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For combustion of 1 mole of benzene at 2...

For combustion of 1 mole of benzene at `25^(@)C`, the heat of reaction at constant pressure is `-780.9" kcal."`
What will be the heat of reaction at constant volume ?
`C_(6)H_(6(l))+7(1)/(2)O_(2(g))rarr 6CO_(2(g))+3H_(2)O_((l))`

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Verified by Experts

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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