Question

For the mechanism

$A + B \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} C; C \xrightarrow{k_3} D$

The equilibrium step is fast.

The reaction rate, $\frac{d[D]}{dt}$ is

Answer 1

$\frac{k_1 k_3}{k_2}[A][B]$

Answer 2

The rate-determining step is $C \xrightarrow{k_3} D$, so the rate is $k_3[C]$. The fast equilibrium $A + B \underset{k_2}{\stackrel{k_1}{\rightleftharpoons}} C$ implies $k_1[A][B] = k_2[C]$, thus $[C] = \frac{k_1}{k_2}[A][B]$. Substituting this into the rate equation yields $\frac{d[D]}{dt} = k_3 \left(\frac{k_1}{k_2}[A][B]\right) = \frac{k_1 k_3}{k_2}[A][B]$.