Heat of reaction for `C_(6)H_(12)O_(6(s))+6O_(2(g)) rarr 6CO_(2(g))+6H_(2)O_(v)` at constant pressure is `-651 kcal` at `17^(@)C`. Calculate the heat of reaction at constant volume at `17^(@)C`.

Explanation

`DeltaH = DeltaU +DeltanRT….(i)`
Given `DeltaH =- 651 xx10^(3) cal, R = 2 cal`
Substituting all value in equation (i), we get
`T = 290 K`,
`Deltan = 6+6 -6 = 6`
`:. =- 651 xx 10^(3) = DeltaU +6 xx 2xx 290`
and `DeltaU =- 654480 cla or -654.48 kcal `
Note: For `C_(6)H_(12)O_(6)(g) +6O_(2)(g) rarr 6CO_(2)(g) +6H_(2)O(v) Deltan =` Number of products molecules -Number of reactants molecules (only in gaseous phase) `=6 +6 -6 = 6`
Here, `H_(2)O(v)` is also in gaseous state.