Question

At \(298 \mathrm{K},\) the standard enthalpies of formation for \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\) and \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\) are \(227 \mathrm{kJ} / \mathrm{mol}\) and \(49 \mathrm{kJ} / \mathrm{mol},\) respectively. a. Calculate \(\Delta H^{\circ}\) for $$\mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow 3 \mathrm{C}_{2} \mathrm{H}_{2}(g)$$ b. Both acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) can be used as fuels. Which compound would liberate more energy per gram when combusted in air?

Short answer

The standard enthalpy change for the reaction \(C_6H_6(l) \longrightarrow 3C_2H_2(g)\) is 632 kJ/mol. Acetylene, with a specific combustion enthalpy of approximately -48.25 kJ/g, releases more energy per gram upon combustion compared to benzene, which releases approximately -41.83 kJ/g. Therefore, acetylene is the better fuel in terms of energy release per gram when combusted in air.

Step-by-step solution

Finding ΔH° for the given reaction

To find the standard enthalpy change for \(C_6H_6(l) \longrightarrow 3C_2H_2(g)\), we will use the formula: \[\Delta H^\circ = \sum^n_{i=1} \Delta H^\circ_{\text{products}} - \sum^n_{i=1} \Delta H^\circ_{\text{reactants}}\] In our case, the standard enthalpy of formation for \(C_6H_6 (l)\) is 49 kJ/mol, and for \(C_2H_2 (g)\) is 227 kJ/mol. Plugging in the values, we get \[\Delta H^\circ = 3\times (227) - 1 \times(49)\]

Calculate ΔH° for the reaction

Now let's calculate the value of ΔH° from the expression above: \[\Delta H^\circ = (681) - (49)\] \[\Delta H^\circ = 632 \mathrm{kJ/mol}\] So, the standard enthalpy change for the reaction \(C_6H_6(l) \longrightarrow 3C_2H_2(g)\) is 632 kJ/mol.

Comparing energy release per gram upon combustion

In this step, we'll first calculate the energy released per gram upon combustion for each compound by finding their specific combustion enthalpies. After that, we'll compare which compound releases more energy per gram. For acetylene: The combustion reaction for acetylene is: \(2 C_2H_2(g) + 5 O_2(g) \longrightarrow 4 CO_2(g) + 2 H_2O(g)\) It has a ΔH° = -1256 kJ/mol (based on the given enthalpy of formation of acetylene). Molar mass of acetylene: \(M_{C_2H_2} = 2 \times (12.01) + 2 \times (1) = 26.04 \mathrm{g/mol}\) Specific combustion enthalpy for acetylene: \(-1256\, \mathrm{kJ/mol} / 26.04\, \mathrm{g/mol} \approx -48.25\, \mathrm{kJ/g}\) For benzene: The combustion reaction for benzene is: \(C_6H_6(l) + \frac{15}{2} O_2(g) \longrightarrow 6 CO_2(g) + 3 H_2O(g)\) It has a ΔH° = -3268 kJ/mol (based on the given enthalpy of formation of benzene). Molar mass of benzene: \(M_{C_6H_6} = 6 \times (12.01) + 6 \times (1) = 78.12 \mathrm{g/mol}\) Specific combustion enthalpy for benzene: \(-3268\, \mathrm{kJ/mol} / 78.12\, \mathrm{g/mol} \approx -41.83\, \mathrm{kJ/g}\)

Conclusion

Comparing both specific combustion enthalpies, we can see that acetylene, with a specific combustion enthalpy of approximately -48.25 kJ/g, releases more energy per gram upon combustion compared to benzene, which releases approximately -41.83 kJ/g. Therefore, acetylene is the better fuel in terms of energy release per gram when combusted in air.

Key concepts

Standard Enthalpy of Formation

Understanding standard enthalpy of formation is essential when delving into enthalpy changes in chemical reactions. This concept refers to the change in enthalpy when one mole of a compound is formed from its elements in their standard states.
For example, the formation of acetylene \(\mathrm{C}_2\mathrm{H}_2(g)\) from carbon and hydrogen.Standard conditions are typically at 298 K and 1 atm pressure, allowing for consistent comparison. The values are generally given in kJ/mol.

• Example: \(\mathrm{C}_2\mathrm{H}_2(g)\) has an enthalpy of formation of 227 kJ/mol.
• Example: Benzene \(\mathrm{C}_6\mathrm{H}_6(l)\) has an enthalpy of formation of 49 kJ/mol.

These values enable us to calculate the energy changes involved in reactions like benzene converting to acetylene.

Combustion Enthalpy

Combustion enthalpy refers to the energy change when a substance completely burns in the presence of oxygen. It is a crucial factor in assessing the energy efficiency of fuels.
Combustion reactions involve reactants turning into products like \(\mathrm{CO}_2(g)\) and \(\mathrm{H}_2\mathrm{O}(g)\).Each substance has a specific combustion reaction and associated enthalpy change.

• Acetylene: \(2 \mathrm{C}_2\mathrm{H}_2(g) + 5 \mathrm{O}_2(g) \rightarrow 4 \mathrm{CO}_2(g) + 2 \mathrm{H}_2\mathrm{O}(g)\)
• Benzene: \(\mathrm{C}_6\mathrm{H}_6(l) + \frac{15}{2} \mathrm{O}_2(g) \rightarrow 6 \mathrm{CO}_2(g) + 3 \mathrm{H}_2\mathrm{O}(g)\)

By comparing these reactions, one can determine which fuel releases more energy.

Energy Release Per Gram

To evaluate a fuel's efficiency, compare the energy released per gram. This involves dividing the combustion enthalpy by the molar mass.For acetylene, the combustion enthalpy is \(-1256\) kJ/mol. The molar mass is 26.04 g/mol, leading to a specific energy release of approximately \(-48.25\) kJ/g.For benzene, it is \(-3268\) kJ/mol with a molar mass of 78.12 g/mol, giving around \(-41.83\) kJ/g.

• Acetylene releases more energy: \(-48.25\) kJ/g.
• Benzene releases less energy: \(-41.83\) kJ/g.

In practical terms, acetylene is a more potent energy source per gram.

Enthalpy Calculation

Calculating enthalpy changes for reactions involves applying the standard enthalpies of formation. The general formula is:\[ \Delta H^\circ = \sum \Delta H^\circ_{\text{products}} - \sum \Delta H^\circ_{\text{reactants}} \]For the reaction \(\mathrm{C}_6\mathrm{H}_6(l) \rightarrow 3 \mathrm{C}_2\mathrm{H}_2(g)\), use the given formation enthalpies.

• Products: 3 moles of \(\mathrm{C}_2\mathrm{H}_2(g)\), each with 227 kJ/mol.
• Reactant: 1 mole of \(\mathrm{C}_6\mathrm{H}_6(l)\) with 49 kJ/mol.

Plugging in these values:\[ \Delta H^\circ = 3 \times 227 - 49 = 632 \, \mathrm{kJ/mol} \]This calculation reveals the energy involved, helping to quantify fuel reactions and understand energy exchanges.