Question

Half life of a first order reaction is $4s$ and the initial concentration of the reactant is $0.12M$. The concentration of the reactant left after $16s$ is

• A. $0.0075M$
• B. $0.06M$
• C. $0.03M$
• D. $0.015M$

Answer 1

Answer: (A)

$t_{1/2}=4sT=16s$
$n=\frac{T}{t_{1/2}}=\frac{16}{4}=4$
$\left(\therefore T=n\times t_{1/2}\right)$
$[A]=\left[A_0\right]{\left(\frac{1}{2}\right)}^n$
$=0.12\times {\left(\frac{1}{2}\right)}^4=\frac{0.12}{16}=0.0075M$
Where $\left[A_0\right]=$ initial concentration and $[A]$
$=$ concentration left after time $t$.