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Two reactions r_(1) and R_(2) have indentical prexponential factors.Activation energy of R_(1) exceeds that of R_(2) by 10 KJ mol^(-1) .If K_(1) and k_(2) are rate constants for reaction R_(1) and R_(2) respectively at 300 K,then ln (K_(2)//K_(1)) is equal to ..........(R=8.314 J mol^(-1)K^(-1))

Answer»

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Solution :`k_(1)=AE^((-Ea_(1))/(RT)),k_(2)Ae^((-(Ea_(1)-10))/(RT)`
in `[(k_(2))/(k_(1))]=(10)/(RT)=(10)/(8.314xx10^(-3)xx300)=4`


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