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For a chemical reaction at `27^(@)C` , the activation energy is 600 R . The ratio of the rate constants at `327^(@)C` to that of at `27^(@)C` will beA. 2B. 40C. eD. `e^(2)` |
Answer» Correct Answer - c ln `(K_(2))/(K_(1)) = (E_(a))/(R) ((1)/(T_(1)) - (1)/(T_(2)))` or , ln `(K_(2))/(K_(1)) = (600R)/(R) ((1)/(300) - (1)/(600))` or , ln `(K_(2))/(K_(1)) = (600R)/(R) ((2-1)/(600)) = 1 ` ln `(K_(2))/(K_(1))` = ln e `" " (K_(2))/(K_(1)) = e` |
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