Question

The rate of reaction increase 4-fold when a concentration of reactant is increased 16 times. If the rate of reaction is $4\times {10}^{-6}$ $mole{L}^{-1}{s}^{-1}$ when concentration of the reactant is $4\times {10}^{-4}$ $mole{L}^{-1}$ , the rate constant of the reaction will be:

• A. $2\times {10}^{-4}$ $mol{e}^{1/2}{L}^{-1/2}{s}^{-1}$
• B. $1\times {10}^{-2}$ $s^{-1}$
• C. $2\times {10}^{-2}$ $mol{e}^{-1/2}{L}^{1/2}{s}^{-1}$
• D. $25mol{e}^{-1/2}$ $L^{-1}$

Answer 1

The correct option is A $2\times {10}^{-4}$ $mol{e}^{1/2}{L}^{-1/2}{s}^{-1}$

Rate, R=$A\to B\frac{-d\left[A\right]}{dt}=\frac{-d\left[B\right]}{dt}=k{\left[A\right]}^x$

$R^{\prime }=4Rand\left[A\right]=16\left[A\right]$

$4={16}^{x}x=0.5$

$R=k{\left[A\right]}^x$

Given $R=4\times {10}^{-6}$ and $\left[A\right]=4\times {10}^{-4}$

$\therefore 4\times {10}^{-6}=k\times \sqrt{4\times {10}^{-4}}$

$k=\frac{4\times {10}^{-6}}{2\times {10}^{-2}}=2\times {10}^{-4}mol{e}^{1/2}{L}^{-1/2}{s}^{-1}$

Option A is correct.