Question

The half-lives of two samples are $0.1s$ and $0.8s$. Their respective concentration are $400$ and $50mol{L}^{-1}$ respectively. The order of the reaction is

• A. 0
• B. 2
• C. 1
• D. 4

Answer 1

Answer: (B)

We know that
$\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}={\left[\frac{a_2}{a_1}\right]}^{\left(n-1\right)}$
where, $n=$ order of the reaction
Given, ${\left(t_{1/2}\right)}_1=0.1s,a_1=400$
${\left(t_{1/2}\right)}_2=0.8s,a_2=50$
On putting the values,
$\frac{0.1}{0.8}={\left[\frac{50}{400}\right]}^{\left(n-1\right)}$
Taking log on both sides
$\log \frac{0.1}{0.8}=(n-1)\log \frac{50}{400}$
$\log \frac{1}{8}=(n-1)\log \frac{1}{8}$
$0.90=(n-1)0.90$
$n-1=1\Rightarrow n=2$