Determine the enthalpy change of the reaction. `C_(3)H_(5) (g) + H_(2) (g) rarr C_(2)H_(6) (g) + CH_(4) (g)` at `25^(@)C`, using the given heat of combustion value under standard conditions: Compound `H_(2) (g) CH_(4) (g) C_(2) H_(6) (g) C` (graphite)
`Delta H^(@) (kJ//mol) -285.8 =890.0 - 1560.0 -393.5` The standard heat of formation of `C_(3)H_(8) (g)` is `-103.8 kJ//mol`

Answer

`-55.7 kJ//mol`

Explanation

From the given data, we can get following equations.
(i) `H_(2) + (1)/(2) O_(2) rarr H_(2) O, Delta H_(1) = - 285.8 kJ`
(ii) `CH_(4) + 2O_(2) rarr CO_(2) + 2H_(2) O, Delta H_(2) = -890 kJ`
(iii) `C_(2)H_(6) + (7)/(2)O_(2) rarr 2CO_(2) + 3H_(2) O, Delta H_(3) = - 1560 kJ`
(iv) `C (s) + O_(2) rarr CO_(2), Delta H_(4) = - 393.5 kJ`
(v) `3C (s) + 4H_(2) rarr C_(3) H_(8) , Delta H_(4) = - 103.8 kJ`
The required equation is
`C_(3)H_(8) (g) + H_(2) (g) rarr C_(2) H_(6) (g) + CH_(4) (g) Delta H = ?`
We can get the desired equation using the manipulations given below
`[3 xx (iv) + 5 xx (i) ] - [(v) + (iii) + (ii)]`
`Delta H = (3 Delta H_(4) + 5 Delta H_(1)) - (Delta H_(5) + Delta H_(3) + Delta H_(2))`
`= [3 xx (-393.5) + 5 xx (-285.8)] - [-103.8 - 1560 - 590]`
`= -55.7 kJ` mole