1.

Two reactants R_(1) and R_(2) have identical pre-exponential 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^(-E_(a_(1)) //RT) , k_(2) = Ae^(-(Ea_(1) - 10) //RT)`
Ln `((k_(2))/(k_(1))) = (10)/(RT) = (10)/(8.314 xx 10^(-3) xx 300) = 4`


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