Question

In a reaction, $A+B\to \text{Product}$, rate is doubled when the concentration of B is doubled and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled. Rate law for the reaction can be written as

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> AIPMT 2012 (PRE)

• A. $\text{Rate}=k\left[A\right][B{]}^2$
• B. $\text{Rate}=k\left[A{]}^2\right[B]$
• C. $\text{Rate}=k\left[A{]}^2\right[B{]}^2$
• D. $\text{Rate}=k\left[A\right]\left[B\right]$

Answer 1

The correct option is B $\text{Rate}=k\left[A{]}^2\right[B]$

Let the order of reaction with respect to A and B is x and y respectively. So, the rate law can be given as
$R=k\left[A{]}^x\right[B{]}^y..\left(i\right)$
When the concentration of only B is doubled, the rate is doubled, so
$2R=k\left[A{]}^x\right[2B{]}^y..\left(ii\right)$
If concentrations of both the reactants A and B are doubled, the rate increases by a factor of 8, so
$8R=k\left[2A{]}^x\right[2B{]}^y..\left(iii\right)$
From equation (i) and (ii)
$2=2^{y}y=1$
From equation (ii) and (iii)
$4=2^{x}x=2$
Substitution of the value of x and y in Eq. (i) gives,
$R=k\left[A{]}^2\right[B{]}^1$