Notifications
Clear all
1
24/01/2022 4:03 pm
Topic starter
The activation energy for the reaction
2HI(g) → H2 + I2(g) is 209.5 kJ mol-1 at 581K. C
calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
1 Answer
1
24/01/2022 4:06 pm
In the given case:
Ea = 209.5 kJ mol-1 = 209500 J 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 = e- Ea/RT ⇒ In x = -E