The activation energy for the reaction 2HI(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?

Answer

Given that
Activation energy, E_a = 209.5 kJ mol⁻¹
Multiply by 1000 to convert in j
E_a= 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

Find the value of 1 / anti ln(43.4 ) we get
1.47 x 10^-19
Do you know any other method of solving this problem ?
The activation energy for the reaction 2HI_(g) → H₂ + 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?
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