Question

Acetylene burns in air according to the following equation: $$\begin{array}{r}\mathrm{C}_{2} \mathrm{H}_{2}(g)+{ }_{2}^{5} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \Delta H_{\mathrm{ru}}^{\circ}=-1255.8 \mathrm{~kJ}\end{array}$$Given \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{CO}_{2}(g)=-393.5 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{\mathrm{f}}^{\circ}\) of \(\mathrm{H}_{2} \mathrm{O}(g)=$$-241.8 \mathrm{~kJ} / \mathrm{mol},\) find \(\Delta H_{i}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\)

Short answer

The standard enthalpy of formation of \( \mathrm{C}_{2}\mathrm{H}_{2} \) is 227 kJ/mol.

Step-by-step solution

Write down the given reaction

Write down the provided combustion reaction of acetylene (C\textsubscript{2}H\textsubscript{2}): \( \mathrm{C}_{2}\mathrm{H}_{2} (g) + \frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(g) \) The enthalpy change of reaction, \( \Delta H_{\mathrm{rxn}}^{\circ} = -1255.8 \mathrm{~kJ} \)

Identify the given standard enthalpies of formation

Given data:\( \Delta H_{\mathrm{f}}^{\circ} \) of \( \mathrm{CO}_{2}(g) = -393.5 \mathrm{~kJ} / \mathrm{mol}\)\( \Delta H_{\mathrm{f}}^{\circ} \) of \( \mathrm{H}_{2}\mathrm{O}(g) = -241.8 \mathrm{~kJ} / \mathrm{mol} \)

Write the formula for enthalpy change of the reaction

The enthalpy change of the reaction \( \Delta H_{\mathrm{rxn}}^{\circ} \) can be found using the formula:\( \Delta H_{\mathrm{rxn}}^{\circ} = \sum \Delta H_{\mathrm{f (products)}}^{\circ} - \sum \Delta H_{\mathrm{f (reactants)}}^{\circ} \)

Substitute the values of enthalpies of formation into the formula

Substitute the known values into the formula:\( -1255.8 = [2 \times (-393.5) + (-241.8)] - [ \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2} ) + \frac{5}{2} \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{O}_{2})] \) Note: \( \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{O}_{2}) = 0 \mathrm{~kJ} / \mathrm{mol} \)

Simplify and solve for \( \Delta H_{\mathrm{f}}^{\circ} (\mathrm{C}_{2}\mathrm{H}_{2}) \)

Simplify the equation:\( -1255.8 = [2 \times (-393.5) + (-241.8)] - \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2} ) \)\( -1255.8 = [-787 - 241.8] - \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2} ) \)\( -1255.8 = -1028.8 - \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2} ) \)Rearrange to solve for \( \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2}) \):\( \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2} ) = -1028.8 + 1255.8 \)\( \Delta H_{\mathrm{f}}^{\circ} ( \mathrm{C}_{2}\mathrm{H}_{2} ) = 227 \mathrm{~kJ} / \mathrm{mol} \)

Key concepts

enthalpy of formation

Enthalpy of formation, often symbolized as \(\therm{H_{\text{f}}^{\text{o}}}\), refers to the change in enthalpy when one mole of a compound is formed from its constituent elements. This process always involves substances in their standard states.

Standard state means the physical state of a substance under standard conditions, which are typically 25 degrees Celsius and 1 atmosphere of pressure. For example:

• The enthalpy of formation for \(\therm{CO_2} (g)\) is \(\therm{-393.5 \text{ kJ/mol}}\).
• For \(\therm{H_2O} (g)\), it's \(\therm{-241.8 \text{ kJ/mol}}\).

The enthalpy of formation values are particularly useful for calculating the enthalpy changes of various reactions using Hess's Law.

combustion reactions

Combustion reactions are a type of chemical reaction where a substance mixes with oxygen to release heat and produce oxides, generally causing what we know as fire or burning. The general equation for a combustion reaction is:

\(\therm{Fuel + O_2 \rightarrow Oxides + Energy}\)
Here, the fuel such as acetylene (\(\therm{C_2H_2}\)) reacts with oxygen to produce carbon dioxide (\(\therm{CO_2}\)) and water (\(\therm{H_2O}\)): \(\therm{C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(g)}\)
This process is exothermic, implying it releases heat. The amount of heat released during combustion can be significant and is usually denoted by \(\therm{\bigtriangleup H_{\text{rxn}}}\). For the combustion of acetylene in the given exercise, the heat released is \(\therm{-1255.8 \text{ kJ}}\).

Hess's Law

Hess's Law states that the total enthalpy change of a chemical reaction is the same, no matter how many steps it takes to complete the reaction. The law is based on the fact that enthalpy is a state function.

According to Hess's Law, one can determine the enthalpy change of a reaction by summing the enthalpy changes of individual steps that lead to the final reaction. This is particularly useful when the direct measurement of \(\therm{\bigtriangleup H_{\text{rxn}}}\) is difficult. The formula for applying Hess's Law is:

\(\therm{\bigtriangleup H_{\text{rxn}}^{\text{o}} = \bigtriangleup H_{\text{f (products)}}^{\text{o}} - \bigtriangleup H_{\text{f (reactants)}}^{\text{o}} }\)

For the given combustion reaction of acetylene, Hess's Law allows us to find the enthalpy of formation (\