Question

step by step

Rates of Non-elementary reactions Class Activity Consider the following mechanism: 2NO N202 k-1 NzOz + C12-2+2 NOCI a. What is the net reaction? b. Which of the above species is an intermediate? Why? c. Find the rate of the overall reaction using the steady-state approximation. d. Repeat part (c) using the rate-determining step method, assuming the first-step is fast and at equilibrium and the second step is slow. 2NO N,02 (ast,at equilibrium) k-1 N202 + Cl2-22 NOCI (slow) e. At what conditions the rate laws predicted by the steady-state approximation and by the rate-determining step become identical.

Answer 1

a. Net reaction: 2NO + Cl₂$\rightarrow$ 2NOCl (adding the aove given reaction equations)

b. N₂O₂ is the intermediate species because the amount of N₂O₂ produced = amount of N₂O₂ consumed and so no product can be obtained after the completion of the reaction

c. The elementary reactions,

2NO  $\mathbf{\overset{k_{1}}{\rightarrow}}$  N₂O₂

N₂O₂ 2NO

K-1

N₂O₂ + Cl₂$\mathbf{\overset{k_{2}}{\rightarrow}}$ 2NOCl

Overall rate of change, r_NOCl = k₂ [N₂O₂][Cl₂]

r_N2O2 = 0.5 k₁[NO]² - k_-1[N₂O₂] - k₂[ [N₂O₂][Cl₂]

Applying steady-state approximation,

0.5 k₁[NO]² - k_-1[N₂O₂] - k₂[ [N₂O₂][Cl₂] = 0

[N₂O₂] = 0.5 * (k₁ [NO]²) / (k_-1 + k₂[Cl₂])

$\therefore$ r_NOCl = k₂ [N₂O₂][Cl₂] = 0.5 * k₂ * (k₁[NO]²[Cl₂]) / (k_-1 + k₂[Cl₂])

d. Slowest step determines the rate of the reaction.

r_NOCl = k₂ [N₂O₂][Cl₂]

e. when [NO] and [N₂O₂] approximated as 1 and k_-1 >> k₂ then the rate laws can be identical