For first order reaction involving `2A rarr` products the specific reaction rate is `0.0084 s^(-1)`. If `2.50` moles of A are taken in a `5.0` litre flask, then moles of A remaining after 60 seconds will be

Options

A. `0.637`
B. `0.786`
C. `0.555`
D. `0.913`

Answer

D

Explanation

For a first order reaction of the type
`aArarr`products
the integrater rate law is
`k=(2.303)/(at)log([A]_(0))/([A]_(t))`
If we assume that `'x'` moles of reactant `A` will remain after `60seconds, then
`0.0084s^(-1)=(2.303)/(2(60s))log(2.50//5M)/(x//5M)`
`=(2.303)/(2(60s))log(2.50)/(x)`
log `(2.50)/(x)=((0.0084s^(-1))(2)(60s))/(2.303)`
`x=0.9125` mole