Calculate the enthalpy of the reacation .
`C_(2)H_(2)(g) + H_(2)(g) to C_(2)H_(6)(g)`.
Given enthalpy of combustion of ethaene , ethane, and hydrogen are - 140 kJ , -1550 kJ and - 286 kJ, respectively .

Explanation

`C_(2)H_(2) + 3O_(2) to 2O_(2) (g) + 2H_(2) O(g) " " Delta H = -1401 kJ " " (1)`
`C_(2)H_(6)(g) + ((7)/(2)) O_(2) (g) to 2O_(2)(g) + 3H_(2)O(l) " " DeltaH = -1550 k J " "(2)`
`H_(2)(g) + ((1)/(2)) O_(2)(g) to =H_(2)O(l) Delta H = - 286 kJ " " (3)` Reversing eq.(2), we get
`2CO_(2) (g) + 3H_(2)O(l) to C_(2)H_(6)(g) + (7)/(2)O_(2)(g) " " DeltaH = 1550 kJ" "4`
(1) + (2) + (4) , we get
`C_(2)H_(2)(g) + H_(2)(g) to C_(2)H_(5)(g) " " Delta H = - 137 KJ`
`therefore` The enthlpy of the reaction is ` - 137 kJ`