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The rate constant for a first order reaction is `60s^(-1)`. How much time will it take to reduce the initial concentrationof the reactant to its `1//16th` value ? |
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Answer» For the first order reaction `t=2.303/k log a/(a-x)` Let the initial concentration =a , Final concentration (a-x) `= a//16, k=60s^(-1)` `t=2.303/(60s^(-1)) log a/(a//16) = 2.303/(60s^(-1)) log 16` `=(2.303xx1.2041)/(60s^(-1)) = 0.0462s = 4.62 xx 10^(-2)s` |
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