InterviewSolution
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InterviewSolution
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`Zn+2H^(o+)rarrZn^(2+)+H_(2)` The half-life periof is independent of the concentration of zinc at constant `pH`. For the constant concentration of `Zn`, the rate becomes `100` times when `pH` is decreased form `3` to `2`. Hence,A. `(dx)/(dt) = k[Zn]^(0)[H^(o+)]^(2)`B. `(dx)/(dt) = k[Zn][H^(o+)]^(2)`C. Rate is not affected if the concentration of zinc is made four times and that of `H^(o+)` ion is halved.D. (d) Rate becomes four times if the concentration of `H^(o+)` ion is doubled at constant `Zn` concentration. |
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Answer» Correct Answer - B::C::D (i) Since `t_(1//2)` is independent of concentration of `Zn` at constant `pH` means that the order w.r.t. `[Zn] =1`. (ii). Le `r_(1) prop [H^(o+)]^(x) rArr r_(1) prop [10^(-3)]^(x)` , [when `pH = 3`] `r_(2) = 100r_(1) prop[10^(-2)]^(x)` , [when `pH = 2`] Thus, `(r_(2))/(r_(1)) = [(10^(-2))/(10^(-3))]^(x)` `100 = (10)^(2) = [10]^(x) rArr x = 2` Hence, order w.r.t. `[H^(o+)] = 2` (b) Hence, the rate `= ((dx)/(dt)) = k[Zn][H^(o+)]^(2)` So (b) is the correct answer. ( c) `((dx)/(dt)) = r_(1) = k[Zn][H^(o+)]^(2)` `r_(2) = k[Zn][(H^(o+))/(2)]^(2)` Thus, `r_(2) = r_(1)` (d) `r_(1) = k[Zn][H^(o+)]^(2)` ...(i) `r_(2) = k[Zn][2H^(o+)]^(2)` ...(ii) Divide (ii) by (i), `r_(2) = 4r_(1)` |
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