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?

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: \[ 2\,\mathrm{C}_2\mathrm{H}_2(g) + 5\,\mathrm{O}_2(g) \longrightarrow 4\,\mathrm{CO}_2(g) + 2\,\mathrm{H}_2\mathrm{O}(l) \]

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 \(2095.388\,\mathrm{kJ}\).

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 \(2272.559\,\mathrm{kJ}\).