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 C6H6(l) → 3C2H2(g) is ΔH° = 632 kJ/mol. Comparing the energy liberated per gram when combusted in air, acetylene (C2H2) would liberate more energy with -8.73 kJ/g, compared to benzene (C6H6) which only liberates -0.63 kJ/g.

Step-by-step solution

Write down the formula for calculating the reaction enthalpy

The formula used to calculate the reaction enthalpy, ΔH°, is as follows: \[ ΔH°_{reaction} = Σ nΔH°_{products} - Σ nΔH°_{reactants} \] where \(Σ\) represents the sum, \(n\) is the stoichiometric coefficient of each species, and \(ΔH°\) is the standard enthalpy of formation of each substance involved in the reaction.

Identify the enthalpies of formation for each species in the reaction

We are given the standard enthalpy of formation for C2H2(g) and C6H6(l) at 298 K: - ΔH° (C2H2) = 227 kJ/mol - ΔH° (C6H6) = 49 kJ/mol

Calculate the reaction enthalpy using the formula

Now we can use the formula to calculate the reaction enthalpy ΔH° for the given reaction. \[ ΔH°_{reaction} = 3 × ΔH°_{C_2H_2} - ΔH°_{C_6H_6} \] \[ ΔH°_{reaction} = 3 × 227 - 49 = 681 - 49 = 632\text{ kJ/mol} \] The standard enthalpy change for the reaction C6H6(l) → 3C2H2(g) is ΔH° = 632 kJ/mol. **Part b: Compare the energy liberated per gram when combusted in air**

Write down the balanced combustion reactions for both compounds

For C2H2(g) and C6H6(l), the combustion reactions are: - C2H2(g) + 2.5O2(g) → 2CO2(g) + H2O(g) - C6H6(l) + 7.5O2(g) → 6CO2(g) + 3H2O(g)

Calculate the energy per gram for both combustion reactions

To calculate the energy per gram for each combustion reaction, we need to compute the energy per mole (ΔH°) divided by the molar mass of the compound. The molar mass of C2H2 is approximately 26 g/mol, whereas the molar mass of C6H6 is approximately 78 g/mol. Energy per gram for C2H2 combustion: \[ \text{Energy per gram} = \frac{ΔH°}{\text{molar mass}} = \frac{-227\text{ kJ/mol}}{26\text{ g/mol}} = -8.73\text{ kJ/g} \] Energy per gram for C6H6 combustion: \[ \text{Energy per gram} = \frac{ΔH°}{\text{molar mass}} = \frac{-49\text{ kJ/mol}}{78\text{ g/mol}} = -0.63\text{ kJ/g} \] Based on the calculations, acetylene (C2H2) would liberate more energy per gram (-8.73 kJ/g) when combusted in air, compared to benzene (C6H6), which only liberates -0.63 kJ/g.

Key concepts

Reaction Enthalpy

Every chemical reaction either absorbs or releases energy, commonly measured as heat. The quantity we use to understand this heat exchange is called reaction enthalpy. It tells us how much heat would be transferred if a chemical reaction proceeded under standard conditions. These conditions include a pressure of 1 atm and a temperature of 298 K.
To calculate the reaction enthalpy (\( \Delta H^\circ \)), we use the formula:

• \[\Delta H^\circ_{reaction} = \Sigma n \Delta H^\circ_{products} - \Sigma n \Delta H^\circ_{reactants}\]

Here, \( \Delta H^\circ \) represents the standard enthalpy of formation of each species, and \( n \) is its stoichiometric coefficient. By applying this formula, we assess the heat changes of the given chemical reaction.

Combustion Reaction

A combustion reaction is a chemical process where a compound reacts rapidly with oxygen, releasing energy in the form of heat and light. It’s what occurs when fuels like gas, coal, or wood burn. For our case study, we're looking at the combustion of acetylene (\( \text{C}_2\text{H}_2 \)) and benzene (\( \text{C}_6\text{H}_6 \)).
The balanced equations for these reactions are:

• For acetylene:\( \text{C}_2\text{H}_2(g) + 2.5\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(g) \)
• For benzene:\( \text{C}_6\text{H}_6(l) + 7.5\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(g) \)

These reactions are exothermic, meaning they release energy as they proceed. It’s the reason why combustion is a popular means of energy generation worldwide.

Energy Per Gram

When comparing the energy efficiency of fuels, it’s helpful to look at the energy released per gram of fuel. This measure tells us how much energy is obtained from a specific mass of the substance upon combustion.
To find the energy per gram, you divide the energy change per mole by the molar mass of the compound:

• For acetylene:\[\text{Energy per gram} = \frac{-227 \text{ kJ/mol}}{26 \text{ g/mol}} = -8.73 \text{ kJ/g}\]
• For benzene:\[\text{Energy per gram} = \frac{-49 \text{ kJ/mol}}{78 \text{ g/mol}} = -0.63 \text{ kJ/g}\]

Acetylene offers more energy per gram compared to benzene, making it a more potent fuel when considering mass efficiency.

Standard Enthalpy Change

The standard enthalpy change (\( \Delta H^\circ \)) reflects the heat of a reaction under standard conditions. It is the difference in the total enthalpy between the products and the reactants. The "standard" term refers to a set of conditions that allow us to consistently measure and compare enthalpy changes between different reactions.
In our exercise, calculating the standard enthalpy change for the reaction: \( \text{C}_6\text{H}_6(l) \rightarrow 3\text{C}_2\text{H}_2(g) \), involves the transformation of benzene to acetylene. The outcome tells us the amount of heat absorbed by this endothermic reaction, calculated as 632 kJ/mol.
Understanding standard enthalpy changes is vital for making informed choices about energy usage, especially in industries where efficiency and cost are crucial.