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 \(\mathrm{H}_{2} \mathrm{O}(l)\). (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?

Short answer

The balanced chemical equation for the combustion of acetylene gas (C2H2) is \(2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO_{2}(g) + 2H_{2}O(l)\). The heat produced in burning 1 mol of C2H2 under standard conditions is 1257 KJ/mol. The maximum amount of useful work that can be accomplished under standard conditions by this reaction is 1182.48 KJ/mol.

Step-by-step solution

Write the balanced equation for the combustion of acetylene gas

The combustion of acetylene gas (C2H2) in the presence of oxygen (O2) forms carbon dioxide (CO2) and water (H2O). We can write the unbalanced chemical equation as: C2H2(g) + O2(g) → CO2(g) + H2O(l) Now we need to balance the equation: 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l) So, the balanced chemical equation for the combustion of acetylene gas is: 2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)

Calculate the heat produced in burning 1 mol of C2H2

To calculate the heat produced in the reaction, we can use the standard enthalpy of combustion values, which do not depend on temperature. The standard enthalpy of combustion (ΔH°) is the enthalpy change when 1 mole of a substance reacts completely with oxygen under standard conditions. From the standard enthalpy of combustion values table, we have: ΔH°(C2H2, g) = -1300 KJ/mol ΔH°(O2, g) = 0 KJ/mol (reason: oxygen is in standard conditions) ΔH°(CO2, g) = -393.5 KJ/mol ΔH°(H2O, l) = -286 KJ/mol Using the balanced equation, we can calculate the heat produced in burning 1 mol of C2H2: ΔH°(reaction) = [(4 * ΔH°(CO2) + 2 * ΔH°(H2O)) - (2 * ΔH°(C2H2) + 5 * ΔH°(O2))] ΔH°(reaction) = [(4 * -393.5) + (2 * -286) - (2 * -1300) - (5 * 0)] ΔH°(reaction) = -1257 KJ/mol Therefore, 1257 KJ of heat is produced in burning 1 mol of C2H2 under standard conditions.

Calculate the maximum amount of useful work accomplished

To calculate the maximum amount of useful work that can be accomplished under standard conditions by this reaction, we have to determine the Gibbs free energy change (ΔG°). The formula to calculate ΔG° is: ΔG° = ΔH° - TΔS° To find the value of ΔS°, we need to use standard entropy values (S°) from the standard entropy values table. We have: S°(C2H2, g) = 200.94 J/mol K S°(O2, g) = 205.2 J/mol K S°(CO2, g) = 213.6 J/mol K S°(H2O, l) = 69.9 J/mol K Now, we can calculate the entropy change (ΔS°) for the reaction: ΔS°(reaction) = [(4 * S°(CO2) + 2 * S°(H2O)) - (2 * S°(C2H2) + 5 * S°(O2))] ΔS°(reaction) = [(4 * 213.6) + (2 * 69.9) - (2 * 200.94) - (5 * 205.2)] ΔS°(reaction) = -250.08 J/mol K Now, we have both ΔH°(reaction) and ΔS°(reaction). We can calculate ΔG° using the formula ΔG° = ΔH° - TΔS° at T = 298 K: ΔG° = (- 1257) - (298 * -250.08 / 1000) ΔG° = - 1257 + 74.52 ΔG° = -1182.48 KJ/mol The maximum amount of useful work that can be accomplished under standard conditions by this reaction is equal to ΔG°, which is 1182.48 KJ/mol.

Key concepts

Balanced Chemical Equation

Understanding the balanced chemical equation is fundamental to grasping chemical reactions, such as the combustion of acetylene. A balanced equation has equal numbers of each type of atom on both sides of the equation, conserving mass and adhering to the law of conservation of matter. The reaction for the combustion of acetylene ($C\_2H\_2$) involves acetylene reacting with oxygen ($O\_2$) to produce carbon dioxide ($CO\_2$) and water ($H\_2O$).

The equation initially starts unbalanced and requires adjusting the coefficients to ensure the number of atoms of each element are equal on both reactant and product sides. In the case of acetylene combustion:
2 C2H2(g) + 5 O2(g) → 4 CO2(g) + 2 H2O(l)

This indicates that two molecules of acetylene react with five molecules of diatomic oxygen to produce four molecules of carbon dioxide and two molecules of liquid water. By balancing the chemical equation, we also set the stage for further analysis, such as calculating the heat released during the combustion or determining the Gibbs free energy change associated with the reaction.

Enthalpy of Combustion

The enthalpy of combustion (ΔH^°comb) is the energy released as heat when a substance completely reacts with oxygen under standard conditions. It is a crucial concept in thermodynamics and serves as a measure of how much energy is released in the form of heat during a combustion reaction. For acetylene ($C\_2H\_2$), the reaction's enthalpy change can be calculated using the standard enthalpies of formation for the reactants and products.

By utilizing the balanced chemical equation, we can determine the heat released when burning 1 mol of acetylene gas, which is represented by the formula:
ΔH°(reaction) = [(4 * ΔH°($CO\_2$) + 2 * ΔH°($H\_2O$)) - (2 * ΔH°($C\_2H\_2$) + 5 * ΔH°($O\_2$))]

With the given standard enthalpies of combustion:$\Delta H°($ C\_2H\_2$,\; g)\; =\; -1300\; KJ/mol$,$\Delta H°($ CO\_2$,\; g)\; =\; -393.5\; KJ/mol$, and $\Delta H°($ H\_2O$,\; l)\; =\; -286\; KJ/mol$, the total heat produced during the combustion of 1 mol of acetylene gas at standard conditions is -1257 KJ/mol. This quantity is significant as it indicates the potential energy that could be harnessed from acetylene in applications such as welding.

Gibbs Free Energy

Gibbs free energy (ΔG^°) is a thermodynamic property that can be used to predict the direction of a chemical reaction and quantify the maximum amount of non-expansion work that can be performed by the reaction at constant pressure and temperature. It combines the system's enthalpy and entropy to provide a useful measure of the 'usefulness' or 'work potential' of a reaction.

The formula for calculating the Gibbs free energy change is:
ΔG° = ΔH° - TΔS°

where ΔH° is the change in enthalpy, T is the temperature in Kelvin, and ΔS° is the change in entropy associated with the reaction. During the combustion of acetylene, the entropy change is calculated considering the standard molar entropies of the reactants and products, while adjusting for the coefficients from the balanced equation.
Given that at standard conditions (298 K), the ΔH°(reaction) is -1257 KJ/mol, and the calculated ΔS°(reaction) is -250.08 J/mol·K, the Gibbs free energy change for the reaction can be determined:ΔG° = -1257 KJ/mol - (298 K * -250.08 J/mol·K / 1000)
ΔG° = -1182.48 KJ/mol

This negative value for ΔG° indicates that the combustion of acetylene is a spontaneous process under standard conditions and would be capable of performing 1182.48 KJ/mol of useful work. This concept is highly relevant in not only theoretical chemistry but also in practical applications where energy efficiency is of concern.