Question

For a reaction $2A+B\to 2AB$, it is found that doubling the concentration of both the reactants the rate increases to $8$ times that of initial rate but doubling the concentration of B alone doubles the rate. Then the order of the reaction with respect to $A$ and $B$ are :

• A. $0,3$
• B. $0,2$
• C. $2,1$
• D. $2,2$

Answer 1

Answer: (C)

$2,1$

For the given reaction,

$2A+B\to 2AB$
Rate $=r=$ $k\left[A{]}^m\right[B{]}^n$ $.........\left(i\right)$, where $m=$ order with respect to $A$ & $n=$ order with respect to $B$

Given that rate $=8r=$ $k\left[2A{]}^m\right[2B{]}^n$ $........\left(ii\right)$

Also, rate $=2r=$ $k\left[A{]}^m\right[2B{]}^n$ $......\left(iii\right)$

On dividing $\left(ii\right)$ by $\left(iii\right)$, we get

$\frac{8r}{2r}=\frac{k\left[2A{]}^m\right[2B{]}^n}{k\left[A{]}^m\right[2B{]}^n}$

$\therefore 2^2=2^m$ & hence, $m=2$

Next divide $\left(i\right)$ by $\left(iii\right)$, we get

$\frac{r}{2r}=\frac{k\left[A{]}^m\right[B{]}^n}{k\left[A{]}^m\right[2B{]}^n}$

$\therefore [\frac{1}{2}{]}^1=[\frac{1}{2}{]}^n$ & hence, $n=1$.

Hence, order with respect to $A$ and $B$ is $2$ and $1$ respectively.