Given the following standard heats of reactions:
(a) heta 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

The required equation is
`C_(2)H_(2)(g) +H_(2)(g) rarr C_(2)H_(4)(g), DeltaH^(Theta) = ? `
Given
a. `H_(2)(g) +(1)/(2) O_(2)(g) rarr H_(2)O(l), DeltaH^(Theta) =- 68.3 kcal`
b. `C_(2)H_(2)(g) +(5)/(2) O_(2)(g) rarr 2CO_(2)(g) +H_(2)O(l), DeltaH^(Theta) =- 310.6 kcal`
c. `C_(2)H_(4)(g) +3O_(2)(g) rarr 2CO_(2)(g) +2H_(2)O(l), DeltaH^(Theta) = - 337.2 kcal`
The required equation cna be achived by adding equaitons (a) and (b) and subtracting (c).
`C_(2)H_(2)(g) +H_(2)(g) +3O_(2)(g) - C_(2)H_(4)(g) - 3O_(2)(g) rarr 2CO_(@) +2H_(2)O(l) -2CO_(2)(g) - 2H_(2)O(l)`
or `C_(2)H_(2)(g) +H_(2)(g)rarr C_(2)H_(4)(g)`
`DeltaH^(Theta) =- 68.3 - 310.6 -(-337.2)`
`=- 378.9 + 337.2 =- 41.7 kcal`
We know that
`DeltaH^(Theta) = DeltaU^(Theta) + DeltanRT`
or `DeltaU^(Theta) DeltaH^(Theta) - DeltanRT`
`Deltan = (1-2) =- 1,R =2 xx 10^(-3) kcal mol^(-1)K^(-1)` and `T = (25 +273) = 298K`
Substituting the values in above equation,
`DeltaU^(Theta) =- 41.7 -(-1) (2 xx 10^(-3)) (298)`
`=- 41.7 + 0.596 =- 41.104 kcal`