Given below are some standard heats of reaction (at constant pressure)
(A) Heat of formation of water = -68.3 kcal
(B) Heat of combustion of acetylene = - 310.6 kcal
(C ) Heat of combustion of ethylene = - 337.2 kcal
Calculating the heat of reaction for the hydrogenation of acetylene at constant pressure and at constant volume (at `25^(@)C`)

Options

A. at constant pressure `-41.104` kcal
B. at constant volume `= -41.7` kcal
C. at constant pressure = -41.7 kcal
D. at constant volume `= - 42.104` kcal

Answer

C

Explanation

`C_(2)H_(2(g)) + H_(2(g)) rarr C_(2)H_(4(g))`
`Delta H_(r) = - 310.6 + (-68.3) - [-337.2]`
`= - 3789 + 337.2 = - 41.7`
`Delta n_(g) = -1`
`-41.7 = Delta U + (-1) xx (2 xx 298)/(1000) rArr Delta U = - 41.014`