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# 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
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
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