Consider the reaction $H_2(g)+B{r}_2(g)2HBr(g),$ which proceeds via the mechanis

below. The compound $M(g)$ is an unspecified monatomic gas.

$B_2(g)+M(g)2Br(g)+M(g)$

$Br(g)+H_2(g)HBr(g)+H(g)$

$H(g)+B{r}_2(g)HBr(g)+Br(g)$

a$.$ Starting a rate expression based on Rate $=-\frac{d[H_2]}{dt}$ and step $2$ of the mechanism, appl

the steady$-$state approximation for the intermediates $H$ and $Br$ to show that the rate law

for the net reaction can be expressed as

Rate $=\frac{k_1^{\frac{1}{2}}{k}_2[H_2][B{r}_2{]}^{\frac{1}{2}}}{k_{-1}^{\frac{1}{2}}(\frac{k_{-2}[HBr]}{k_3[B{r}_2]}+1)}$

b$.$ What is the order of the reaction in the various species in the limit that $HBr$ goes to

zero?

c$.$ Substitute your steady state approximation for $B{r}_2$ back into your expression from pa

$b$ appropriately and then identify the rate determining step of the mechanism in this lin

by comparing your resulting rate law to the elementary steps of the mechanism.

d$.$ What is the order of the reaction in the various species in the limit of large $HBr?$

e$.$ Apply the reversibility of steps $1$ and $2($assume equilibrium has been reached$)$ to

simplify your expression from part d by substituting for $\frac{k_1}{k_{-1}}$ and $\frac{k_2}{k_{-2}},$ respectively.

f$.$ Identify the rate determining step of the mechanism in this limit by comparing your

resulting rate law to the elementary steps of the mechanism.