Calculate the enthalpy change for the reaction
`C_(2)H_(4)(g) + H_(2)(g) to C_(2)H_(6)(g)`
using the data given below :
`C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(l) DeltaH = -1415 kJ`
`C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l) DeltaH = -1566 kJ`
`H_(2)(g) + 1/2 O_(2)(g) to H_(2)O(l) DeltaH = -286 kJ`

Options

A. `-437` kJ
B. `+35` kJ
C. `-135` kJ
D. none of these

Answer

C

Explanation

(i) `C_(2)H_(4)(g) + 3O_(2)(g) to 2CO_(2)(g) + 2H_(2)O(l) DeltaH = -1415 kJ`
(ii) `C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l) DeltaH = -1566 kJ`
(iii) `H_(2)(g) + 1/2O_(2)(g) to H_(2)O(l) DeltaH = -286 kJ`
Adding eq.(i) and (iii)
`C_(2)H_(4)(g) + 7/2O_(2)(g) + H_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l) DeltaH = -1701 kJ`
Subtracting eqn (ii) from eqn (iv)
`C_(2)H_(4)(g) + H_(2)(g) to C_(2)H_(6)(g) DeltaH = -135 kJ`