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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 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» Correct Answer - B::D 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` |
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