For the reversible elementary reaction $A B$ the activation energies are $E_{f} = 90$ and $E_{r} = 60 kJ mo l^{- 1}$ (both Arrhenius). The ratio $K_{320} / K_{300}$ of the equilibrium constants $K = k_{f} / k_{r}$ is (use $R = 8.314 J mo l^{- 1} K^{- 1}$ )

Answer: A

K = k_{f} / k_{r} \propto e^{- (E_{f} - E_{r}) / RT}

, so

ln \frac{K _{2}}{K _{1}} = \frac{E _{f} - E _{r}}{R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{30000}{8.314} (0.000208) = 0.7517

. The ratio is

2.12

(endothermic, K rises).