Question

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

Answer 1

Correct Answer - C
Rate constant of first of this order of first order reaction
`K=(2.303)/(t) Log""((A)_(0))/((A)_(t))`
`or k=(2.303)/(t)xxlog_(10)"" (0.8)/(0.2)`
because 0.6 mole of B is formed
suppose `t_(1) ` hour are reguireed for changing the concentration of A from 0.9 mole to 0.675 mole of B remaining mole of A =0.9-0.675=0.225
`therefore k=(2.303)/(t_(1))log_(10)(0.9)/(0.225)`
From Eqs. (i) and(ii)
`(2.303)/(1) log_(10)(0.8)/(0.2)=(2.303)/(t_(1)) log_(10)(0.9)/(0.225)`
`2.303 log _(10 ) 4=(2.303)/(t_(1))log _(10)4`
`t_(1) =1h`