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 = $k[A]^2[B]^2$
• B. Rate = k[A] [B]
• C. Rate = $k[A]^2 [B]$
• D. Rate = $k[A] [B^2]$

Answer 1

Answer: (C)

Let the orders of reaction with respect to A and B be x and y respectively. Hence the rate law will be
R = $k[A]^x [B]^y$ .....(i)
When the concentration of B only is doubled, the rate is also doubled.
Hence $R_1 = k [A]^x[B]^y$ .....(ii)
When concentration of both the reactants is doubled, the rate is 8 times
Hence $R_2 = k[2A]^x [2B]^y = 8 \, R$ .....(iii)
or $8R = k \times 2^x \times 2^y \, [A]^x [B]^y$ .....(iv)
From eq. $(i)$ and eq. $(ii)$ we get
$\frac{2R}{R} = \frac{[A]^x [2B]^y}{[A]^x [B]^y}$
$\therefore $ $2 = 2y$
or $y = 1$
From eq. $(i)$ and $(iv)$, we get
$\frac{8R}{R} = \frac{2^x.2^y[A]^x [2B]^y}{[A]^x [B]^y}$
or $8= 2^x.2y$
Substituting the value of y
8 = $2^x . 2^1$
or 4 = $2^x$ or $(2)^2 = (2)^x$
$\therefore $ $x = 2$
Substituting the values of x and y in eqn. $(i)$
or R = $k[A]^2 [B]$