InterviewSolution
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InterviewSolution
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A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction . (Given : log 2 = `0.3010` , log 4 = `0.6021` , R = `8.314 J K^(-1) mol^(-1)`) . |
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Answer» `t_(1//2)` for first order reaction : `t_(1//2) = (0.693)/(K)` For 300 K , `" " K_(1) = (0.693)/(t_(1//2)) = (0.693)/(40) S^(-1)` For 320 K , `" " K_(2) = (0.693)/(t_(1//2)) = (0.693)/(20) S^(-1)` `log ((K_(2))/(K_(1))) = (E_(a))/(2.303R) ((1)/(T_(1)) - (1)/(T_(2)))` log `(((cancel0.693)/(20))/((cancel0.693)/(40))) = (E_(a))/(2.303 xx 8.314) [(1)/(300) - (1)/(200)]` `log(2) = (E_(a))/(2.303 xx 8.314)xx ((320 - 300)/(300 xx 320))` `0.3010 = (E_(a))/(2.303 xx 8.314) xx (cancel20)/(cancel200_(150) xx cancel320)` `E_(a) = 0.3010 xx 2.303 xx 8.314 xx 150 xx 32` `E_(a) = 27663, 8 J//"mol". |
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