# What is the standard enthalpy of formation of methane, given that the average C-H bond enthalpy is 414 kj/mol and the reactions: C(s) -->C(g) H=716 kj/mol 2H2(g)--> 4H(g) H= 872.8 kj/mol?

Jan 7, 2016

$\Delta {H}_{\text{f"^@ = -"67.2 kJ/mol}}$

#### Explanation:

The standard enthalpy of formation, $\Delta {H}_{\text{f}}^{\circ}$, for a given compound is defined as the enthalpy change of reaction when one mole of said compound is formed from its constituent elements in their most stable form.

In your case, the standard enthalpy of formation for methane, ${\text{CH}}_{4}$, is calculated for the reaction

${\text{C"_text((s]) + 2"H"_text(2(g]) -> "CH}}_{\textrm{4 \left(g\right]}}$

Your goal here will be to find a way to get the enthalpy change of reaction for the above reaction by using Hess' Law.

So, you know that the average $\text{C"-"H}$ bond enthalpy is equal to $\text{414 kJ/mol}$. Notice that this value carries a positive sign.

As you know, bond breaking is an endothermic process because it requires energy. This means that the value given to you corresponds to how much energy is needed, on average, in order to break a $\text{C"-"H}$ bond.

However, you're interested forming methane, so you can say that the enthalpy change of reaction for

$\text{C"_text((g]) + 4"H"_text((g]) -> "CH"_text(4(g])" " " } \textcolor{p u r p \le}{\left(1\right)}$

will carry a negative sign, since this time you're making bonds, which is an exothermic process.

Since you're making four $\text{C" - "H}$ bonds, you will have

$\Delta {H}_{\text{rxn 1"^@ = 4 xx (-"414 kJ/mol") = -"1656 kJ/mol}}$

The other two reactions given to you are

$\text{C"_text((s]) -> "C"_text((g]), " "DeltaH_text(rxn 2)^@ = "716 kJ/mol"" " " } \textcolor{p u r p \le}{\left(2\right)}$

and

$2 \text{H"_text(2(g]) -> 4"H"_text((g])," " DeltaH_text(rxn 3)^@ = "872.8 kJ/mol" " " " } \textcolor{p u r p \le}{\left(3\right)}$

Notice that equation $\textcolor{p u r p \le}{\left(3\right)}$ represents the breaking of two $\text{H"-"H}$ bonds, which is why it carries a positive sign.

Since Hess' Law tells you that the overall enthalpy change for a reaction is independent of the pathway or the number of steps taken, you can add equations $\textcolor{p u r p \le}{\left(1\right)}$, $\textcolor{p u r p \le}{\left(2\right)}$, and $\textcolor{p u r p \le}{\left(3\right)}$ to get

$\textcolor{w h i t e}{\times \times \times x} \text{C"_text((s]) -> color(red)(cancel(color(black)("C"_text((g])))), " " " "DeltaH_text(rxn 2)^@ = "716 kJ/mol}$
color(red)(cancel(color(black)("C"_text((g])))) + color(blue)(cancel(color(black)(4"H"_text((g])))) -> "CH"_text(4(g]), " "color(white)(x)DeltaH_text(rxn 1)^@ = -"1656 kJ/mol"
$\textcolor{w h i t e}{\times \times x} 2 \text{H"_text(2(g]) -> color(blue)(cancel(color(black)(4"H"_text((g]))))," " " "DeltaH_text(rxn 3)^@ = "872.8 kJ/mol}$
$\frac{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times \times}}{\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times \times \times \times x}}$

$\text{C"_text((s]) + 2"H"_text(2(g]) -> "CH"_text(4(g])," " " } \Delta {H}_{\textrm{r x n}}^{\circ} = \Delta {H}_{\textrm{f}}^{\circ}$

You will thus have

$\Delta {H}_{\text{f"^@ = DeltaH_"rxn 1"^@ + DeltaH_"rxn 2"^@ + DeltaH_"rxn 3}}^{\circ}$

$\Delta {H}_{\text{f"^@ = "716 kJ/mol" + (-"1656 kJ/mol") + "872.8 kJ/mol}}$

DeltaH_"f"^@ = color(green)(-"67.2 kJ/mol")