For the isomerization X Y,

CHEMISTRY · Single correct · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry

Problem

For the isomerization X →\rightarrow Y, ΔG∘=−40 kJ mol−1\Delta G^{\circ} = -40\,\text{kJ mol}^{-1}, the forward activation energy is 140 kJ mol−1140\,\text{kJ mol}^{-1}, and the reverse activation energy is 180 kJ mol−1180\,\text{kJ mol}^{-1}. After several hours at 298 K, essentially no Y is detected. Which modification would most effectively accelerate the appearance of Y while leaving the equilibrium position essentially unchanged? (a) Increase the reaction temperature by 50 K50\,\text{K}. (b) Introduce a catalyst that lowers only the forward activation energy by about 60 kJ mol−160\,\text{kJ mol}^{-1}. (c) Introduce a catalyst that lowers the activation energy of both the forward and reverse steps by about 60 kJ mol−160\,\text{kJ mol}^{-1}. (d) Cool the reaction mixture to 250 K250\,\text{K} to make ΔG∘\Delta G^{\circ} more negative.

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