Question

The activation energy for the reaction is 210 kj/mol at 550k. Calculate tha fraction of molecules of reactants having energy equal to or greater rha activation energy.

Answer 1

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?

Answer

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



 

Answer 2

The fraction of molecules of reactants having energy equal to or greater rha activation energy is equal to 1.135×10^-20.

• We are given that the activation energy for the reaction is 210 KJ/mole and the temperature is equal to 550K.
• The fraction of molecules of reactants having energy equal to or greater rha activation energy is given by e^(-EA/RT).
• Now, substituting the values we get the fraction of molecules to be 1.135×10^-20.