Given that : (i) heat of formation of water = -68.3 kcal (ii) heat of combustion of acetylene = - 310.6 kcal (iii) heat of combustion of ethylene = - 337.2 kcal. Calculate the heat of reaction for the hydrogenation of acetylene at constant volume `(25^(@)C)`.

Explanation

`H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l) , Delta H^(0)= - 68.3 ` kcal ..(1)
`C_(2)H_(2)(g)+(5)/(2)O_(2)(g) to 2CO_(2)(g)+H_(2)O(l) , Delta H^(0)= -310.6 ` kcal ..(2)
`C_(2)H_(4)(g)+3O_(2)(g) to 2CO_(2)(g)+2H_(2)O(l), Delta H^(0) = - 337.2 ` kcal .. (3)
Subtracting equation (3) from the equation obtained by adding equation (1) and (2), we have `C_(2)H_(2)(g)+H_(2)(g) to C_(2)H_(4)(g), Delta H^(0)= (-68.3-310.6+337.2) " kcal " = -41.7 " kcal "` .. (4)
We know, `Delta h^(0)= Delta U^(0)+Delta nRT`. As per reaction (4), `Delta n= 1 - (1+1)+- 1`. Given, T= 273+25 = 298 K.
`therefore - 41.7xx10^(3) " cal " = Delta U^(0)- Rxx 298 K`
`= Delta U^(0) -1.987xx298 ` cal
`therefore Delta U^(0)= -41.10` kcal.