Given the following standard heats of reactions:
(a) heat of formation of water `= -68.3 kcal`, (b) heat of combustion of `C_(2)H_(2) =- 310.6 kcal`, (c ) heat of combustion of ethylene `=- 337.2 kcal`. Calculate the heat of reaction for the hydrogenation of acetylene at constant volume and at `25^(@)C`.

Explanation

Standard heats of reaction means `DeltaH` at `1atm` and `25^(@)C`. Given,
`H_(2) +1//2O_(2)(g) rarr H_(2)O, DeltaH =- 68.3 kcal …(i)`
`C_(2)H_(2) +5//2O_(2) rarr 2CO_(2) +H_(2)O, DeltaH =- 310.6 kcal …(ii)`
`C_(2)H_(4) +3O_(2) rarr 2CO_(2) +2H_(2)O, DeltaH =- 337.2 kcal ...(iii)`
`C_(2)H_(2)rarr C_(2)H_(4),DeltaH =? `
Add equations (i) and (ii) and Subtract equation (iii)
`C_(2)H_(2) +H_(2) rarr C_(2)H_(4),DeltaH =- 41.7 kcal`
Now, `DeltaH = DeltaU +DeltanRT`
`Deltan =1 -2 =-1, R =2 cal, T = 298 K`
`-41.7 xx 10^(3) = DeltaU +(-1) xx2xx 298`
`DeltaE =- 41104 cal or DeltaU =- 41.104 kcal`