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. `1 hr`
B. `0.5 hr`
C. `0.25 hr`
D. `2 hr`

Answer

A

Explanation

For the reaction
`2A + B rarr 3C + D`
Rate constant of first-order reaction
`k = (2.303)/(t) "log"_(10)((A)_(0))/((A)_(t))` or `k = (2.303)/(1) xx"log"_(10) (0.8)/(0.2)` …(i)
(because `0.6` mole of `B` is formed)
Suppose `t_(1)` and hour is required 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`
`:. k = (2.303)/(t) "log"_(10)(0.9)/(0.225)`
Form Eqs (i) and (ii), we get
`k = (2.303)/(t) "log"_(10)(0.8)/(0.2) = (2.303)/(t_(1)) "log"_(10)(0.9)/(0.225)`
`2.303 log_(10) 4 = (2.303)/(t) log_(10) 4`
`t_(1) = 1 hr`