A reactant can follow two independent one-step pathways at 298 K to form products A or B

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

Problem

A reactant can follow two independent one-step pathways at 298 K to form products A or B. • Pathway A: Ea=80 kJ mol−1E_a = 80\,\text{kJ\,mol}^{-1}, ΔGA∘=−20 kJ mol−1\Delta G^{\circ}_{\text{A}} = -20\,\text{kJ\,mol}^{-1} • Pathway B: Ea=35 kJ mol−1E_a = 35\,\text{kJ\,mol}^{-1}, ΔGB∘=−10 kJ mol−1\Delta G^{\circ}_{\text{B}} = -10\,\text{kJ\,mol}^{-1} If the reaction mixture is kept at a low temperature immediately after mixing, which product is most likely to predominate and why? (a) Product A, because its pathway releases more free energy despite the higher activation energy. (b) Product B, because its pathway has the lower activation energy even though it is less exergonic. (c) Product B, because both its activation energy and its ΔG∘\Delta G^{\circ} are lower than those of Pathway A. (d) Product A, because at low temperature reactions are under thermodynamic control, so the more stable product predominates.

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