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Find out two-third `(2//3)` life of a first order reaction in which `k=5.48 xx 10^(-14)s^(-1)` |
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Answer» For the first order reaction, `t=2.303/k log ([A]_(0))/[A]` Let `[A]_(0) = a, [A] = (a-2//3a) = 1//3a, k=5.48 xx 10^(-14)s^(-1)` `t_(2//3) = 2.303/k log ([A]_(0)/[A]) = 2.303/(5.48 xx 10^(-14)s^(-1))log a/(1//3a) = 2.303 / (5.48 xx 10^(4)s^(-1))log3` `(2.303)/(5.48 xx 10^(-14)s^(-1)) xx 0.4771= 2.0 xx 10^(13)`s |
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