Question

In a reaction, $A+B\to$ 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 :

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

Answer 1

The correct option is D

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
$R=k\left[A{]}^x\right[B{]}^y$ …(i)
When the concentration of only B is doubled, the rate is doubled, so
$R^{\prime }=k\left[A{]}^x\right[2B{]}^y=2R$ ….(ii)
If concentrations of both the reactants A and B are doubled, the rate increases by a factor of 8, so $R^{\prime \prime }=k\left[2A{]}^x\right[2B{]}^y=8R$ ….(iii)
$\Rightarrow k{2}^x{2}^y\left[A{]}^x\right[B{]}^y=8R$ ....(iv)
Form Eqs. (i) and (ii), we get
$\Rightarrow \frac{2R}{R}=\frac{\left(\right[A{]}^x\left[2B{]}^y\right)}{\left(\right[A{]}^x\left[B{]}^y\right)}2=2^y\therefore y=1$
From Eqs.(i) and (iv), we get
$\Rightarrow \frac{8R}{R}=\frac{\left(2^x{2}^y\right[A{]}^x\left[B{]}^y\right)}{\left(\right[A{]}^x\left[B{]}^y\right)}or8=2^x{2}^y$
Substitution of the value of y gives,
$8=2^x{2}^{1}4=2^x(2{)}^2=(2{)}^x\therefore X=2$
Substitution of the value of x and y in Eq. (i) gives,
$R=k\left[A{]}^2\right[B]$