Question

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.

Answer 1

Approximately
`\rm 810\; kJ \cdot mol^(-1)`.

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

Step-by-step 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
`\rm \text{H-C $\equiv$ C-H}` in oxygen
`\rm O_2\, (g)`.

`\displaystyle \rm 2 \, C_2H_2\, (g) + 5\, O_2\, (g) \to 4\, CO_2\, (g) + 2\, H_2O\, (g)`.

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

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

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
`\Delta H` should be negative.

In each mole of this reaction, bonds broken include:

• 
  `2 * \rm C - H` bonds,
• 
  `1 * \rm C\equiv C` bond, and
• 
  `(5)/(2)* \rm O=O` bonds.

In each mole of this reaction, bonds formed include:

• 
  `2 * 2* \rm C=O` bonds (each
  `\rm CO_2` molecule contains two such bonds,) and
• 
  `2 * \rm H-O` bonds.

Hence

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

Rewrite the equation to isolate the unknown
`E(\rm C\equiv C)`:

`\begin{aligned}& E(\rm C\equiv C) \cr = &\Delta H - 2* E(\rm C-H) - (5)/(2) * E(\rm O=O) \cr &\phantom{=} + 2 * E(\rm C=O) + E(\rm H-O) \end{aligned}`.

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

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