Question

The gas-phase reaction shown, between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), was run in an apparatus designed to maintain a constant pressure. (a) Write a balanced chemical equation for the reaction depicted and predict whether \(w\) is positive, negative, or zero. (b) Using data from Appendix \(\mathrm{C}\), determine \(\Delta H\) for the formation of one mole of the product. Why is this enthalpy change called the enthalpy of formation of the involved product? [Sections 5.3 and 5.7\(]\)

Short answer

The balanced chemical equation for the gas-phase reaction between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) is \[ \mathrm{N}_{2} (g) + 2\,\mathrm{O}_{2}(g) \rightarrow 2\,\mathrm{NO}_{2}(g).\] The work done during the reaction is zero (\(w = 0\)) as there is no change in the number of moles of gas. The enthalpy change (ΔH) for the formation of 1 mole of the product, nitrogen dioxide, is 66.4 kJ/mol.

Step-by-step solution

Balanced Chemical Equation

First, we need to write a balanced chemical equation for the reaction between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). We know that nitrogen and oxygen can react to form nitrogen dioxide (\(\mathrm{NO}_{2}\)). So, the balanced chemical equation will be: \[ \mathrm{N}_{2} (g) + 2\,\mathrm{O}_{2}(g) \rightarrow 2\,\mathrm{NO}_{2}(g) \]

Predicting Work

Now, we have to predict whether \(w\) is positive, negative, or zero. During the reaction, if there is an expansion of the system (increase in volume), the work is done by the system, which means \(w\) is negative. If the volume decreases (compression) during the reaction, then the work is done on the system, which means \(w\) is positive. No volume change indicates zero work. In this reaction, 1 mole of \(\mathrm{N}_{2}\) and 2 moles of \(\mathrm{O}_{2}\) combine to produce 2 moles of \(\mathrm{NO}_{2}\), which doesn't cause any change in the number of moles of gas. Thus, the volume remains constant, and the work done is zero (\(w = 0\)).

Determining ΔH for the formation of 1 mole of the product using Appendix C

To calculate the enthalpy change (ΔH) for the reaction, we'll need the enthalpies of formation (ΔHf) for the reactants and products, which can be found in Appendix C. The equation for ΔH for the given reaction is shown below: \[ \Delta H = \sum_{products} n\,\Delta H_{f}^{0}(products) - \sum_{reactants} n\,\Delta H_{f}^{0}(reactants) \] Using the values from Appendix C, we can plug the values into the equation. ΔHfº(N2) = 0 kJ/mol (since it is in its elemental form) ΔHfº(O2) = 0 kJ/mol (since it is in its elemental form) ΔHfº(NO2) = 33.2 kJ/mol Plugging these values into the ΔH equation: \[ \Delta H = 2(33.2\,\text{kJ/mol}) - (1(0\,\text{kJ/mol}) + 2(0\,\text{kJ/mol})) \] \[ \Delta H = 66.4\,\text{kJ/mol} \] Hence, the enthalpy change (ΔH) for the formation of 1 mole of the product, nitrogen dioxide, is 66.4 kJ/mol.

Key concepts

Balanced Chemical Equation

When writing a balanced chemical equation, we aim to ensure that the number of atoms for each element is the same on both sides of the equation. For the reaction between nitrogen and oxygen to produce nitrogen dioxide, the balanced equation is:\[ \mathrm{N}_{2} (g) + 2\,\mathrm{O}_{2}(g) \rightarrow 2\,\mathrm{NO}_{2}(g) \]In this equation:

• The left side of the equation has one nitrogen molecule and two oxygen molecules.
• The right side results in two nitrogen dioxide molecules.
• This satisfies the conservation of mass, meaning atoms are neither created nor destroyed.

Balancing chemical equations is crucial to understanding the stoichiometry of a reaction, which in turn helps predict the quantities of reactants needed and products formed.

Enthalpy of Formation

The enthalpy of formation refers to the heat change when one mole of a compound is formed from its elemental components at standard conditions. For our reaction producing nitrogen dioxide, we calculate the enthalpy change using known enthalpies of formation:

• The standard enthalpy of formation for Nitrogen (\( \mathrm{N}_{2} \)) and Oxygen (\( \mathrm{O}_{2} \)) is zero, since they are in their natural elemental state.
• The standard enthalpy of formation for nitrogen dioxide (\( \mathrm{NO}_{2} \)) is 33.2 kJ/mol.

Using these values and the formula:\[ \Delta H = \sum_{products} n\,\Delta H_{f}^{0}(products) - \sum_{reactants} n\,\Delta H_{f}^{0}(reactants) \]We find that the enthalpy change (\( \Delta H \)) is 66.4 kJ/mol for the reaction. This value indicates the energy required or released when forming 1 mole of nitrogen dioxide from nitrogen and oxygen.

Gas-Phase Reaction

A gas-phase reaction like the one between \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \) to form \( \mathrm{NO}_{2} \) involves reactants and products in the gaseous state. Here are some key aspects:

• Gas-phase reactions often occur at constant pressure in laboratory setups to simplify calculations and maintain conditions.
• In this particular reaction, since there is no change in the overall number of gas moles, the volume does not change, resulting in zero work done (\( w = 0 \)).
• Maintaining constant pressure and volume allows us to focus on enthalpy changes, rather than accounting for work performed by or on the system.

Understanding gas-phase reactions is crucial as they are common in both natural and industrial processes, making it important to analyze their behavior under various conditions.