Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

9372462318 / 9987178554

Mon to Sat - 10 AM to 7 PM

If reaction at 500K proceeds at 400 K in presence of catalyst then what is value of Ea ; if catalyst lowers activation barrier by 20kJ/mol???

Asked by Umesh 20th May 2017, 10:04 AM
Answered by Expert
Answer:

Consider Ea and E’a be the energy of activation in presence and absence of catalyst for hydrogenation reaction.

We know,

K = Ae–Ea/RT

So in presence of catalyst,

 K1 = Ae–Ea/(R × 500)

And in absence of catalyst,

K2 = Ae–E’a/(R × 400)

Since the two rates are given to be same,

r1 = r2 => K1 = K2

So, Ae–Ea/(R × 500) = Ae–E’a/(R × 400)

Ea/(R × 500) = E’a/(R × 400)

Ea/500 = (Ea – 20)/400 [As, Ea – E’a = 20]

Hence Ea = 100 kJ mol–1

Answered by Expert 20th May 2017, 6:42 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Free related questions

Chat with us on WhatsApp