A first order reaction is carried out starting with 10 mol `l^(-1)` of the reactant. It is 40% complete in 1h if the same reaction is carried out with an initial concentration of 5 mol `l^(-1)` the percentage of the reaction that is completed in 1h will be

Options

A. 0.4
B. 0.8
C. 0.2
D. 0.6

Answer

A

Explanation

For first order reaction t=`(2.303)/(k)log(a_(0))/(a_(0)-x)`
Case I: `a_(0)=10 mol L^(-1)` for 40% completion of the reaction `t_(1)=(2.303)/(k)log(10/6)=1h`
Case II : `a_(0)=5 mol L^(-1)` if the initinal concentration that is a = 5 mol `L^(-1)` then `t_(2) =(2.303)/(k) log (5)/(5-x)`
Given that `t_(1)=t_(2)` =1h so log `(5/3)=log (5)/(5-x) or 5/3 =(5)/(5-x) or x=2`
Therefore the percentage of reaction completed is `2/5 xx100=40%`