the rate of a first - order reaction is 0.04 mol `L^(-1) s^(-1)` at 10 sec and 0.03 mol `L^(-1) S^(-1)` at 20 sec after initiation of the reaction the half -life period of the reaction is

Options

A. `34.1s`
B. 44.1 s
C. 54.1s
D. 24.1

Answer

D

Explanation

Given order of reaction =1
Rate of reaction at 10 s=0.04 mol `L^(-1)S^(-1)`
rate of reaction at 20s=0.03 `mol L^(-1)S^(-1)`
` therefore ` half - life period `(t_(1//2))=?`
we have the equation for rate - constant 'K' in first order reaction .
`K =(2303)/(t) log""(A_(t))/(A_(0))=(2.303)/(105 ) log ""(0.04)/(0.03)`
` =(2.303)/(105)xx0.124`
`k=0.028s^(-1)`
we know that
`t_(1//2)=(0.693)/(k) =(0.693)/(0.028773391S^(-1))`
`=24.14 s=24.1S`