The activation energy for the reaction 2 HI(g) → H_{2} + I_{2} (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?

Given-

Energy of activation, E_{a}= 209.5 kJ mol^{–}^{1}

E_{a} in joules = 209.5 × 1000 = 209500 J mol^{–}^{1}

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, E_{a}.

∴ 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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