The rate of formation of $HBr$ can be explained by the following mechanism

$B{r}_{2}2Br,$dots$1$

$B{r}_2+H_{2}HBr+H,$dots$2$

$H+B{r}_{2}HBr+Br,$dots$3$

$H+HBr{H}_2+Br,$dots$4$

$2BrB{r}_2,$dots$-1$

a$.$ Derive an expression for the rate of formation of $HBr$ making use of the steady state approximation.

b$.$ What is the effect of $HBr$ concentration on the rate of reaction? Explain your answer.