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

The required equation is
`C_(2)H_(2)(g)+H_(2)(g)toC_(2)H_(4)(g)," "DeltaH=?`
Given, (a). `H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)" "(DeltaH=-68.3kcal)` . . . (i)
(b) `C_(2)H_(2)(g)+(5)/(2)O_(2)(g)to2CO_(2)(g)+H_(2)O(l)" "(DeltaH=-310.6kcal)` . . .(ii)
(c) `C_(2)H_(4)(g)+3O_(2)(g)to2CO_(2)(g)+2H_(2)O(l)" "(DeltaH=-337.2kcal)`
The reqruied equation can be achieved by adding eqs. (i) and (ii) and subtracting (iii)
`C_(2)H_(2)(g)+H_(2)(g)+3O_(2)(g)-C_(2)H_(4)(g)-3O_(2)(g)to2CO_(2)+2H_(2)O(l)-2CO_(2)(g)-2H_(2)O(l)`
or `C_(2)H_(2)(g)+H_(2)(g)toC_(2)H_(4)(g)`
`DeltaH=-68.3-310.6-(-337.2)=-378.9+337.2=-41.7kcal`
We know that,
`DeltaH=DeltaU+DeltanRT`
or `DeltaU=DeltaH-DeltanRT`
`Deltan=(1-2)=-1,R=2xx10^(-3)" kcal mol"^(-1)K^(-1)`
and `T=(25+273)=298K`
Substitutinig the values in above equation,
`DeltaU=-41.7-(-1)(2xx10^(-3))(298)`
`=-41.7+0.596=-41.104kcal`