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A first order reaction is `50%` complete in `30` minutes at `27^(@)C` and in `10` minutes at `47^(@)C`. The rate constant at `47^(@)C` and energy of activation of the reaction in kJ`//`mole will beA. `0.0693, 43.848 kJ mol^(-1)`B. `0.0560,45.621 kJ mol^(-1)`C. `0.0625, 42.926 kJ mol^(-1)`D. `0.0660, 46.189 kJ mol^(-1)` |
Answer» Correct Answer - A We know that `k = (2.303)/(t) log((a)/(a-x))` Subtituting the values at the two given conditions `k_(27^(@)) = (0.0693)/(30) = 0.0231 , k_(47^(@)) = (0.693)/(10)= 0.0693` We also know that `"log" (k_(2))/(k_(1)) = (2.303)/(t)"log"((a)/(a-x))` or `E_(a) = (2.303 R xx T_(1)T_(2))/(T_(2) - T_(1)) "log"(k_(2))/(k_(1))` `= (2.303 xx 8.314 xx 10^(-3) xx 300 xx 320)/(320 - 300) xx "log" (0.0693)/(0.231)` `= 43.848 kJ mol^(-1)` |
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