at 2500 K is $+118 \mathrm{kJ} \mathrm{mol}^{-1}$. The equilibrium constant for the reaction is [Given: $\mathrm{R}=8.314 \mathrm{J} \mathrm{K}^{-1} \mathrm{mol}^{-1}$ ]
(a)0.994
(b)1.006
(c)$3.42 \times 10^{-3}$
(d)292.12
Answer
SELF-PRACTICE — the source book printed no answer.
Nothing is invented here, so this question has no answer on record.