Question

The activation energy for the reaction \[\ce{2 HI_{(g)} -> H2_{(g)} + I2_{(g)}}\] is 209.5 kJ mol⁻¹ at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?

Answer 1

In the given case:

E_a = 209.5 kJ mol⁻¹ 

T = 581 K

R = 8.314 JK⁻¹ mol⁻¹

Now, the fraction of molecules of reactants having energy equal to or greater than activation energy is given as:

x = `"n"/"N"`

x = `e^((-E_a)/"RT")`

In x = `- (E_a)/"RT"`

or, log x = `-E_a/(2.303  "RT")`

or, log x = `-(209.5 xx 10^3  "J mol"^-1)/(2.303 xx 8.314  "JK"^-1  "mol"^-1 xx 581  "K")`

= −18.8323

x = antilog (−18.8323)

= antilog 19.1677

= 1.471 × 10⁻¹⁹