For a reaction, the rate constant is expressed as `k = Ae^(-40000//T)`. The energy of the activation is

Rate constant for a chemical reaction taking place at 500K is expressed as K=A. e^(-1000) The activation energy of the reaction is :

for a given reaction at temeperature T, the velocity constant k, is expressed as : k=ae^(-27000K'//T) (K' = Boltzmann constant) Given , R=2 cal K^(-1) "mol"^(-1) . Calculate the value of energy of activation. Comment on the results.

A reaction takes place in three rate determining steps having rate constants K_(1), K_(2), K_(3) respectively. The overall rate constant K=(K_(1)K_(3))/K_(2) . If energy of activations for the three steps are respectively 10, 20, 40 kJ . The overall energy of activation is:

The rate constant is given by Arrhenius equation. k=Ae^(-E_(a)//RT) Calculate the ratio of the catalysed and uncatalysed rate constants at 25^(@)C if the energy of activation of a catalysed raction is 162 kJ and for the uncatalysed reaction the value is 350 kJ

The rate of a reaction depends upon the temperature and is quantitatively expressed as k=Ae^((-Ea)/(RT) i) If a graph is plotted between log k and 1/T, write the expression for the slope of the reaction? ii) If at under different conditions E_(a1)andE_(a2) are the activation energy of two reactions. If E_(a1) = 40 J// "mol and " E_(a2) = 80 J//" mol" . Which of the two has a larger value of the rate constant?

The rate constant is given by the Arrhenius equation as : k = Ae^(-Ea)//RT Calcualte the ratio of the catalysed and uncatalysed rate constants at 25^(@) C If the energy of activation of a analysed reaction is 162 kJ mol^(-1) and for uncatalysed reaction, the value is 350 kJ mol^(-1) .