Consider the reversible reaction V ⇌ W that proceeds through the two-step mechanism V → X → W for the forward direction and W → X → V for the reverse direction. The energy profile for the forward reaction (all values relative to V) is summarized below.
Forward direction:
• TS₁ (V → X) = +95 kJ·mol⁻¹
• Intermediate X = +25 kJ·mol⁻¹
• TS₂ (X → W) = +60 kJ·mol⁻¹
• Product W = –30 kJ·mol⁻¹
Answer the following.
(a) Calculate the activation energy for each elementary step in the reverse direction (W → X and X → V).
(b) On a single set of axes, qualitatively sketch the potential-energy profile for both the forward and reverse reactions, clearly labeling all transition states, intermediates, reactants, and products.
(c) Using your calculated activation energies, discuss which direction (forward or reverse) is kinetically favored at 298 K, assuming equal pre-exponential factors.
(d) The equilibrium constant, K, for V ⇌ W at 298 K is 4. Using thermodynamic arguments and without performing numerical calculations involving K, explain how the position of equilibrium is consistent with the energy profile you have analyzed.