Acetylene gas (ethyne; HC≡CH) burns with oxygen in an oxyacetylene torch to produce carbon dioxide, water vapor, and the heat needed to weld metals. The heat of combustion for acetylene is −1259 kJ/mol. Calculate the C≡C bond energy.

Assumption: [tex]-1259\; \rm kJ \cdot mol^{-1}[/tex] measures the enthalpy change of the reaction [tex]\displaystyle \rm C_2H_2\, (g) + \frac{5}{2}\, O_2\, (g) \to 2\, CO_2\, (g) + H_2O\, (\textbf{g})[/tex] where the water is in its gaseous state (as "water vapor.")

Explanation:

The heat of combustion of a substance gives the enthalpy change when one mole of that substance reacts with excess oxygen.

Start by balancing the equation for the complete combustion of ethyne [tex]\rm \text{H-C $\equiv$ C-H}[/tex] in oxygen [tex]\rm O_2\, (g)[/tex].

[tex]\displaystyle \rm 2 \, C_2H_2\, (g) + 5\, O_2\, (g) \to 4\, CO_2\, (g) + 2\, H_2O\, (g)[/tex].

Set coefficient of [tex]\rm \text{H-C $\equiv$ C-H}[/tex] should be equal to one. Simply divide all coefficients by the coefficient of [tex]\rm \text{H-C $\equiv$ C-H}[/tex].

[tex]\displaystyle \rm C_2H_2\, (g) + \frac{5}{2}\, O_2\, (g) \to 2\, CO_2\, (g) + H_2O\, (\textbf{g})[/tex].

The enthalpy change of a reaction like this one is equal to

The energy of bonds broken, minus
The energy of bonds formed.

If the energy of bonds formed is greater than that of the bonds broken, the reaction would be exothermic and [tex]\Delta H[/tex] should be negative.

In each mole of this reaction, bonds broken include:

[tex]2 \times \rm C - H[/tex] bonds,
[tex]1 \times \rm C\equiv C[/tex] bond, and
[tex]\dfrac{5}{2}\times \rm O=O[/tex] bonds.

In each mole of this reaction, bonds formed include:

[tex]2 \times 2\times \rm C=O[/tex] bonds (each [tex]\rm CO_2[/tex] molecule contains two such bonds,) and
[tex]2 \times \rm H-O[/tex] bonds.

Hence

[tex]\begin{aligned}& \Delta H \cr =&E(\text{Bonds Broken}) - E(\text{Bonds Formed}) \cr =& 2\times E(\rm C-H) + E(\rm C\equiv C) + \dfrac{5}{2} \times E(\rm O=O)\cr & \phantom{=} - 2 \times E(\rm C=O) - E(\rm H-O) \end{aligned}[/tex].

Rewrite the equation to isolate the unknown [tex]E(\rm C\equiv C)[/tex]:

[tex]\begin{aligned}& E(\rm C\equiv C) \cr = &\Delta H - 2\times E(\rm C-H) - \dfrac{5}{2} \times E(\rm O=O) \cr &\phantom{=} + 2 \times E(\rm C=O) + E(\rm H-O) \end{aligned}[/tex].

Look up the bond energy for these bonds (except for the unknown carbon-carbon triple bond)

[tex]\begin{aligned}& E(\rm C\equiv C) \cr = &\Delta H - 2\times E(\rm C-H) - \dfrac{5}{2} \times E(\rm O=O) \cr &\phantom{=} + 2 \times E(\rm C=O) + E(\rm H-O) \cr =& (-1259) + (2 \times 2 \times (804) + 2 \times (463)) \cr &\phantom{=}- (2 \times 414 + 5/2 \times 498)\cr =& 810\, \rm kJ \cdot mol^{-1}\end{aligned}[/tex].