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 (\(\mathrm{C}_{2} \mathrm{H}_{2}(g)\)) is: \[ \mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \] The heat produced when burning 1 mol of acetylene gas under standard conditions is -1290.1 kJ/mol. The maximum amount of useful work that can be accomplished under standard conditions by the combustion of acetylene gas is -1235.0 kJ/mol.

Step-by-step solution

Write the balanced chemical equation for the combustion of acetylene gas

We need to create a balanced chemical equation for the combustion of acetylene gas. Acetylene gas is given as \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\). Upon combustion, it produces carbon dioxide (\(\mathrm{CO}_{2}(g)\)) and water (\(\mathrm{H}_{2} \mathrm{O}(l)\)). The balanced chemical equation will be: \[ \mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l) \]

Calculate the heat produced in burning 1 mol of acetylene gas

Under standard conditions, we can use the enthalpy change of the reaction (\(\Delta H^\circ\)) to calculate the heat produced when 1 mol of acetylene gas is burned. First, we need to determine the enthalpy changes of formation (\(\Delta H_f^\circ\)) of the reactants and products. The standard enthalpy change of formation for acetylene gas \(\mathrm{C}_{2} \mathrm{H}_{2}\) is +226.7 kJ/mol, for oxygen gas \(\mathrm{O}_{2}\) is 0 kJ/mol, for carbon dioxide gas \(\mathrm{CO}_{2}\) is -393.5 kJ/mol, and for liquid water \(\mathrm{H}_{2} \mathrm{O}\) is -285.8 kJ/mol. To calculate the enthalpy change of the reaction (\(\Delta H^\circ\)), we can use the formula: \[ \Delta H^\circ = [\Sigma n \Delta H_f^\circ(\mathrm{products})] - [\Sigma n \Delta H_f^\circ(\mathrm{reactants})] \] Where \(n\) represents the stoichiometric coefficients and \(\Delta H_f^\circ\) represents the standard enthalpy changes of formation. Using the stoichiometric coefficients and the given enthalpy changes of formation, we get: \[ \Delta H^\circ = [(2(-393.5 \mathrm{kJ/mol}) + 1(-285.8 \mathrm{kJ/mol})] - [(1(+226.7 \mathrm{kJ/mol}) + \frac{5}{2}(0 \mathrm{kJ/mol})] \] \[ \Delta H^\circ = -1290.1 \mathrm{kJ/mol\, of\, C_{2}H_{2}} \] So, the amount of heat produced in burning 1 mol of acetylene gas under standard conditions is -1290.1 kJ/mol.

Calculate the maximum amount of useful work under standard conditions

Under standard conditions, the maximum amount of useful work that can be accomplished by a reaction is given by the change in Gibbs free energy (\(\Delta G^\circ\)). We can find the change in Gibbs free energy using the equation: \[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \] Where \(T\) is the temperature in Kelvin and \(\Delta S^\circ\) is the entropy change of the reaction. In this problem, both reactants and products are at 298 K. The standard entropy values of \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), \(\mathrm{O}_{2}(g)\), \(\mathrm{CO}_{2}(g)\), and \(\mathrm{H}_{2} \mathrm{O}(l)\) are 200.9, 205.0, 213.6, and 69.9 J/mol*K, respectively. To calculate the entropy change of the reaction, we can use a similar formula as in step 2: \[ \Delta S^\circ = [\Sigma n S^\circ(\mathrm{products})] - [\Sigma n S^\circ(\mathrm{reactants})] \] Using the entropy values and stoichiometric coefficients, we get: \[ \Delta S^\circ = [(2(213.6) + 69.9) - (200.9 + \frac{5}{2}(205.0))] \, \mathrm{J/mol\,*\, K} \] \[ \Delta S^\circ = -188.1 \, \mathrm{J/mol\, *\, K} \] Now we can calculate the change in Gibbs free energy: \[ \Delta G^\circ = -1290.1\, \mathrm{kJ/mol} - 298\, \mathrm{K} \times (-188.1 \,\mathrm{J/mol\, *\, K}) \] \[ \Delta G^\circ = -1290.1\, \mathrm{kJ/mol} - (-55.1\, \mathrm{kJ/mol}) \] \[ \Delta G^\circ = -1235.0\, \mathrm{kJ/mol\, of\, C_{2}H_{2}} \] So, the maximum amount of useful work that can be accomplished under standard conditions by the combustion of acetylene gas is -1235.0 kJ/mol.

Key concepts

Balanced Chemical Equation

In chemical reactions such as combustion, writing a balanced chemical equation is essential. It ensures that we conserve the mass and reflect the actual chemical process happening. For acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), the balanced equation when it combusts with oxygen, producing carbon dioxide and water, is: \[\mathrm{C}_{2} \mathrm{H}_{2}(g) + \frac{5}{2} \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{CO}_{2}(g) + \mathrm{H}_{2} \mathrm{O}(l)\]In this equation, we balance the number of carbon, hydrogen, and oxygen atoms from reactants to products. Always remember that the coefficients in a balanced equation denote the relative amounts of reactants and products, not just numbers placed for convenience. It reflects how many molecules of each substance partake in the reaction.

Enthalpy Change

Enthalpy change \(\Delta H\) is a crucial concept in thermochemistry and represents the heat absorbed or released during a reaction at constant pressure. In the combustion of acetylene gas, we calculate the enthalpy change by considering the enthalpy of formation of each reactant and product:\[\Delta H^\circ = [\Sigma n \Delta H_f^\circ(\mathrm{products})] - [\Sigma n \Delta H_f^\circ(\mathrm{reactants})]\]This formula takes into account the energy needed to form the products from the reactants. In our problem, burning 1 mol of acetylene releases a significant amount of heat, \(-1290.1 \mathrm{kJ/mol}\), indicating it's an exothermic reaction where energy is released to the surroundings.

Gibbs Free Energy

Gibbs Free Energy \(\Delta G\) is a measure of the maximum reversible work that can be performed by a thermodynamic process at constant pressure and temperature. It's a vital concept in determining whether a reaction is spontaneous. It combines enthalpy and entropy changes:\[\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\]For acetylene combustion, \(\Delta G\) helps us find out the feasible work the reaction can do. Calculating it gives us \(-1235.0 \mathrm{kJ/mol}\), showing the energy available for work as a result of this chemical transformation. A negative \(\Delta G\) implies spontaneous reactions under the given conditions.

Thermochemistry

Thermochemistry is the branch of chemistry dealing with the heat involved in chemical reactions. It covers concepts like enthalpy, entropy, and energy changes during reactions. When looking at combustion, thermochemical calculations tell us about the energy profiles of reactions. By understanding the heat transfer, we gain insight into whether a reaction is endothermic or exothermic. In this context, the combustion of acetylene releasing \(-1290.1 \mathrm{kJ/mol}\) shows it's an exothermic process, a common trait in combustion, where energy released can be harnessed for applications like welding.

Entropy Change

Entropy \(\Delta S\) quantifies the degree of disorder or randomness in a system, playing a critical role in the thermodynamics of reactions. While enthalpy tells us about the heat exchange, entropy gives insight into the molecular organization. When calculating entropy change for the combustion of acetylene, we use:\[\Delta S^\circ = [\Sigma n S^\circ(\mathrm{products})] - [\Sigma n S^\circ(\mathrm{reactants})]\]For this reaction, \(\Delta S\) is \(-188.1 \mathrm{J/mol\, *\, K}\), indicating a decrease in randomness from reactants to products. This decrease, along with other factors, affects the Gibbs Free Energy and thus the spontaneity and usability of the reaction's energy.