Question

For the reaction $2A+3B\to$ products, A is taken in excess and on changing the concentration of B from 0.1 M to 0.4 M, the rate becomes doubled. Thus, the rate law of the reaction is:

• A. $\frac{dx}{dt}=k\left[A{]}^2\right[B{]}^2$
• B. $\frac{dx}{dt}=k\left[A\right]\left[B\right]$
• C. $\frac{dx}{dt}=k\left[A{]}^0\right[B{]}^2$
• D. $\frac{dx}{dt}=k[B{]}^{1/2}$

Answer 1

The correct option is D $\frac{dx}{dt}=k[B{]}^{1/2}$

2A + 3B $\to$ product

Since 'A' is taken in excess, so the rate of reaction doesn't depend on 'A'.

Rate $=K[B{]}^N$ given Rate${}_1=K[0.1{]}^n$ .....[1]

given Rate${}_2=2\times$Rate${}_1=K[0.4{]}^n$ ......[2]

From equation (1) and (2),

n =$\frac{1}{2}$

Rate law is given by,

$\frac{dx}{dt}=K[B{]}^{\frac{1}{2}}$

Hence, option $D$ is correct.