A certain reaction `A rarr B` follows the given concentration (Molarity)-time graph. Which of the following statements is/are true?

(a)The reaction is second order with respect to A .
(b) The rate for this reaction at 20 s will be 7 × 10 − 3 M s − 1
(c) The rate for this reaction at 80 s will be 1.75 × 10 − 3 M s − 1
(d) The [ B ] will be 0.35 M at t = 60 s .

Options

A. The reaction is second order with respect to `A`.
B. The rate for this reaction at `20 s` will be `7 xx 10^(-3) M s^(-1)`
C. The rate for this reaction at `80 s` will be `1.75 xx 10^(-3) M s^(-1)`
D. The `[B]` will be `0.35 M` at `t = 60 s`.

Answer

B, D

Explanation

Use initial rate law methof, let the reaction be first order,
So, `k = (2.303)/(t) log.(a)/(a-x)`
(b) At `t =20 s, k_(1) = (2.303)/(20)log.((0.4)/(0.2))`
and at `t = 40 s, k_(1) = (2.303)/(40)s, k_(2) = (2.303)/(40) log. ((0.4)/(0.1)) = k_(1)`
`rArr` Assumption is correct `(because k_(1) = k_(2))`
Rate at `20 s = k[A] = (0.693)/(20) xx 0.2`
`= 0.0063 ~~ 7 xx 10^(-3) M s^(-1)`
Clearly, half life `t_(1//2) = 20 s`
(d) In `60 s`, number of life `= (60)/(20) =3`
`rArr [B]` at `60 s`
`= 0.4 xx 0.4 ((1)/(2))^(3) = 0.35 M`