Two competing mechanisms convert reactant R to product P

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

Problem

Two competing mechanisms convert reactant R to product P. • Mechanism 1 (two-step): step 1

E_{a1} = 40\,\text{kJ}\,\text{mol}^{-1}

, step 2

E_{a2} = 35\,\text{kJ}\,\text{mol}^{-1}

, overall

\Delta H = -50\,\text{kJ}\,\text{mol}^{-1}

. • Mechanism 2 (single-step):

E_{a} = 60\,\text{kJ}\,\text{mol}^{-1}

, overall

\Delta H = -110\,\text{kJ}\,\text{mol}^{-1}

. At a given temperature, a catalyst is introduced that lowers

E_{a1}

of Mechanism 1 by

15\,\text{kJ}\,\text{mol}^{-1}

but leaves

E_{a2}

unchanged. No change occurs to Mechanism 2. Which statement is most consistent with the likely outcome after the catalyst is added? (a) Mechanism 2 dominates because its larger exothermicity overrides kinetic considerations. (b) Both mechanisms proceed equally fast because the catalyst lowers the energy of the transition states in both pathways. (c) Mechanism 1 will dominate because its highest activation barrier becomes lower than that of Mechanism 2. (d) Mechanism 1 remains slower overall because its rate is limited by the unaltered

E_{a2}

Related theory

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