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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 |
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Answer» `-437` 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` |
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