Question

For the reaction $2A+3B\to products$, 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. $\frac{dx}{dt}=k\left[A{]}^2\right[B{]}^2$
  
• 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 $\frac{dx}{dt}=k[B{]}^{1/2}$

$2A+3B\to products$
Since A is in Excess, the order wrt to A is zero. Therefore, Rate law becomes
$Rate=k[B{]}^n$
Given,
$Rat{e}_1=k[0.1{]}^n$...........(1)
Given $Rat{e}_2=2Rat{e}_1=k[0.4{]}^n$......(2)
From (1) and (2) $n=\frac{1}{2}$
Therefore, the Rate law becomes $\frac{dx}{dt}=k[B{]}^{1/2}$