Please wait...
1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

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 toll free number

022-62211530

Mon to Sat - 11 AM to 8 PM

1. Enthalpy of combastion of benzen ,carbon,hydrogen,are 3264.5kj,-393.5kj, -285.8kj respectively? calculate the enthalpy of formation of benzen.

Asked by rahmani 27th January 2011, 12:00 AM
Answered by Expert
Answer:
Dear Student
 
The reaction for the formation of benzene is
 
6C(s)  + 3H2(g) -------> C6H6
 
In general we can understand the reaction path as follows:
 
 
 1. C6H6 + O2(g) ----->  CO2(g) + H2O  [delta]H = - 3264.5 kJ/mol
2. C(s)  +  O2(g) ----->  CO2(g)         [delta]H = - 393.5 kJ/mol
3.  H2(g)  +  1/2 O2(g)  ----->   H2O(l)  [delta]H = - 285.8 kJ/mol

According to the Hess's Law:

ΔH - 3264.5 = 6(-393.5) + 3(-285.8)

Rearranging and solving:

ΔH = 3264.5 + 6(-393.5) + 3(-285.8)

= 3264.5- (2361 +857.4)

ΔH = +46.1 kJ mol-1

We hope that clarifies your query.

Regards

Team

Topperlearning

Answered by Expert 28th January 2011, 10:59 AM
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!