Question

The half-lives of two samples are $0.1\,s$ and $0.8\, s$. Their respective concentration are $400$ and $50\,mol\,L^{-1}$ 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, $\left(t_{1/2}\right)_{1} = 0.1\, s,\, a_{1} = 400$
$\left(t_{1/2}\right)_{2} = 0.8\, s,\, a_{2} = 50$
On putting the values,
$\frac{0.1}{0.8} = \left[\frac{50}{400}\right]^{\left(n-1\right)} $
Taking log on both sides
$\log \frac{0.1}{0.8} =(n-1) \log \frac{50}{400}$
$\log \frac{1}{8} =(n-1) \log \frac{1}{8}$
$0.90 =(n-1) 0.90$
$n-1=1 \Rightarrow n=2$