For the combustion of 1 mole of liquid benzene at `27^(@)C` the heat of reaction at constant pressure is given by:
`C_(6)H_(6)(l)+(15)/(2)O_(2)(g)to6CO_(2)(g)+3H_(2) O(l),`
`DeltaH=-78kcal`
What would the be heat of reaction at constant volume?

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 ?

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))

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))

The difference between heats of reaction at constant pressure and constant volume for reaction: 2C_(6)H_(6)(l)+15O_(2)(g)to12CO_(2)(g)+6H_(2)O(l) at 25^(@)C in kJ is

From the following data at constant volume for combustion of benzene, calculate the heat of this reaction at constant pressure condition. C_(6)H_(6(1))+71//2O_(2(g))rarr 6CO_(2(g))+13H_(2)O_((l))