Question

The standard enthalpy of combustion of ethene gas, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\), is \(-1411.1 \mathrm{~kJ} / \mathrm{mol}\) at \(298 \mathrm{~K}\). Given the following enthalpies of formation, calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{4}(g)\). $$ \begin{array}{ll} \mathrm{CO}_{2}(g) & -393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \mathrm{O}(l) & -285.8 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

Short answer

The short answer is: The standard enthalpy of formation for ethene gas (C2H4) is \(52.5 \text{ kJ/mol}\).

Step-by-step solution

Write down the balanced equation for combustion reaction

The balanced equation for the combustion reaction of ethene gas is: \(C2H4(g) + 3O2(g) -> 2CO2(g) + 2H2O(l)\)

Calculate the enthalpy change for reactants and products of combustion reaction

We have been provided with the enthalpies of formation for CO2(g) and H2O(l). We need to multiply them by their respective stoichiometric coefficients from the balanced equation and then sum them up for reactants and products. \(Σ[ΔH_{f}^o (\text{reactants})] = 0\), since O2 (g) as a pure element has a standard enthalpy of formation of 0. \(Σ[ΔH_{f}^o (\text{products})] = 2 × −393.5 + 2 × (−285.8) = \) \((-787 \text{ kJ/mol}) + (-571.6 \text{ kJ/mol}) = -1358.6 \text{ kJ/mol}\)

Calculate the standard enthalpy of formation for ethene gas (C2H4)

Now we can use the formula for calculating the standard enthalpy of formation: \(ΔH_{f}^o (C2H4) = Σ [ΔH_{f}^o (\text{products})] - Σ[ΔH_{f}^o (\text{reactants})] - ΔH_{combustion}\) \(ΔH_{f}^о (C2H4) = (-1358.6 \text{ kJ/mol}) - (0) - (-1411.1 \text{ kJ/mol})\) \(ΔH_{f}^о (C2H4) = 52.5 \text{ kJ/mol}\) The standard enthalpy of formation for ethene gas (C2H4) is \(52.5 \text{ kJ/mol}\).

Key concepts

Standard Enthalpy of Combustion

The standard enthalpy of combustion is a crucial concept in chemical thermodynamics. It refers to the heat released when one mole of a substance completely burns in oxygen under standard conditions, usually at a temperature of 298 K (25°C) and a pressure of 1 atmosphere. This value is always negative because combustion is an exothermic process, meaning it releases energy. For example, in the exercise, the standard enthalpy of combustion of ethene gas is given as ewline ewline ewline ewline ewline −1411.1 kJ/mol. Understanding this property aids in assessing energy content of fuels and in calculating enthalpies of formation, as demonstrated in the problem at hand. Biochemically, it plays a role in understanding how much energy organisms can obtain from metabolizing certain substances.

Chemical Thermodynamics

Chemical thermodynamics involves the study of energy changes in chemical reactions, which is pivotal for understanding reaction spontaneity, equilibrium, and efficiency. It encompasses laws of thermodynamics that explain how energy is conserved and transferred. The first law, for instance, indicates that energy cannot be created or destroyed. In our exercise, we apply these principles to find the enthalpy of formation. By calculating the energy changes during the combustion of a molecule like ethene gas, we can derive the energy content of the molecule itself, a method deeply rooted in thermodynamic principles.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. A balanced chemical equation is central to these calculations, as it provides the ratio in which substances react and form products. Using the coefficients from the balanced equation, we can determine the amount of heat released or absorbed in the reaction.In the given example, stoichiometry is key for multiplying the standard enthalpies of formation of products by their coefficients to calculate the total energy change during the reaction. It's also necessary for ensuring that the equation adheres to the law of conservation of mass, meaning the number of atoms of each element is the same on both sides of the equation.

Balanced Chemical Equation

A balanced chemical equation is a written representation of a chemical reaction with an equal number of atoms for each element present on both the reactant and product sides. Balancing equations is fundamental in chemical equations because it reflects the conservation of matter and allows for correct stoichiometric calculations. In our exercise, the combustion of ethene gas is represented by a balanced equation: ewline ewline ewline ewline ewline ewline \(C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)\). Correctly balancing this equation is essential for accurately calculating the enthalpy change using the provided enthalpies of formation for carbon dioxide and water.