Question

From the following heats of combustion, \(\begin{aligned} \mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{C}(\text { graphite })+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) calculate the enthalpy of formation of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) from its elements: $$ \mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) $$

Short answer

The enthalpy of formation of methanol is \(-238.4\,\mathrm{kJ/mol}\).

Step-by-step solution

Understand the Problem

We need to find the enthalpy of formation for methanol, which is the enthalpy change when one mole of methanol is formed from its elements in their standard states. We are given the heats of combustion for methanol, graphite, and hydrogen and need to use these to calculate the desired enthalpy.

Write the Chemical Equations

We have the following chemical equations:1. \[\mathrm{CH}_{3}\mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\]\[\Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \mathrm{~kJ} / \mathrm{mol}\]2. \[\mathrm{C(\text{graphite})}+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)\]\[\Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol}\]3. \[\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\]\[\Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\]

Define the Target Equation

The target reaction for which we need the enthalpy change is:\[\mathrm{C(\text{graphite})}+2 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CH}_{3}\mathrm{OH}(l)\]

Key concepts

Heats of Combustion

Heats of combustion is an important concept in chemistry used to determine the energy released during the combustion of a substance. The combustion process involves a chemical reaction between a fuel and an oxidant, typically oxygen from the air, resulting in the formation of combustion products such as carbon dioxide and water, and the release of heat energy.

In this context, every chemical equation provided has an associated enthalpy change, labeled as \( \Delta H_{\text{rxn}}^{\circ}\). This symbol represents the standard enthalpy change of the combustion reaction in kilojoules per mole. For example, the combustion of methanol is associated with a heat release of \( -726.4 \, \mathrm{kJ} / \mathrm{mol}\). Heats of combustion are measured in controlled environments to ensure that the pressure and temperature remain constant, typically at standard conditions (298 K and 1 atm).

Understanding the heat of combustion allows us to calculate other related thermodynamic quantities, such as the enthalpy of formation. By applying conservation of energy principles and Hess's Law, we can deduce the desired enthalpy values from known heats of combustion.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions that provide insight into the reactants and products involved in a transformation. They are an essential tool for chemists, allowing for the precise description of compounds and the stoichiometry of the reaction.

The chemical equations used in the problem include:

• The combustion of methanol: \[\mathrm{CH}_{3}\mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\]
\[\Delta H_{\mathrm{rxn}}^{\circ}=-726.4 \, \text{kJ/mol}\]

This shows that methanol reacts with oxygen to form carbon dioxide and water, releasing energy.

Another equation represents the combustion of graphite to form carbon dioxide:

• \(\mathrm{C(\text{graphite})}+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \)
\[\Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \, \text{kJ/mol}\]

This highlights the importance of standard states such as graphite for carbon.

Finally, the formation of water from hydrogen indicates the exothermic nature of this reaction:

• \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) \)
\[\Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \, \text{kJ/mol}\]

These equations are fundamental in understanding the reactions involved and in setting the stage for applying Hess's Law to calculate the desired enthalpies.

Standard Enthalpy Change

Standard enthalpy change, represented as \( \Delta H^{\circ} \, \text{or} \, \Delta H_{\text{rxn}}^{\circ}\), is a thermodynamic property that measures the heat change during a chemical reaction at standard conditions. These conditions include a temperature of 298 K (25°C) and a pressure of 1 atm.

When discussing standard enthalpy change, it is crucial to remember that it is defined per mole of reaction as described by a balanced equation. This property is especially useful for calculating changes involved in formation reactions, such as forming one mole of a compound from its constituent elements in their standard states.

In the given problem, we aim to calculate the standard enthalpy of formation of methanol using the provided heats of combustion. Hess's Law allows us to exploit the additive nature of enthalpy to rearrange given equations into our target equation:

• Combustion of methanol to yield carbon dioxide and water with the associated enthalpy change: \( \Delta H = -726.4 \, \text{kJ/mol}\)
• Combustion of graphite to produce carbon dioxide: \( \Delta H = -393.5 \, \text{kJ/mol}\)
• Formation of water from hydrogen: \( \Delta H = -285.8 \, \text{kJ/mol}\)

By rearranging these equations and using the principle of conservation of energy, we can solve for the desired enthalpy of methanol's formation.