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The activation energy of a reaction is zero. Its rate constant at 280 K is `16 xx 10^(-6) s^(-1)` the rate constant at 300k isA. `3.2 xx 10^(-6) s^(-1)`B. zeroC. `1.6 xx 10^(-6)s^(-1)`D. `1.6 xx 10^(-5)s^(-1)` |
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Answer» Correct Answer - C `"log" (k_(2))/(k_(1)) = (E_(a))/(2.303 R) [(1)/(T_(1)) - (1)/(T_(2))]` Given, Activation energy of a reaction, `E_(A) = 0` Rate constant, `k_(1) = 1.6 xx 10^(-6) s^(-1)` Temperature, `T_(1) = 280 K, T_(2) = 300 K` According to Arrhenius equation `"log" (k_(2))/(1.6 xx 10^(-6)) = (0)/(2.303 R) [(1)/(280) - (1)/(300)]` `"log" (k_(2))/(1.6 xx 10^(-6)) = 0` `(k_(2))/(1.6 xx 10^(-6))` = antilog 0 `(k_(2))/(1.6 xx 10^(-6)) = 1` `:. k_(2) = 1.6 xx 10^(-6) s^(-1)` |
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