Question

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right)\), propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\), and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are \(+185.4,+20.4\), and \(-103.8 \mathrm{~kJ} / \mathrm{mol}\), respectively. (a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) (b) Calculate the heat evolved on combustion of \(1 \mathrm{~kg}\) of each substance.

Short answer

The heat evolved per mole on combustion of propyne, propylene, and propane are -1849.5 kJ/mol, -1885.5 kJ/mol, and -2043.9 kJ/mol, respectively. The heat evolved on combustion of 1 kg of propyne, propylene, and propane are -46200 kJ/kg, -44790 kJ/kg, and -46340 kJ/kg, respectively.

Step-by-step solution

Write the balanced chemical equations

Write the balanced chemical equations for the combustion of propyne, propylene, and propane: (a) Propyne: \(C_3H_4(g) + \frac{5}{2}O_2(g) \rightarrow 3CO_2(g) + 2H_2O(g)\) (b) Propylene: \(C_3H_6(g) + \frac{9}{2}O_2(g) \rightarrow 3CO_2(g) + 3H_2O(g)\) (c) Propane: \(C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(g)\)

Calculate ΔH for each reaction using Hess's Law

Use Hess's law to find the enthalpy change for each reaction. Hess's law states that the enthalpy change for a reaction is equal to the sum of the enthalpy of formation of the products minus the sum of the enthalpy of formation of the reactants (ΔH = ΔH_products - ΔH_reactants). Given the enthalpies of formation for propyne, propylene, and propane are +185.4, +20.4, and -103.8 kJ/mol, respectively. The enthalpies of formation for CO2(g) and H2O(g) are -393.5 kJ/mol and -241.8 kJ/mol, respectively. (a) ΔH_propyne_combustion = (3ΔH_CO2+ 2ΔH_H2O) - (ΔH_propyne + 0) = ([3 x (-393.5)] + [2 x (-241.8)]) - (185.4) (b) ΔH_propylene_combustion = (3ΔH_CO2+ 3ΔH_H2O) - (ΔH_propylene + 0) = ([3 x (-393.5)] + [3 x (-241.8)]) - (20.4) (c) ΔH_propane_combustion = (3ΔH_CO2+ 4ΔH_H2O) - (ΔH_propane + 0) = ([3 x (-393.5)] + [4 x (-241.8)]) - (-103.8)

Calculate the heat evolved in each reaction

Calculate the heat evolved (ΔH) in each reaction: (a) ΔH_propyne_combustion = ([-1180.5] + [-483.6]) - (185.4) = -1849.5 kJ/mol (b) ΔH_propylene_combustion = ([-1180.5] + [-725.4]) - (20.4) = -1885.5 kJ/mol (c) ΔH_propane_combustion = ([-1180.5] + [-967.2]) - (-103.8) = -2043.9 kJ/mol

Calculate the heat evolved per kilogram for each substance

For each substance, calculate the heat evolved per kilogram using the molar mass. (a) Propyne: Molar mass of C3H4 = 3(12.01) + 4 (1.008) = 40.048 g/mol Heat evolved per kg of propyne = -1849.5 kJ/mol / (40.048 g/mol) = -46.20 kJ/g * 1000 = -46200 kJ/kg (b) Propylene: Molar mass of C3H6 = 3(12.01) + 6(1.008) = 42.080 g/mol Heat evolved per kg of propylene = -1885.5 kJ/mol / (42.080 g/mol) = -44.79 kJ/g * 1000 = -44790 kJ/kg (c) Propane: Molar mass of C3H8 = 3(12.01) + 8(1.008) = 44.096 g/mol Heat evolved per kg of propane = -2043.9 kJ/mol / (44.096 g/mol) = -46.34 kJ/g * 1000 = -46340 kJ/kg. Therefore, the heat evolved on combustion of 1 kg of propyne, propylene, and propane are -46200 kJ/kg, -44790 kJ/kg, and -46340 kJ/kg, respectively.

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation, denoted as \( \Delta H_{f}^{\circ} \), is a crucial concept in thermodynamics that measures the heat change when one mole of a compound is formed from its elements in their standard states. For example, the formation of water from hydrogen and oxygen gases releases or absorbs energy, represented as the enthalpy of formation for water.

In chemistry homework, you might encounter exercises asking to calculate the heat evolved during combustion. To find this, you'd need to know the enthalpy of formation of the products and reactants involved in the reaction. Specifically, in our exercise, the enthalpy of formation for propyne, propylene, and propane are given as +185.4, +20.4, and -103.8 kJ/mol, respectively. These values indicate the amount of heat absorbed or released when one mole of each substance is formed under standard conditions.

It is important to differentiate between exothermic and endothermic reactions in these problems. An exothermic reaction, such as combustion, releases heat, resulting in a negative enthalpy change. Conversely, an endothermic reaction absorbs heat, leading to a positive enthalpy change. Understanding this concept aids in better interpreting the outcomes of various chemical processes.

Hess's Law

Hess's Law is a fundamental principle that states the total enthalpy change for a chemical reaction is the same, no matter how many steps the reaction takes. This law is based on the concept that enthalpy is a state function—it depends only on the initial and final states of the system, not on the path taken.

Students can leverage Hess's Law to solve challenging problems where the pathway of a reaction is not one step but several. For example, when direct measurement of the enthalpy change of a reaction is complicated, Hess's Law allows the use of alternate, simpler reactions with known enthalpy changes to calculate the enthalpy of the target reaction.

The law simplifies the process by ensuring you can add or subtract known enthalpy changes of reactions to determine the enthalpy change of a different, related reaction. In the context of our exercise, Hess's Law enables us to calculate the enthalpy of combustion for substances like propyne, propylene, and propane, even if we only know their standard enthalpies of formation and those of the combustion products, CO_2 and H_2O.

Combustion Reactions

Combustion reactions are exothermic chemical reactions that occur when a substance, typically a fuel, reacts with an oxidant, releasing heat and often light. Commonly, oxygen gas serves as the oxidant and products include substances like carbon dioxide and water.

In a combustion reaction, the exact amount of heat released can be measured or calculated, and this is often expressed as the enthalpy of combustion. Calculations involving combustion reactions are integral to fields like combustion engineering, environmental science, and energy policy-making.

For complete combustion, ensuring that there is no leftover fuel or oxygen, you can use stoichiometry to write a balanced chemical equation. This equation is pivotal for calculating the enthalpy change during combustion, which accounts for both the energy needed to break bonds in the reactants and the energy released when new bonds are formed in the products. The heat evolved during a combustion reaction, as shown in the exercise for propyne, propylene, and propane, gives insights into the energy efficiency of different fuels—an essential aspect in both industrial applications and daily life.