Question

A proposed mechanism for the oxidation of nitric oxide to nitrogen dioxide is shown below. 2 no (g) ⇄ n2o2 (g) fast, reversible step n2o2 (g) + o2 (g) → 2 no2 (g) slow step what rate law is consistent with this mechanism?

Answer 1

Rate   law  is  determined  the  rate  of  the  slowest  chemical  reaction  since   it   determine  the  rate  of  overall  reaction  .  but   this  applies  when  the  slowest  step  come  before  the  fast  step.  For  the    equation  above   the  fast  step  come  before  the  slowest  step.  The  rate  law  is  therefore.
2NO  to  n2o2  (fast)
N2O2  +O2  -->2NO2(SLOW)
NO2+O2 ---> NO +O3(FAST)
rare   law  = k2(N2O2)(O2)

Answer 2

What rate law is consistent with this mechanism? $k_2[N_2O_2][O_2]$

Further explanation

Rate law is the equation that relates the reaction rate with the concentrations of the reactants and constant parameters.

Nitric oxide is  is a colorless gas with the formula NO. It is free radical and has an unpaired electron. While nitrogen dioxide is the chemical compound with formula NO₂. It is an intermediate in the industrial synthesis of nitric acid.

The mechanism for the oxidation of nitric oxide to nitrogen dioxide is shown below:

$2NO (g)$⇄ $N_2O_2 (g)$ fast, reversible step $N_2O_2 (g) + O_2 (g)$ → $2NO_2 (g)$

rate = rate of slow step = $k_2[N_2O_2][O_2]$

Therefore $N_2O_2$ is NOT a reactant or a product. So then we should eliminate it from the rate law.

For equilibrium of $2NO (g)$⇄ $N_2O_2 (g)$

The conversion of products and  reactants are an example of equilibrium

rate forward = rate backward

$k_f [NO]^2 = k_r [N_2O_2]$

So the Equilibrium Constant $K_{eq}$is:

$\frac{k_f}{k_e}  = \frac{[N_2O_2]}{[NO]^2} =K_{eq}$

While for Reversible Step of  $2NO (g)$⇄ $N_2O_2 (g)$

Forward reaction rates and backward reaction rates will become equal quickly

rate forward = rate backward

$k_f [NO]^2 = k_r [N_2O_2]$

Therefore it will be

$[N_2O_2] = \frac{k_f}{k_r} [NO]^2$

rate = rate of slow step = $k_2[N_2O_2][O_2]$

$[N_2O_2] = \frac{k_f}{k_r}[NO]^2$

Therefore, rate = $k_2 [N_2O_2][O_2] = k_2 \frac{k_f}{k_r} [NO]^2[O_2]$

rate = $k_{observed} [NO]^2[O_2] = k_2 K_{eq}[NO]^2[O_2]$

Learn more

1. Learn more about rate law https://brainly.com/question/1398869
2. Learn more about nitric oxide https://brainly.com/question/10493606
3. Learn more   about nitrogen dioxide https://brainly.com/question/10727615

Answer details

Grade:    9

Subject:  chemistry

Chapter:  oxidation

Keywords: rate law, nitric oxide, nitrogen dioxide, reversible step, oxidation