Question

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

• A. $\frac{dx}{dt}=k\left[A{]}^2\right[B{]}^3$
• B. $\frac{dx}{dt}=k\left[A\right]\left[B\right]$
• C. $\frac{dx}{dt}=k\left[A{]}^0\right[B{]}^2$
• D. $\frac{dx}{dt}=k[B{]}^{\frac{1}{2}}$

Answer 1

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

Since A is present in large excess, it does not affect the rate of the reaction and rate of reaction will depends on concentration of B only.

$\therefore r=k[B{]}^n$

$r=\frac{dx}{dt}=k[0.1{]}^n$.....(1)

$r^{\prime }=2r=\frac{dx}{dt}=k[0.4{]}^n$......(2)

Divide equation (2) with (1)
$\frac{2r}{r}=\frac{k[0.4{]}^n}{k[0.1{]}^n}$

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

Hence, the rate law expression is $\frac{dx}{dt}=k[B{]}^{\frac{1}{2}}.$