Problem 73 - Chemistry: The Central Science in SI Units, Global Edition

Question

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathbf{b})\) How much heat is produced in burning \(1 \mathrm{~mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to \(298 \mathrm{~K} ?\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

Step-by-Step Solution

Answer

(a) The balanced equation for the combustion of acetylene gas is: \[ 2\,\mathrm{C}_2\mathrm{H}_2(g) + 5\,\mathrm{O}_2(g) \longrightarrow 4\,\mathrm{CO}_2(g) + 2\,\mathrm{H}_2\mathrm{O}(l) \] (b) The heat produced when one mole of acetylene is burned under standard conditions is \(2095.388\,\mathrm{kJ}\). (c) The maximum useful work that can be accomplished under standard conditions for this reaction is \(2272.559\,\mathrm{kJ}\).

Step 1: (a) Write a balanced equation for the combustion of acetylene gas

The combustion of acetylene involves the reaction between acetylene, \(\mathrm{C}_2\mathrm{H}_2(g)\), and oxygen, \(\mathrm{O}_2(g)\), to produce carbon dioxide \(\mathrm{CO}_2(g)\) and water \(\mathrm{H}_2\mathrm{O}(l)\). The balanced equation for this combustion is: \[ 2\,\mathrm{C}_2\mathrm{H}_2(g) + 5\,\mathrm{O}_2(g) \longrightarrow 4\,\mathrm{CO}_2(g) + 2\,\mathrm{H}_2\mathrm{O}(l) \]

Step 2: (b) Calculate heat produced in the combustion of 1 mol of acetylene

To calculate the heat produced in burning one mole of acetylene, we need to use the heat of formation (\(\Delta H_f\)) values for the reactants and products. We will use the following standard heat of formation values at \(298\,\mathrm{K}\): \[ \Delta H_f^{\circ}[\mathrm{C}_2\mathrm{H}_2(g)] = 226.73\,\mathrm{kJ}/\mathrm{mol} \] \[ \Delta H_f^{\circ}[\mathrm{O}_2(g)] = 0\,\mathrm{kJ}/\mathrm{mol} \] \[ \Delta H_f^{\circ}[\mathrm{CO}_2(g)] = -393.5\,\mathrm{kJ}/\mathrm{mol} \] \[ \Delta H_f^{\circ}[\mathrm{H}_2\mathrm{O}(l)] = -285.829\,\mathrm{kJ}/\mathrm{mol} \] To calculate the heat produced for the reaction, \(\Delta H_{rxn}\), we use the equation: \( \Delta H_{rxn} = \sum{n\Delta H_f^{\circ}}(\text{products}) - \sum{n\Delta H_f^{\circ}}(\text{reactants}) \) where \(n \Delta H_f^{\circ}\) is the enthalpy of formation of each compound, multiplied by its stoichiometric coefficient. (1) Calculate heats of formation for reactants: \( \,\mathrm{C}_2\mathrm{H}_2 = 226.73\,\mathrm{kJ}/\mathrm{mol} \) (2) Calculate heats of formation for products: \[ 4\,\mathrm{CO}_2 = 4 \times -393.5\,\mathrm{kJ}/\mathrm{mol} = -1574\,\mathrm{kJ}/\mathrm{mol} \] \[ 2\,\mathrm{H}_2\mathrm{O} = 2 \times -285.829\,\mathrm{kJ}/\mathrm{mol} = -571.658\,\mathrm{kJ}/\mathrm{mol} \] (3) Calculate the heat of reaction, \(\Delta H_{rxn}\): \[ \Delta H_{rxn} = (-1574 + (-571.658)) - 226.73 \] \[ \Delta H_{rxn} = -2095.388\,\mathrm{kJ}/\mathrm{mol} \] Therefore, the heat produced when one mole of acetylene is burned under standard conditions is \(2095.388\,\mathrm{kJ}\).

