# 2-step reaction mechanism for the decomposition of NO2Br

I need to propose a 2-step reaction mechanism for the following reaction: $$\ce{2NO2Br -> 2NO2 +Br}$$ And prove the mechanism is consistent with: $$v=k[\ce{NO2Br}]^2$$ For the mechanism to be consistent with the rate equation, the first equation would have to be the limiting one and also be in this form: $$\ce{2NO_2Br -> Products}$$ I've tried using: $$\ce{2NO2Br -> N2O4 +Br2}$$ $$\ce{N2O4 -> 2NO2}$$ But the problem with this mechanism is that $$\ce{N2O4}$$ is not unstable enough to be consumed immediately. Any help with this problem?
Would also appreciate some advice on proposing reaction mechanisms. Thank you.

• Note that there is ongoing equilibrium $\ce{N2O4 <=> 2 NO2}$. Also, there is no rule the 2nd reaction must be fast. Apr 21, 2019 at 9:58
• Can a mechanism be found so that the reaction rate depends solely on $\ce{[NO2Br]}$? Apr 21, 2019 at 10:18
• The reaction kinetics of multiple reagents is usually based on the slowest reaction of the reaction chain.But as there is no other reagent, it is quite possible. But I am not currently aware of the actual reaction mechanism. Apr 21, 2019 at 10:22

It is more probable like \begin{align} \ce{NO2Br &-> NO2 + Br} \\ \ce{NO2Br + Br &-> NO2 + Br2} \\ \ce{2 Br &-> Br2} \\ \end{align}

The last reaction is a minor one in case concentration of $$\ce{Br}$$ is low.

The reaction rate order can be concentration dependent and need not be the integer.

In fact, it is rather mathematical parameter, related to solution of differential equations for a complex reaction system.

If the 2nd reaction is fast enough, the overall reaction rate is given by the slow rate of generation of $$\ce{Br}$$, which fast reacts to form $$\ce{Br2}$$

If the 2nd reaction is slow enough, it's rate $$k_{\rm 2}\cdot [\ce{NO2Br}][\ce{Br}]$$ can be written as $$k_{\rm 2a}\cdot [\ce{NO2Br}]^2$$

The exact solution is to solve system of differential equations for the rates of the concentration changes.

$$\frac{\mathrm{d}[\ce{Br}]}{\mathrm{d}t}=k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}] - k_3 [\ce{Br}]^2$$

For the dynamic equilibrium of the steady concentration of $$\ce{Br}$$:

\begin{align} 0&=-k_1.[\ce{NO2Br}] + k_2.[\ce{NO2Br}][\ce{Br}] + k_3 [\ce{Br}]^2 \\ [\ce{Br}]&=[ -k_2.[\ce{NO2Br}]+\sqrt((k_2.[\ce{NO2Br}])^2+4.k_3.k_1.[\ce{NO2Br}])]/(2.k_3) \\ \frac{\mathrm{d}[\ce{NO2Br}]}{\mathrm{d}t}&=-k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}]\\ \end{align}

• How is the order of reaction related to which reaction is the fast one? Apr 21, 2019 at 10:58
• @jcgereda See the update and the link. Apr 21, 2019 at 12:02
• Why does the constant change when [Br] is substituted to [NO2Br]? Also: How can I relate mathematically [Br] consumed by the second reaction with [NO2Br] consumed by the first one? Apr 21, 2019 at 13:21
• 1/ Because concentrations of NO2Br and Be are not the same. 2/ see the update provided in a while.. Apr 21, 2019 at 13:36