Calculate the standard enthalpy of the reaction,
`2C_(("graphite")) + 3 H_(2(g)) to C_(2) H_(6(g)) Delta H^(@) = ?`
from the following `Delta H^(@)` values:
(a) `C_(2) H_(6(g)) + (7)/(2) O_(2(g)) to 2CO_(2(g)) + 3H_(2) O_((1)) Delta H^(@) = - 1560 kJ`
(b) `H_(2(g)) + (1)/(2) O_(2(g)) to H_(2) O_((1)) Delta H^(@) = - 285.8 kJ`
(c ) `C_(("graphite")) + O_(2(g)) to CO_(2(g)) Delta H^(@) = - 393.5 kJ`

Explanation

We need two moles of graphite as a reactant. So, we multiply equation (a) by to otbain:
`2C ("graphite") + 2O_(2(g)) to 2CO((g))`
`Delta H = - 2(-393.5 kJ//mol) = - 787.0 kj//mol`
We need three moles of `H_(2)` as a reactant. So, we multiply equation (b) by three to obtain:
`3H_(2(g)) + (3)/(2) O_(2(g)) to 3H_(2) O_((l))`
`Delta H = 3 - (285.8 kj//mol) = - 857.4 kj//mol)`
We need one mole of `C_(2)H_(6)` as product. Equation (c ) has two mole of `C_(2) H_(6)` as a reactant, so, we need to reverse the equation and divide it by 2.
`2CO_(2(g)) + 3H_(2)O_((l)) to C_(2)H_(2(g)) + (7)/(2) O_(2(g))`
`Delta H = (1)/(2) (311.6 kJ//mol) = 1559.8 kJ//mol`
Adding equations together, we have
`2C ("graphite") + 2O_(2(g)) to 2CO_(2(g)), Delta H = - 787.0 kj//mol`
`3H_(2(g)) + (3)/(2) O_(2(g)) to 3H_(2) O_((l)), Delta H = - 857.4 kK//mol`,
`2CO_(2(g)) + 3H_(2)O_((l)) to C_(2) H_(6(g)) + (7)/(2) O_(2(g)), Delta H = 1559.8 kJ//mol`
`overline(2C ("graphite") + 3H_(2(g)) to C_(2)H_(6(g)), Delta H^(@) = - 84.6 kJ//mol)`