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 standard enthalpy of formation for ethene gas, \(C_2H_4(g)\), is \(52.5 \text{ kJ/mol}\).

Step-by-step solution

Write the balanced combustion equation for ethene

The combustion of ethene gas can be represented as: \(C_2H_4(g) + O_2(g) \rightarrow CO_2(g) + H_2O(l)\) Since there are 2 carbon atoms in ethene, the reaction will produce 2 moles of CO2. Also, 4 hydrogen atoms in ethene will produce 2 moles of H2O. To balance the equation, we need to add appropriate coefficients. Therefore, the balanced equation is: \(C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)\)

Write down the enthalpy change relationship

The standard enthalpy change of the reaction can be calculated using the enthalpies of formation of the products and reactants. The relationship can be written as: \(\Delta H_{rxn}^\circ = \sum \Delta H_{f, products}^\circ - \sum \Delta H_{f, reactants}^\circ\) For our reaction: \(\Delta H_{comb}^\circ = [2 \times \Delta H_{f, CO_2}^\circ + 2 \times \Delta H_{f, H_2O}^\circ] - [\Delta H_{f, C_2H_4}^\circ + 3 \times \Delta H_{f, O_2}^\circ]\) Since the enthalpy of formation of elemental oxygen gas is zero (\(\Delta H_{f, O_2}^\circ = 0\)), the equation simplifies to: \(-1411.1 \text{ kJ/mol} = [2 \times (-393.5 \text{ kJ/mol}) + 2 \times (-285.8 \text{ kJ/mol})] - [\Delta H_{f, C_2H_4}^\circ]\)

Solve for the enthalpy of formation of ethene

Now, we can solve for the enthalpy of formation of ethene gas \(\Delta H_{f, C_2H_4}^\circ\). We first calculate the term inside the brackets: = \(2 \times (-393.5 \text{ kJ/mol}) + 2 \times (-285.8 \text{ kJ/mol})\) = \(-787 \text{ kJ/mol} - 571.6 \text{ kJ/mol}\) = \(-1358.6 \text{ kJ/mol}\) Now, plug this value into the equation: \(-1411.1 \text{ kJ/mol} = -1358.6 \text{ kJ/mol} - \Delta H_{f, C_2H_4}^\circ\) Rearrange the equation and solve for \(\Delta H_{f, C_2H_4}^\circ\): \(\Delta H_{f, C_2H_4}^\circ = -1358.6 \text{ kJ/mol} + 1411.1 \text{ kJ/mol}\) \(\Delta H_{f, C_2H_4}^\circ = 52.5 \text{ kJ/mol}\) The standard enthalpy of formation for ethene gas is \(52.5 \text{kJ/mol}\).

Key concepts

Enthalpy of Combustion

The enthalpy of combustion refers to the heat released when a substance completely burns in oxygen. For ethene (\(C_2H_4(g)\)), the enthalpy of combustion is \(-1411.1 \text{ kJ/mol}\). This means that when one mole of ethene is burned in oxygen, \(1411.1 \text{ kJ}\) of energy is released.This is an exothermic reaction, meaning it gives off heat. In general, the higher the enthalpy of combustion, the more energy the compound releases upon burning.Understanding the enthalpy of combustion is crucial in fields like energy production, where fuels are evaluated based on how much energy they release upon combustion.

Balanced Chemical Equation

A balanced chemical equation is essential to represent a chemical reaction accurately. It ensures that the number of each type of atom is the same on both sides of the equation, fulfilling the law of conservation of mass.For ethene's combustion:\(C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)\)This equation shows:

• 1 mole of ethene reacts with 3 moles of oxygen
• Produces 2 moles of carbon dioxide and 2 moles of water

Balancing chemical equations is a fundamental skill in chemistry, as it provides the precise stoichiometry needed for calculating enthalpy changes and other reaction metrics.

Standard Enthalpy Change

The standard enthalpy change (\(\Delta H_{rxn}^\circ\)) of a reaction measures the heat exchanged under standard conditions (1 atm pressure and 298 K temperature).For the combustion of ethene, it's calculated using the standard enthalpies of formation of products and reactants:\(\Delta H_{comb}^\circ = \sum \Delta H_{f, products}^\circ - \sum \Delta H_{f, reactants}^\circ\)This formula is crucial because it allows chemists to determine the heat change of reactions without directly measuring every scenario, ensuring predictions and calculations are accurate and viable for real-world applications.

Enthalpy Calculations

Enthalpy calculations involve determining the heat of reactions, such as using enthalpies of formation to find reaction enthalpies. For ethene, the standard enthalpy of formation \(\Delta H_f^\circ\) can be deduced as follows:1. Combine the known enthalpies of formation for \(CO_2\) and \(H_2O\) to calculate the total enthalpy of products.2. Apply the enthalpy relationship: \(\Delta H_{comb}^\circ = [2 \times -393.5 \text{ kJ/mol} + 2 \times -285.8 \text{ kJ/mol}] - \Delta H_{f, C_2H_4}^\circ\)3. Solve the equation for \(\Delta H_{f, C_2H_4}^\circ\) to find its value (\(52.5 \text{ kJ/mol}\)).Understanding how to perform such calculations is pivotal in chemistry, allowing the prediction and understanding of how much heat each reaction will typically release or absorb.