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The slope of a line in the graph of log k versus `1/T` for a reaction is `-5841`K. Calculate energy of activation for the reaction.

Answer» Slope = `-E_(a)/(2.303R)` or `E_(a) = -2.303 R xx "slope"`
Slope =`-5841 K, R=8.314 JK^(-1)mol^(-1)`
`therefore E_(a) = -2.303 xx (8.314 JK^(-1) mol^(-1)) xx (-5841 K) = 1.118 xx 10^(5)J mol^(-1)`


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