The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to produce `0.6 mol` of `B` is `1 hr`. What is the time taken for the conversion of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?

Options

A. `0.25 h`
B. `2 h`
C. `1h`
D. `0.5h`

Answer

C

Explanation

First case
`overset(ArarrB)underset(t=1h(0.8-x)molxmol)(t=0h0.8mol0mol)`
We are given
`x=0.6mol`
:. `(0.8-x)mol=(0.8-0.6)mol=0.2mol`
Integrater rate law for first order reaction is
`k=(2.303)/(t)log([A]_(0))/([A])`
Substituting the ressult, we get
`k=(2.303)/(1h)log(0.8mol)/(0.2mol)`
`overset(ArarrB)underset(t=th(0.9-x)molxmol)(t=0h0.9mol0mol)`
We are given
`x=0.675mol`
:. `(0.9-x)mol=(0.9-0.675)mol`
Substituting the result into integrater rate law, we get
`k=(2.303)/(th)log(0.9mol)/(0.225mol)`
Since the value of rate constant does not depend upon concentration, we can equate the Eqs `(1)` and `(2)`
`(2.303)/(1h)log(0.8mol)/(0.2mol)=(2.303)/(th)log(0.9mol)/(0.225)`
`(2.303)/(1h)log4=(2.303)/(th)log4`
Therefore, `t=1h`
Alternatively, in both the cases, the fraction of reaction completed is the same:
`f_(1)=(0.6)/(0.8)=00.75`
`f_(2)=(0.675)/(0.9)=0.75`
For a first order reaction, the required to complete the same fraction of reaction is indepedent of intial concentrations.