Question

The following mechanism is proposed for the reduction of by

a. What is the intermediate?

b. Derive an expression for the rate law (rate for the overall reaction using the steady-state approximation.

Short answer

a. MoCl5- is the intermediate.

b.$R=k_2\left[N{O_3}^{-}\right]+\frac{k_1\left[MoC{l_6}^{2-}\right]}{k_{-1}\left[MoC{l_5}^{-}\right]\left[C{l}^{-}\right]}{k}_2\left[N{O_3}^{-}\right]$

Step-by-step solution

Write down the rate law of the reaction with the help of reaction mechanism.

The rate law = d[NO2-]/dt is for the 2nd step

R = k2 [NO3]- [MoCl5-] in the steady state approximately the assumption is that the concentration of the intermediate stays constant or the rate of the intermediate produced is equal to the rate of the intermediate consumed.

K1 [MoCl62-] = k-1 [MoCl5-] [Cl-] + k2 [NO3-][MoCl5-]

Solving for [MoCl5-]  

Solving for [MoCl5-] = $\frac{k_1\left[MoC{l_6}^{2-}\right]}{k_{-1}\left[MoC{l_5}^{-}\right]\left[C{l}^{-}\right]}$+ k2 [NO3-]

R = k2 [NO3-] k2 [NO3-]