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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?

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?

Grade:8

1 Answers

Gaurav
askIITians Faculty 164 Points
9 years ago

Given that
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−1mol−1
According to Arhenious equation
K = A e–Ea/RT
In this formula term e–Ea/RTrepresent the number of molecules which have energy equal or more than activation energy
Number of molecules = e–Ea/RT
Plug the values and solving we get
Number of molecules
1.47 x 10-19

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