Question
The activation energy for the reaction
2HI (g) → H2 + I2(g)
is 209.5 kJ mol–1 at 581 k. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Solution
2HI (g) → H + I2(g)
Activation energy, Ea = 209.5 kJ mol−1
Multiply by 1000 to convert in j
Ea= 209500 J mol−1
Temperature, T = 581 K
Gas constant, R = 8.314 JK−1 mol−1
According to Arhenious equation
K = A e –Ea/RT
In this formula term e –Ea/RT represent the number of molecules which have energy equal or more than activation energy
Number of molecules = e –Ea/RT
Plug the values we get
Number of molecules
taking antilog of we get
1.47 x 10-19
