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)

Given below are some standard heats of reaction (at constant pressure)

1.	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)
Answer» at CONSTANT PRESSURE `-41.104` kcal at constant volume `= -41.7` kcal at constant pressure = -41.7 kcal at constant volume `= - 42.104` kcal Solution :`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`