4.9 The activation energy for the reaction 2 HI(g) → H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
4.9 The activation energy for the reaction 2 HI(g) → H2 + I2 (g) is 209.5 kJ mol–1 at 581K.Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
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1 Answer
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4.9 Given-
Energy of activation, Ea= 209.5 kJ mol–1
Ea in joules = 209.5 × 1000 = 209500 J mol–1
Temperature, T = 581K
Gas constant, R = 8.314 J K-1 mol-1
Using the Arrhenius equation,
k = Ae-Ea/RT
In this equation, the term e-Ea/RT represents the number of molecules which have kinetic energy greater than the activation energy, Ea.
∴The number of molecules = e-Ea/RT → Equation 1. Substituting all the known values in equation 1, we get,
= e-209500/8.314X581
=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 activ
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