In a second order reaction, when the concentration of both the reactants are equal, the reaction is completed `20%` in 500 s. How long would it take for the reaction to go to `60%` completion?

Options

A. 3000s
B. 5000s
C. 1000 s
D. 2000s

Answer

A

Explanation

a) For the second order reaction,
`k=1/(t xx a). (X/(a-x))`
According to available data, in the first case
t=500s , a=100
a-x `=80%` of 100 = 80
`k=(1 xx 20)/(500s xx 100 xx 80)= 5 xx 10^(-6)s^(-1)`
In the second case, a=100
`(a-x) = 40%` of 100 = 40
`t=(1 xx 60)/((5 xx 10^(-6)s^(-1)) xx (100 xx 40))=3000s`.