1.

For the combustion of 1 mole of liquid benzene at 25^(@)C , the heat of reaction at constant pressure is given by, C_(6)H_(6)(l)+7(1)/(2)O_(2)(g) to 6CO_(2)(g)+3H_(2)O(l),DeltaH=-780980 "cal" What would be the heat of reaction at constant volume ?

Answer»

SOLUTION :We have,
`DeltaH=DeltaU+Deltan_(g)RT` …(Eqn.8)
Here, `Deltan_(g)=6-7.5=-1.5`
Thus, `DeltaU=DeltaH-Deltan_(g)RT`
`=-780980-(-1.5)xx2xx298`
`=-780090 "CALORIES"`.


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