A reaction proceeds by three elementary steps with the following activation energies…

CHEMISTRY · Single correct · A Level · Chemistry · Kinetics · Kinetics

Problem

A reaction proceeds by three elementary steps with the following activation energies relative to the reactants: • Step 1: 45 kJ mol−145\,\text{kJ}\,\text{mol}^{-1} • Step 2: 80 kJ mol−180\,\text{kJ}\,\text{mol}^{-1} • Step 3: 30 kJ mol−130\,\text{kJ}\,\text{mol}^{-1} The overall enthalpy change is ΔHrxn=−20 kJ mol−1\Delta H_{\text{rxn}} = -20\,\text{kJ}\,\text{mol}^{-1}. A catalyst offers an alternative pathway with activation energies of 3030, 5050, and 35 kJ mol−135\,\text{kJ}\,\text{mol}^{-1} for Steps 1, 2, and 3, respectively. Which statement about the catalyzed energy profile is correct? (a) The highest transition state lies 65 kJ mol−165\,\text{kJ}\,\text{mol}^{-1} above the reactants, and the overall ΔHrxn\Delta H_{\text{rxn}} is unchanged. (b) The highest transition state lies 50 kJ mol−150\,\text{kJ}\,\text{mol}^{-1} above the reactants, and the overall ΔHrxn\Delta H_{\text{rxn}} remains −20 kJ mol−1-20\,\text{kJ}\,\text{mol}^{-1}. (c) The highest transition state remains 80 kJ mol−180\,\text{kJ}\,\text{mol}^{-1} above the reactants, and the overall ΔHrxn\Delta H_{\text{rxn}} is unchanged. (d) The highest transition state lies 50 kJ mol−150\,\text{kJ}\,\text{mol}^{-1} above the reactants, and the overall ΔHrxn\Delta H_{\text{rxn}} becomes more negative.

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