Discussion Forum

The half-life of two samples are 0.1 and 0.8 s. Their respective concentration are 400 and 50 respectively . The order of the reaction is 

Answer:

Explanation:

It is known

 $\frac{(t_{1/2})_{1}}{(t_{1/2})_{2}}=\left[\frac{a_{2}}{a_{1}}\right]^{(n-1)}$

 Here, n= order of reaction

 Given, $ (t_{1/2})_{1}=0.1 s, a_{1}=400$

 $(t_{1/2})_{2}=0.8 s, a_{2}=50$

 On putting the values 

 $\frac{0.1}{0.8}=\left[\frac{50}{400}\right]^{(n-1)}$

 Taking log on both sides,

 $\log \frac{0.1}{0.8}=(n-1) \log \frac{50}{400}$

  $\log \frac{1}{8}=(n-1)\log \frac{1}{8}$

$n-1=1 \Rightarrow n=2$