In the mechanism $A k X_{1} k X_{- 1} I$ , $I k X_{2} P$ with $k_{1} = 2.0 \times 1 0^{- 3} s^{- 1}$ , $k_{- 1} = 3.0 \times 1 0^{- 2} s^{- 1}$ , $k_{2} = 1.0 \times 1 0^{- 2} s^{- 1}$ , the steady-state approximation for I (used exactly, with no further simplification) gives rate $= k_{eff} [A]$ with $k_{eff}$ equal to (in $s^{- 1}$ )

Answer: C

Steady state:

[I] = k_{1} [A] / (k_{- 1} + k_{2})

, so rate

= k_{2} [I] = \frac{k _{1} k _{2}}{k _{- 1} + k _{2}} [A]

. Numerically

k_{eff} = (2.0 \times 1 0^{- 3}) (1.0 \times 1 0^{- 2}) / (3.0 \times 1 0^{- 2} + 1.0 \times 1 0^{- 2}) = 5.0 \times 1 0^{- 4} s^{- 1}

. The value

6.7 \times 1 0^{- 4}

drops

k_{2}

from the denominator (valid only if

k_{2} ≪ k_{- 1}