Mechanism: step 1, $A + B C$ , fast, with forward rate constant $k_{f} = 1.2 L mo l^{- 1} s^{- 1}$ and reverse rate constant $k_{r} = 0.3 s^{- 1}$ ; step 2, $C + A P$ , slow, with $k_{2} = 0.05 L mo l^{- 1} s^{- 1}$ . Assume step 1 stays at equilibrium. Calculate the rate of formation of $P$ , in units of $1 0^{- 3} M s^{- 1}$ , when $[A] = 0.3 M$ and $[B] = 0.2 M$ (two decimals).

Numerical answer type, answer in 10^-3 M s^-1, tolerance 0.02.

Answer: 3.6 10^-3 M s^-1

The slow step gives rate

= k_{2} [C] [A]

and the equilibrium gives

[C] = K [A] [B]

with

K = k_{f} / k_{r} = 1.2/0.3 = 4.0 L mo l^{- 1}

. So rate

= k_{2} K [A]^{2} [B] = 0.05 \times 4.0 \times 0. 3^{2} \times 0.2 = 3.60 \times 1 0^{- 3} M s^{- 1}