CBSE Class 11-science Answered
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 | 27 Jan, 2011, 12:00: AM
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
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 | 28 Jan, 2011, 10:59: AM
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