Question

In a reaction, A + B $\rightarrow$ 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[A]$[B]^2$
• B. rate = K$[A]^2 [B]^2$
• C. rate = $k[A][B]$
• D. rate = k$[A]^2$[B]

Answer 1

The Correct Option is D

Let the order of reaction with respect to A and B is x and y respectively. So, the rate law can be given as
$\hspace20mm R=k[A]^x [B]^y \hspace25mm ...(i)$
When the concentration of only B is doubled, the rate is doubled, so
$\hspace20mm R_1=k[A]^x [2B]^y=2R \hspace25mm ...(ii)$
If concentrations of both the reactants A and B are doubled, the rate increases by a factor of 8, so
$\hspace10mm R"=k[2A]^x [2B]^y=8R \hspace25mm ...(iii)$
$\Rightarrow \hspace15mm k2^x 2^y [A]^x [B]^y=8R \hspace25mm ...(iv)$
From Eqs. (i) and (ii), we get
$\Rightarrow \hspace15mm \frac{2R}{R}=\frac{[A]^x [2B]^y}{[A]^x [B]^y}$
$\hspace20mm 2=2^y$
$\therefore \hspace20mm y=1$
From Eqs. (i) and (iv), we get
$\Rightarrow \hspace20mm \frac{8R}{R}=\frac{2^x 2^y [A]^x [B]^y}{[A]^x [B]^y}$
or $\hspace20mm 8=2^x 2^y$
Substitution of the value ofy gives,
$\hspace25mm 8=2^x 2^1$
$\hspace25mm 4=2^x$
$\hspace25mm (2)^2=(2)^x$
$\therefore \hspace25mm x=2$
Substitution of the value of x and y in E (i) gives,
$\hspace25mm R=k[A]^2 [B]$