What is the common uses of bases

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>Enthalpy change for the reaction in this case</h3>

In this case you want to calculate the enthalpy change of:

2 C (graphite) + 3 H₂(g) → C₂H₆(g)

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: C₂H₆(g) + $\frac{7}{2}$ O₂(g) → 2 CO₂(g) + 3 H₂O(l) ; ΔH° = –1560 kJ

Equation 2: H₂(g) + $\frac{1}{2}$ O₂(g) → H₂O(l) ; ΔH° = –285.8 kJ

Equation 3: C(graphite) + O₂(g) → CO₂(g) ; ΔH° = –393.5 kJ

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 2 moles of C(graphite) on reactant side and it is present in third equation. In this case it is necessary to multiply it by 2 to obtain the necessary amount. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 2, the variation of enthalpy also.

Now, you need 3 moles of H₂(g) on reactant side and it is present in second equation. In this case it is necessary to multiply it by 3 to obtain the necessary amount and the variation of enthalpy also is multiplied by 3.

Finally, 1 mole of C₂H₆(g) must be a product and is present in the first equation. Since this equation has 1 mole of C₂H₆(g) on the reactant side, it is necessary to locate the C₂H₆(g) on the reactant side (invert it). When an equation is inverted, the sign of delta H also changes.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 CO₂(g) + 3 H₂O(l) → C₂H₆(g) + $\frac{7}{2}$ O₂(g); ΔH° = 1560 kJ

Equation 2: 3 H₂(g) + $\frac{3}{2}$ O₂(g) → 3 H₂O(l) ; ΔH° = –857.4 kJ