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

Given that
(i) `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(1),DeltaH=-68.3kcal`
(ii) `C_(2)H_(2)(g)+2(1)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O(1),DeltaH=-310.6kcal`
(iii) `C_(2)H_(4)(g)+3O_(2)(g) to 2CO_(2)(g)+2H_(2)O(1),DeltaH=-337.2kcal.`
we have to calculate `Deltamu` for the equation,
(iv) `C_(2)H_(2)(g)+H_(2)(g) to C_(2)H_(4)(g),Deltamu=?`
`C_(2)H_(2)` in eqns. (ii) and (iv), and `H_(2)` in eqns. (i) and (iv) are on the same sides, whereas `C_(2)H_(4)` in eqns. (iii) and (iv) is on opposite sides. Thus applying [Eqn. (i) +Eqn. (ii) - Eqn. (iii)] we get
`H_(2)(g)+(1)/(2)O_(2)(g)+C_(2)H_(2)(g)+2(1)/(2)O_(2)(g)-C_(2)H_(2)(g)-H_(2)(g) to `
`H_(2)O(1)+2CO_(2)(g)+H_(2)O(1)-2CO_(2)(g)-2H_(2)O(1),`
`DeltaH=-68.3+(-310.6)-(-337.2)`
or `C_(2)H_(2)(g)+H_(2)(g) to C_(2)H_(4)(g),DeltaH=-41.7kcal.`
`DeltaH` is the heat of reaction at constant pressure and to calculate heat at constant volume, we determine `Deltamu` by using equation,
`DeltaH=Deltamu+Deltan_(g)RT...(Eqn. 8)`
`Deltan_(g)= "moles of gaseous product" - "moles of gaseous reactant" `
`=1-2=-1`
`R=0.002kcal//K// "mole" `
`T=(273+(25)=298K`
Thus we have,
`Deltamu=-41.7-(-1xx0.002xx298)`
`=-41.104kcal.`
[Note: if the heat of reaction is given in kj, R should be taken as ` 8.314xx10^(-3)kJ//K// "mole"` ]