Question

For the non-stoichiometric reaction :

$2 A+B \rightarrow C+D$, the following kinetic data were obtained in three separate experiments, all at $298 {K}$.

The rate law for the formation of $C$ is

(a) $\frac{d C}{d t}=k[A]$

(b) $\frac{d C}{d t}=k[A][B]$

(c) $\frac{d C}{d t}=k[A]^2[B]$

(d) $\frac{d C}{d t}=k[A][B]^2$

Answer 1

Correct Option : (a)

Explanation :

For the reaction, $2 A+B \rightarrow C+D$

Rate of reaction $=-\frac{1}{2} \frac{d[A]}{d t}=-\frac{d[B]}{d t}=\frac{d[C]}{d t}=\frac{d[D]}{d t}$

Now, rate of reaction, $\frac{d[C]}{d t}=k[A]^x[B]^y$

From table,

$1.2 \times 10^{-3}=k(0.1)^x(0.1)^y$ ----(i)

$1.2 \times 10^{-3}=k(0.1)^x(0.2)^y$ ----(ii)

$2.4 \times 10^{-3}=k(0.2)^x(0.1)^y$ ----(iii)

On dividing equation (i) by (ii), we get

$\frac{1.2 \times 10^{-3}}{1.2 \times 10^{-3}}=\frac{k(0.1)^x(0.1)^y}{k(0.1)^x(0.2)^y} \Rightarrow 1=\left(\frac{1}{2}\right)^y \Rightarrow y=0$

On dividing equation (i) by (iii), we get

$\frac{1.2 \times 10^{-3}}{2.4 \times 10^{-3}}=\frac{k(0.1)^x(0.1)^y}{k(0.2)^x(0.1)^y} \Rightarrow\left(\frac{1}{2}\right)^1=\left(\frac{1}{2}\right)^x \Rightarrow x=1$

Hence, $\frac{d[C]}{d t}=k[A]^1[B]^0=k[A]$