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The enthalpy change for the reaction C(3...

The enthalpy change for the reaction `C_(3)H_(8)(g)+H_(2)(g) rarr C_(2)H_(6)(g)+CH_(4)(g)` at `25^(@)C` is -55.7 kJ/mol. Calculate the enthalpy of combustion of `C_(2)H_(6)(g)`. The enthalpy of combustion of `H_(2)`, & `CH_(4)` are -285.8 & -890.0 kJ/mol respectively. enthalpy of combustion of propane is `-2220" kJ mol"^(-1)`.

Text Solution

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As we know any thermochemical `eq^(n)` can be written in terms of heat of combustion as follows
`Delta H_(r)^(@) = Delta H_(c)^(@) ` (Reactant) - `Delta H_(c)^(@)` (Product)
`DeltaH_(r)^(@) = DeltaH_(c)^(@) (C_(3) H_(8)) + DeltaH_(c)^(@) (H_(2)) - {Delta H_(c)^(@) (C_(2)H_(6)) + Delta H_(c)^(@) (CH_(4))} - 55.7 = (-2220 - 285.8) - {- 890 + Delta H_(c)^(@) (C_(2) H_(6))}`
`implies DeltaH_(c)^(@) (C_(2) H_(6)) (g) = -1560 .1 kJ mol^(-1)`
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