The half-life of `2` sample are `0.1` and `0.4` seconds. Their respctive concentration are `200` and `50` respectively. What is the order of the reaction

Options

A. 0
B. 2
C. 1
D. 4

Answer

B

Explanation

we know ,
`(t_(1))/t_(2)=[a^(2)/a^(1)]^(n-1)`where t1 and t2 are the half-life periods when the initial concentration are a1 and a2 respectively according to question,
`t_(1)=0.1`seconds,
`a^(1)=200`
`t_(2)=0.4`seconds,
`a^(2)=50`On substituting the values
`0.1/0.4=[50/200]^(n-1)`On taking both sides,
`log""0.1/0.4=(n-1)log"50/200`
`0.60=(n-1)xx0.60/0.60=1`
`(n-1)=1`
`implies" "n=1+1=2`