For a reaction with $Δ H^{\circ} = + 60.0 kJ mo l^{- 1}$ (independent of temperature), the equilibrium constant is $K_{1}$ at $300 K$ and $K_{2}$ at $330 K$ . Using $R = 8.314 J mo l^{- 1} K^{- 1}$ and $2.303$ for $ln 10$ , $lo g_{10} (K_{2} / K_{1})$ is

Answer: B

Van't Hoff:

lo g_{10} (K_{2} / K_{1}) = \frac{Δ H ^{\circ}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}}) = \frac{60000}{2.303 \times 8.314} (0.0033333 - 0.0030303) = 0.95

. The positive sign agrees with an endothermic reaction whose

K

increases on heating.