Consider the following two-step mechanism for the conversion of reactant S into two…

CHEMISTRY · Multi-part · AP · Chemistry · Thermodynamics and Electrochemistry · Thermodynamics and Electrochemistry

Problem

Consider the following two-step mechanism for the conversion of reactant S into two possible products,

P_x

and

P_y

. Step 1 S ⇌ I (equilibrium constant

K_1

) Step 2a I →

P_x

(rate constant

k_a

) Step 2b I →

P_y

(rate constant

k_b

) Both Step 2a and Step 2b are irreversible and first-order in I. The standard Gibbs free energies of formation of the products at 298 K are

\Delta G_f^{\circ}(P_x) = -15\,\text{kJ}\,\text{mol}^{-1}

and

\Delta G_f^{\circ}(P_y) = -25\,\text{kJ}\,\text{mol}^{-1}

. (a) Derive an expression for the instantaneous product ratio

\frac{[P_x]}{[P_y]}

in terms of

K_1

k_a

, and

k_b

under the steady-state approximation for the intermediate I. (b) Reduce your expression from part (a) to the form

\frac{[P_x]}{[P_y]} = \frac{k_a}{k_b}

under conditions where Step 1 lies far to the right. Explain the physical significance of this limiting case. (c) Suppose that at 298 K the values of

k_a

and

k_b

are

3.0\times10^{-2}\,\text{s}^{-1}

and

1.0\times10^{-3}\,\text{s}^{-1}

, respectively, while

K_1 = 0.10

. Calculate the expected product ratio after a short reaction time and identify whether kinetic or thermodynamic control predominates. Justify fully.