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The rate of a first order reaction is 0.04 mol `"litre"^(-1)s^(-1)` after 10 minutes and 0.03 mol `"litre"^(-1)s^(-1)` after 20 minutes. Find the half life period of the reaction. |
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Answer» Step. I. Calculation of rate constant (k). For first order reaction, rate( r) `propto C` or kC Rate `(r_(1)) = kC_(1)` and Rate `(r_(2)) =kC_(2)` `therefore` Rate `(r_(1))= kC_(1)` and Rate`(r_(2) = kC_(2)` `therefore` `(r_(1)("at 10 min"))/(r_(2)("at 20 min")) = C_(1)/C_(2)= (0.04 mol L^(-1))/(0.03 mol L^(-1)) =0.04/0.03` Now, for first order reaction, `t=(2.303)/k log(C_(1))/C_(2), k = 2.303/t log C_(1)/C_(2) = 2.303/(10 min) log (0.04)/(0.03)` `=2.303/(10 min) xx log(4/3) xx log(4/3) = 2.303/(10 min) xx 0.1249 = 0.0287 min^(-1)` Step II. Calculation of half life period `(t_(1//2))` Half life period `t_(1//2)= 0.693/k = (0.693)/(0.0287 min^(-1)) = 24.15` min |
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