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The rate constant of a first order reaction is 60 s-1 . How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? |
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Answer» Rate constant of reaction, k = 60 s-1 t15/16 = ? Rate constant of first order reaction is given as, \(k\,=\,\frac{2.303}{t}log\frac{a}{a-x}\) or, \(t\,=\,\frac{2.303}{k}log\frac{a}{a-x}\) When \(\frac{15}{16}th\) of reaction is over, if a = 1 M, then a - x = \(\frac{1}{16}\) M Hence, \(t_\frac{15}{16}\) = \(\,\frac{2.303}{60\,s^{-1}}log\frac{1}{1-\frac{15}{16}}\) \(=\,\frac{2.303}{60\,s^{-1}}log\,16\) \(=\,\frac{2.303}{60\,s^{-1}}\) x 1.2041s = 0.046 s |
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