^{-5}}$                B) $7.1\times {{10}^{-9}}$C) $4.2\times {{10}^{-10}}$              D) $1.7\times {{10}^{-19}}$

${{K}_{eq}}=\text{antilog}$ $\Delta {{G}^{o}}=78\,kJ/mol=78000\,J/mol.$ Putting these values in equation 1 we get, ${{K}_{eq}}=\text{antilog}$ ${{K}_{eq}}=\text{anitilog}\,\text{(-4}\text{.0737)}$or ${{K}_{eq}}=8.4\times {{10}^{-5}}.$ Hence, the correct option is [a].