Question

For the non - stoichiometric reaction, 2A+B $\to$ C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.

$\begin{array}{cccc}& \text{Initial}& \text{Initial}& \text{Initial rate of}\\ & \text{concentration}& \text{concentration}& \text{formation of C}\\ & \left[A\right]& \left[B\right]& \left(mol{L}^{-1}{s}^{-1}\right)\\ \left(i\right)& 0.1M& 0.1M& 1.2\times {10}^{-3}\\ \left(ii\right)& 0.1M& 0.2M& 1.2\times 1^{-3}\\ \left(iii\right)& 0.2M& 0.1M& 2.4\times {10}^{-3}\end{array}$ The rate law for the formation of C is

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

Answer 1

The correct option is D

$\frac{dC}{dt}=k\left[A\right]$

This problem can be solved by determining the order of reaction w.r.t each reactant and then writing rate law equation of the given equation accordingly as

$R=\frac{dC}{dt}=k\left[A{]}^x\right[B{]}^y$

where,
x = order of reaction w.r.t A
y = order of reaction w.r.t B

$1.2\times {10}^{-3}=k\left(0.1{)}^x\right(0.1{)}^y$
$1.2\times {10}^{-3}=k\left(0.1{)}^x\right(0.2{)}^y$ .....Doubling [B] had no impact on rate.This implies that change in [B] does not affect the rate. So, order w.r.t [B] is 0
$2.4\times {10}^{-3}=k\left(0.2{)}^x\right(0.1{)}^y$...... Doubling [A] doubled the rate.This implies that change in [A] affects the rate linearly. So, order w.r.t [A] is 1

⇒ $R=k\left[A{]}^1\right[B{]}^0$ is the rate law for the given reaction.

⇒ $\frac{dC}{dt}=k\left[A\right]$