Question

For a first order reaction $t_{0.75}$ is $138.6sec$. Its specific rate constant is (in $s^{-1}$)

• A. ${10}^{-2}$
• B. ${10}^{-6}$
• C. ${10}^{-4}$
• D. ${10}^{-5}$

Answer 1

The correct option is A ${10}^{-2}$

1. As we know that, Rate constant of first order reactions depends on only one reactant.
   

Lets consider rate of reaction be 'k', $A_0$ be the initial concentration of the reactant, $A$ be the concentration of reactant after quarter life time and $t=138.6sec$ is the quarter time.

$A$=$A_0\times e^{-kt}$

After quarter life $A$= $\frac{A_0}{4}$

Therefore,

$\frac{A_0}{4}$=$A_0\times e^{-kt}$

Simplifying equation,

$\frac{1}{4}$=$e^{-kt}$

$4=e^{kt}$

Applying natural log,

$ln\left(4\right)=kt$

Substituting the values,

$1.3865=k\times 138.6$

$k={10}^{-2}$

So the rate constant of the reaction is ${10}^{-2}$