Question

In the following reaction, the initial concentrations of the reactant and initial rates at $298K$ are given:

$2A⟶C+D$

$\begin{array}{cc}{\left[A\right]}_0,mol{L}^{-1}& \text{Initial rate in mol}{L}^{-1}{s}^{-1}\\ 0.01& 5.0\times {10}^{-5}\\ 0.02& 2.0\times {10}^{-4}\end{array}$

The value of the rate constant of this reaction at $298K$ is:

• A. $0.01s^{-1}$
• B. $5\times {10}^{-3}mol{L}^{-1}{s}^{-1}$
• C. $2.0\times {10}^{-2}{mol}^{-1}L{s}^{-1}$
• D. $5\times {10}^{-1}{mol}^{-1}L{s}^{-1}$
• E. $5.0\times {10}^{-1}mol{L}^{-1}{s}^{-1}$

Answer 1

The correct option is D $5\times {10}^{-1}{mol}^{-1}L{s}^{-1}$

Let the order of reaction with respect to $A$ is $n$.

$\therefore$ Rate law is given as,
rate, $r=k{\left[A_0\right]}^n$ ....(i)

On putting the given values in equation (i), we get-
$5.0\times {10}^{-5}=k{\left[0.01\right]}^n$ ....(ii)

$2.0\times {10}^{-4}=k{\left[0.02\right]}^n$ .....(iii)

Dividing equations (iii) by (ii) gives,
$\left[\frac{2.0\times {10}^{-4}}{5.0\times {10}^{-5}}\right]={\left[\frac{0.02}{0.01}\right]}^n$

$\left[\frac{20}{5}\right]={\left(2\right)}^n$

or $\left(4\right)={\left(2\right)}^n$

${\left(2\right)}^2={\left(2\right)}^n$

$\therefore n=2$

On putting the value of $n$ in equation (i), we get-

rate, $r=k{\left[A_0\right]}^2$

$\therefore 5\times {10}^{-5}=k{\left(0.01\right)}^2$

$k=\frac{5\times {10}^{-5}}{{10}^{-4}}=5\times {10}^{-1}{mol}^{-1}L{s}^{-1}$