Step 3: (c) Calculate the maximum useful work

To calculate the maximum useful work that can be accomplished under standard conditions, we need to use the Gibbs free energy change (\(\Delta G\)). This can be calculated from the enthalpy change (\(\Delta H\)) and the change in entropy (\(\Delta S\)) using this equation: \[ \Delta G = \Delta H - T\Delta S \] We first need to find the values of the standard molar entropy (\(S^\circ\)) for all the compounds involved in the reaction: \[ S^\circ[\mathrm{C}_2\mathrm{H}_2(g)] = 200.8\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] \[ S^\circ[\mathrm{O}_2(g)] = 205.03\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] \[ S^\circ[\mathrm{CO}_2(g)] = 213.79\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] \[ S^\circ[\mathrm{H}_2\mathrm{O}(l)] = 69.95\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] Now, we will calculate the change in entropy for the reaction, \(\Delta S\), as the difference between the entropy of products and reactants: \[ \Delta S = \sum{nS^\circ}(\text{products}) - \sum{nS^\circ}(\text{reactants}) \] (1) Calculate the entropy of reactants: \[ 2\,\mathrm{C}_2\mathrm{H}_2 = 2 \times 200.8\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} = 401.6\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] (2) Calculate the entropy of products: \[ 4\,\mathrm{CO}_2 = 4 \times 213.79\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} = 855.16\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] \[ 2\,\mathrm{H}_2\mathrm{O} = 2 \times 69.95\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} = 139.9\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] (3) Calculate the change in entropy, \(\Delta S\): \[ \Delta S = (855.16 + 139.9) - 401.6 \] \[ \Delta S = 593.46\,\mathrm{J}/\mathrm{mol}\cdot\mathrm{K} \] Now, we can use the values of \(\Delta H_{rxn}\) and \(\Delta S\) to find the value of \(\Delta G\): \[ \Delta G = \Delta H_{rxn} - (T \times \Delta S) \] At standard conditions, \(T = 298\,\mathrm{K}\). Therefore, \[ \Delta G = -2095.388\,\mathrm{kJ}/\mathrm{mol} - (298\,\mathrm{K} \times 0.59346\,\mathrm{kJ}/\mathrm{mol}\cdot\mathrm{K}) \] \[ \Delta G = -2095.388 - 177.171\,\mathrm{kJ}/\mathrm{mol} \] \[ \Delta G = -2272.559\,\mathrm{kJ}/\mathrm{mol} \] The maximum useful work that can be accomplished under standard conditions for this reaction is \(2272.559\,\mathrm{kJ}\).

Key Concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a critical factor when studying combustion reactions like that of acetylene gas. This energy term relates to the heat absorbed or released during a chemical reaction. For exothermic reactions, where heat is released, \( \Delta H \) is negative. The combustion equation for acetylene, given earlier as:\[ 2\,\mathrm{C}_2\mathrm{H}_2(g) + 5\,\mathrm{O}_2(g) \longrightarrow 4\,\mathrm{CO}_2(g) + 2\,\mathrm{H}_2\mathrm{O}(l) \]is an exothermic process.

• The heat generated in this reaction is due to the formation of carbon dioxide and water from acetylene.
• Calculations for \( \Delta H \) use standard enthalpy of formation values for products and reactants.

For instance, the enthalpy change for burning 1 mole of acetylene is calculated using:\[ \Delta H_{rxn} = \sum{n\Delta H_f^{\circ}}(\text{products}) - \sum{n\Delta H_f^{\circ}}(\text{reactants}) \]Applying values:

• The enthalpy of formation for CO₂ and H₂O contributes significantly to the large negative \( \Delta H \), affirming the exothermic nature of the reaction.

Gibbs Free Energy

Gibbs free energy, represented as \( \Delta G \), determines the spontaneity of a reaction. It combines enthalpy and entropy changes alongside temperature to predict whether a reaction will proceed. For the acetylene combustion process, the Gibbs free energy can be expressed as:\[ \Delta G = \Delta H - T\Delta S \]where \( T \) is the temperature in Kelvin, \( 298 \). It measures the maximum reversible work that a system can perform.

• If \( \Delta G \) is negative, the reaction can occur spontaneously under standard conditions.
• The calculated \( \Delta G \) for acetylene's combustion indicates a highly spontaneous reaction.

This negative value means the reaction releases free energy, which can be harnessed in the welding process or other uses where acetylene is utilized. Understanding \( \Delta G \) helps engineers and scientists assess energy efficiency and potential work output from chemical processes.

Entropy Change

Entropy change, signified as \( \Delta S \), reflects the degree of disorder or randomness in a system's states. During the combustion of acetylene, the difference in entropy between products and reactants signifies changes in molecular randomness.

• The reaction involves gaseous reactants forming both gases and liquids, influencing the \( \Delta S \) value.
• The computed \( \Delta S \) helps predict the universe's total entropy change, focusing on system orderliness.

Calculating \( \Delta S \) involves using standard molar entropies:\[ \Delta S = \sum{nS^\circ}(\text{products}) - \sum{nS^\circ}(\text{reactants}) \]For this reaction:

• The calculated \( \Delta S \) is positive, indicating an increase in disorder post-reaction.

This growth in entropy, paired with a decrease in enthalpy, contributes to making the Gibbs free energy change \( \Delta G \) negative, thus characterizing the reaction as spontaneous. Understanding the relationship between \( \Delta S \) and other thermodynamic quantities is vital in industrial applications where controlling reaction spontaneity and energy use is crucial.