Question

In the formation of $HBr$ from $H_2\&B{r}_2,$ following mechanism is observed.
1. $B{r}_2\rightleftharpoons 2B{r}^{\bullet }$ Equilibrium step
2. $H_2+B{r}^{\bullet }\to HBr+H^{\bullet }$ Slow step
3. $H^{\bullet }+B{r}^2\to HBr+B{r}^{\bullet }$ Fast step
Calculate the rate of reaction, if concentration of hydrogen is twice that of bromine and the rate constant is equal to 1 $M^{-1/2}Se{c}^{-1}.$ Concentration of bromine is $1M$

• A. $2M{Sec}^{-1}$
• B. $3M{Sec}^{-1}$
• C. $4M{Sec}^{-1}$
• D. $5MSe{c}^{-1}$

Answer 1

Answer: (A)

Rate $=k{\left[H_2\right]}^1{\left[B{r}_2\right]}^{1/2}$

Rate $=1[2{]}^1[1{]}^{1/2}=2M{\mathrm{Sec}}^{-1}$