Chlorine monofluoride (ClF) can be produced through the two-step mechanism shown below

CHEMISTRY · Multi-part · AP · Chemistry · Kinetics · Kinetics

Problem

Chlorine monofluoride (ClF) can be produced through the two-step mechanism shown below, where X represents Cl and Y represents F. Step 1 X₂(g) → 2 X(g) Ea, fwd=250 kJ mol−1E_{a,\,\text{fwd}} = 250\,\text{kJ\,mol}^{-1} ΔH1=+150 kJ mol−1\Delta H_1 = +150\,\text{kJ\,mol}^{-1} Step 2 X(g) + Y₂(g) → XY(g) + Y(g) Ea, fwd=80 kJ mol−1E_{a,\,\text{fwd}} = 80\,\text{kJ\,mol}^{-1} ΔH2=−60 kJ mol−1\Delta H_2 = -60\,\text{kJ\,mol}^{-1} The overall reaction is therefore X₂(g) + Y₂(g) → 2 XY(g) with ΔHoverall=+90 kJ mol−1\Delta H_{\text{overall}} = +90\,\text{kJ\,mol}^{-1}. (a) For Step 1, calculate the activation energy for the reverse direction (2 X → X₂). (b) For Step 2, calculate the activation energy for the reverse direction (XY + Y → X + Y₂). (c) In the reverse reaction sequence that converts 2 XY back to X₂ + Y₂, identify which elementary step is expected to be rate-determining at a fixed temperature. Support your answer quantitatively with the values you obtained in parts (a) and (b). (d) Explain how the enthalpy changes of the two steps influence the relative magnitudes of the forward and reverse activation energies you calculated.

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