Question

Does \(\Delta H_{\mathrm{rxn}}\) for the reaction represented by the following equation equal the standard enthalpy of formation for \(\mathrm{CH}_{3} \mathrm{OH}(l) ?\) Why or why not? [Section 5.7] $$ \mathrm{C}(\text { graphite })+4 \mathrm{H}(\mathrm{g})+\mathrm{O}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l) $$

Short answer

The ΔHrxn for the given reaction is equal to the standard enthalpy of formation for CH₃OH(l), because the reaction involves the formation of one mole of CH₃OH(l) from its constituent elements in their standard states (C(graphite), 4H(g), and O(g)), which meets the definition of standard enthalpy of formation.

Step-by-step solution

Understand the concept of standard enthalpy of formation

The standard enthalpy of formation, ΔHf°, is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions (1 atm pressure and 298.15 K). The standard enthalpy of formation for an element in its standard state is defined to be zero.

Analyze the given reaction with respect to standard enthalpy of formation

The given reaction is: C(graphite) + 4H(g) + O(g) → CH₃OH(l) Here, the product, CH3OH(l), is a single compound formed from its constituent elements in their standard states - graphite for carbon, gaseous hydrogen, and gaseous oxygen.

Compare the given reaction with the definition of standard enthalpy of formation

As per the definition of the standard enthalpy of formation, we need to check if one mole of a product is formed from its constituent elements in their standard states. In the given reaction, one mole of liquid methanol (CH3OH) is formed from one mole of graphite carbon, 4 moles of gaseous hydrogen, and one mole of gaseous oxygen. These are all in their standard states.

Determine if ΔHrxn equals the standard enthalpy of formation for CH3OH(l)

Since the given reaction meets the definition of the standard enthalpy of formation, the enthalpy change of the reaction, ΔHrxn, is equal to the standard enthalpy of formation for CH3OH(l).

Explain the conclusion

The ΔHrxn for the given reaction is equal to the standard enthalpy of formation for CH3OH(l) because the reaction involves the formation of one mole of CH3OH(l) from its constituent elements in their standard states, which is consistent with the definition of standard enthalpy of formation.

Key concepts

Enthalpy Change

Enthalpy change, often represented as \( \Delta H \), is a fundamental concept in thermochemistry. It refers to the heat absorbed or released during a chemical reaction at constant pressure.

• If \( \Delta H \) is negative, the reaction is exothermic, meaning it releases heat to its surroundings.
• If \( \Delta H \) is positive, the reaction is endothermic, meaning it absorbs heat from its surroundings.

Measuring enthalpy change helps us determine how energy is transferred in chemical processes and is vital for predicting reaction behavior. In the context of the given exercise, the enthalpy change \( \Delta H_{\mathrm{rxn}} \) for the reaction forming CH₃OH(l) from its elements indicates how much energy is absorbed or released when one mole of methanol is formed under standard conditions.

Mole Concept

The mole concept is a central idea in chemistry that provides a bridge between the atomic scale and the macroscopic world. A mole is defined as Avogadro's number of particles, approximately \( 6.022 \times 10^{23} \), and serves as a way to express amounts of a chemical substance.

• One mole of any element or compound contains exactly this number of atoms, molecules, or ions.
• The mole allows chemists to count atoms by weighing them, using the atomic or molecular mass.

In the exercise above, the chemical equation characterizes the reaction by showing the conversion of specific amounts of reactants in moles to form one mole of methanol. Understanding this concept is crucial to calculating enthalpy changes and other thermodynamic variables.

Standard States

The concept of standard states is essential in thermochemistry and refers to the most stable form of a substance at 1 atm pressure and a specified temperature, usually 25°C (298.15 K). Standard states are used as a reference to compare the thermodynamic properties of substances in a consistent manner.

• Standard enthalpy values, such as those for formation, use these conditions to ensure comparability across different measurements.
• For example, the standard state of carbon is graphite, and for gases like hydrogen and oxygen, it is their gaseous form at 1 atm.

In the exercise, understanding standard states is key to recognizing that the reactants (graphite, hydrogen gas, and oxygen gas) are all in their standard states, which allows the reaction to align with the definition of standard enthalpy of formation.

Thermochemistry

Thermochemistry is the branch of chemistry that deals with the heat involved in chemical reactions. It focuses on the transfer of energy as heat between a system and its surroundings.

• It includes the study of enthalpy changes, calorimetry experiments, and calculations involving Hess's Law and standard enthalpies of formation.
• Thermochemistry helps us understand how and why energy is released or absorbed in physical and chemical changes.

The exercise reflects thermochemistry principles as it explores the enthalpic behavior of a specific reaction where one mole of methanol is produced from graphite, hydrogen gas, and oxygen gas. By analyzing \( \Delta H_{\mathrm{rxn}} \), students apply thermochemical concepts to verify if the reaction aligns with the standard enthalpy of formation definition.