# Chemical kinetics with the reaction of tert-butyl bromide with azide ion

The reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism:

$\ce{(CH3)3CBr(aq) <=>(CH3)3C+(aq) + Br–(aq)}$

$\ce{(CH3)3C+(aq) + N3–(aq) \to (CH3)3CN3(aq)}$

Assuming that $\ce{(CH3)3C+ (aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants, what is the rate law for this reaction, where $k_1$ is the forward reaction rate of the first equation, $k_{-1}$ is the reverse reaction rate of the first equation, and $k_2$ is the forward reaction rate of the second equation?

(A) $\ce{k_1[(CH3)3CBr]}$

(B) $\ce{k_2[(CH3)3CBr][N3-]}$

(C) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-]}}$

(D) $\ce{\displaystyle\frac{k_1k_2[(CH3)3CBr][N3-]}{k_{-1}[Br-] + k2[N3-]}}$

My thoughts: First of all, I read that $\ce{(CH3)3C+ (aq)}$ is a steady state concentration, which meant that $\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = 0,$ which prompted me to write out differential equations for each reactant. Thus, I got a system of differential equations: $$\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} =-\frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$ $$0=\frac{d\ce{[(CH3)3C+ (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} - \frac{d\ce{[(CH3)3CN3 (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}] -k_2 \frac{d\ce{[(CH3)3CN3 (aq)]}}{dt}$$ $$\frac{d\ce{[Br- (aq)}]}{dt} = -\frac{d\ce{[(CH3)3CBr (aq)}]}{dt} = \frac{k_1}{k_{-1}}\ce{[(CH3)3CBr (aq)}]$$ $$\frac{d\ce{[N3- (aq)}]}{dt} = -k_2\ce{[N3-(aq)]}$$

But I was unsure of how to use these differential equations to actually find the rate law. Could anyone provide any suggestions?

## 1 Answer

You really don't have to write all that differential equations.

Note: I'll write all my steps on a paper, since MathJax formatting is a little time-taking at the moment.

Consider the rate laws for the two reactions:

Since the $\ce{[(CH3)3C+]}$ isn't changing with time, we get

From which, you get

Plugging these in the Rate2 which I mentioned, you will get,

Which is the same as what's given in option (D).

I hope I have satisfied your queries.

• Sorry, but how did you get your first rate law? – Teoc Apr 24 '17 at 5:18
• Those rate laws are directly derived from the reaction. For example, in an elementary reaction: $$\ce{mA + nB -> C}$$ The rate law would be $$Rate=[A]^m+[B]^n$$ – Pritt says Reinstate Monica Apr 24 '17 at 5:22
• According to my book rate would be $k[A]^m [B]^n$ is that just a typo on your part? Additionally, how did you derive the rate law for reactions in equilibrium? – Teoc Apr 24 '17 at 5:32
• You should really take a look at this so you can correctly format your future posts. – ringo Apr 24 '17 at 5:43
• @PrittBalagopal Wouldn't the rate be the negative of what you wrote, because we are measuring the rate of formation of product, not reactant? – Teoc Apr 24 '17 at 15:30