The activation energy for the reaction 2 HI(g) H2 + I2 (g) is 209.5 kJ mol1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?


Given-


Energy of activation, Ea= 209.5 kJ mol1


Ea in joules = 209.5 × 1000 = 209500 J mol1


Temperature, T = 581K


Gas constant, R = 8.314 J K-1 mol-1


Using Arrhenius equation,


k = A


In this equation, the term represents the number of molecules which have kinetic energy greater than the activation energy, Ea.


The number of molecules = Equation 1


Substituting all the known values in equation 1, we get,



= e-43.3708


Finding the value of anti ln(43.3708), we get, 1.47 × 10-19


The fraction of molecules of reactants having energy equal to or greater than activation energy is 1.47 × 10-19


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