Question

The half-life a first order reaction is $24$ hours. If we start with $10M$ initial concentration of the reactant then concentration after $96$ hours will be?

• A. $6.25M$
• B. $1.25M$
• C. $0.125M$
• D. $0.625M$

Answer 1

The correct option is D $0.625M$

Given:

$\left[A_0\right]=10$(initial concentration

$t_{1/2}=24hrs$

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{24}$

$k=0.028875hr{s}^{-1}$

$k=\frac{2.303}{t}Log\frac{\left[A_0\right]}{\left[A\right]}$------(1)

putting all the given values in equation (1)

$0.028875=\frac{2.303}{96}Log\frac{\left[10\right]}{\left[A\right]}$

$Log\frac{10}{\left[A\right]}=1.20364$

on taking antilog we will get:

$\frac{10}{\left[A\right]}=15.9823$

$\left[A\right]=0.62569M$

Option D is correct.