Consider the reversible

CHEMISTRY · Single correct · AP · Chemistry · Kinetics · Kinetics

Problem

Consider the reversible, exothermic reaction A ⇌ B with ΔE=−25 kJ mol−1\Delta E = -25\,\text{kJ\,mol}^{-1}. The forward activation energy is 60 kJ mol−160\,\text{kJ\,mol}^{-1} and the reverse activation energy is 85 kJ mol−185\,\text{kJ\,mol}^{-1}. A catalyst lowers each activation energy by 20 kJ mol−120\,\text{kJ\,mol}^{-1}. Which statement is most consistent with these data? (a) Because both activation energies are lowered by the same amount, the equilibrium constant decreases slightly due to the reaction being exothermic. (b) The activation energies are now equal at 40 kJ mol−140\,\text{kJ\,mol}^{-1}, so ΔE\Delta E becomes 00 and equilibrium is achieved regardless of temperature. (c) The equilibrium constant remains the same because the catalyst does not alter ΔE\Delta E. (d) The equilibrium constant increases because the forward reaction is accelerated more than the reverse as the catalyst lowers its activation energy.

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