Nov 9, 2016

The enthalpy change for a reaction is the difference in enthalpies of the products and reactants.

Explanation:

The first thing to do is to write a properly balanced chemical equation:

$2 C \left(s\right) + 2 {H}_{2} \left(g\right) \to {C}_{2} {H}_{4} \left(g\right)$

Next, use a table of standard thermochemical data to find the standard enthalpy of formation of the reactants and products. The standard enthalpy of formation of any element is defined to be zero, so we have

${\Delta}_{f} {H}^{0} \left(C\right) = 0$ kJ/mol
${\Delta}_{f} {H}^{0} \left({H}_{2}\right) = 0$ kJ/mol

and from the tables , we can look up
${\Delta}_{f} {H}^{0} \left({C}_{2} {H}_{4}\right) = 52.3$ kJ/mol

The standard enthalpy for the balanced reaction is the difference in enthalpies of formation of products and reactants (including stoichiometric coefficients):

${\Delta}_{r x n} {H}^{0} = {\Delta}_{f} {H}^{0} \left({C}_{2} {H}_{4}\right) - \left({\Delta}_{f} {H}^{0} \left(C\right) + 2 \left({\Delta}_{f} {H}^{0} \left({H}_{2}\right)\right)\right)$$= 52.3 - 0 - 2 \cdot 0 = 52.3$ kJ/mol

1. If you have a specific amount of material, the change in enthalpy can be calculated by multiplying by the number of moles of carbon (or the number of moles of the limiting reactant).
2. If the temperature of the system is not the same as the standard temperature of the thermodynamic table, then you can use the correction ${\Delta}_{r x n} H = {\Delta}_{r x n} {H}^{0} + \Delta {C}_{p} \Delta T$, where $\Delta {C}_{p}$ is the difference in heat capacities of products and reactants, and $\Delta T$ is the difference between the system temperature and the standard temperature of the thermodynamic table (usually 298.15K).
3. A pressure correction is also possible, but this is rare.