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For the combustion of 1 mole of liquid b...

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)rarr6CO_(2)(g)+3H_(2)O(l), DeltaH= -780980 "cal"`.
What would be the heat of reaction at constant volume ?

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We have , `DeltaH=DeltaE+Deltan_(s)RT`
Here, `Delta n_(g)=6-7.5= -1.5`.
Thus, `DeltaE=DeltaH-Deltan_(g)RT= -780980-(-1.5)xx2xx298= -780090 "calories"`.
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