Problem 10 - Chemistry: The Central Science in SI Units, Global Edition

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 C, determine \(\Delta H\) for the formation of one mole of the product. [Sections 5.3 and 5.7\(]\)

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between $\mathrm{N}_{2}$ and $\mathrm{O}_{2}$ is: \(\mathrm{N}_{2} + 2\mathrm{O}_{2} \rightarrow 2\mathrm{NO}_{2}\). The work done by the system is positive, and the change in enthalpy for the formation of one mole of the product is \(33.18\, \text{kJ/mol}\).

Step 1: Write the balanced chemical equation

Given that the reaction is between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in the gas phase, we can start by writing the unbalanced equation: \[ \mathrm{N}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{NO}_{2} \] To balance the equation, we add a coefficient to the reactants: \[ \mathrm{N}_{2} + 2\mathrm{O}_{2} \rightarrow 2\mathrm{NO}_{2} \] Now the equation is balanced.

Step 2: Predict whether work is positive, negative or zero

Since the reaction is happening at constant pressure, we know that the work done by the system is given by \(w = -P(\Delta V)\). In this case, there are 3 moles of gas on the reactants side and only 2 moles of gas on the products side. Therefore, the volume of the reactants is more than the volume of the products. In the process of the reaction, the system must contract to decrease its volume. When the system contracts (decreases its volume), it does positive work on the surroundings, meaning the work \(w\) done by the system is positive.

Step 3: Determine ΔH for the formation of one mole of product

To find the change in enthalpy (\(\Delta H\)) for the formation of one mole of the product, we will use Appendix C which provides the enthalpy of formation (\(\Delta H_\mathrm{f}^{\circ}\)) values for different substances. For this reaction, we have: \[ \Delta H^\circ_{\mathrm{reaction}} = \sum \nu \Delta H^\circ_\mathrm{f} \text{(products)} - \sum \nu \Delta H^\circ_\mathrm{f} \text{(reactants)} \] Thus, the change in enthalpy for the reaction is: \[ \Delta H^\circ_{\mathrm{reaction}} = 2 \Delta H^\circ_\mathrm{f}(\mathrm{NO}_2) - \Delta H^\circ_\mathrm{f}(\mathrm{N}_2) - 2 \Delta H^\circ_\mathrm{f}(\mathrm{O}_2) \] Since \(\Delta H^\circ_\mathrm{f}(\mathrm{N}_2)\) and \(\Delta H^\circ_\mathrm{f}(\mathrm{O}_2)\) are both zero (they are elements in their standard states), we have: \[ \Delta H^\circ_{\mathrm{reaction}} = 2 \Delta H^\circ_\mathrm{f}(\mathrm{NO}_2) \] From Appendix C, we find \(\Delta H^\circ_\mathrm{f}(\mathrm{NO}_2) = 33.18 \, \text{kJ/mol}\). Therefore, the change in enthalpy for the reaction is: \[ \Delta H^\circ_{\mathrm{reaction}} = 2(33.18 \, \text{kJ/mol}) = 66.36 \, \text{kJ/mol} \] Since this is the change in enthalpy for the formation of 2 moles of \(\mathrm{NO}_2\), the change in enthalpy for the formation of one mole of product is half this value: \[ \Delta H^\circ_\mathrm{one\, mole} = \frac{66.36 \, \text{kJ/mol}}{2} = 33.18 \, \text{kJ/mol} \] The balanced chemical equation for the reaction is: \[ \mathrm{N}_{2} + 2\mathrm{O}_{2} \rightarrow 2\mathrm{NO}_{2} \] The work \(w\) done by the system is positive. The change in enthalpy for the formation of one mole of the product is \(33.18\, \text{kJ/mol}\).

Key Concepts

Balanced Chemical Equation

In chemical reactions, it is crucial to express the reaction with a balanced chemical equation. This is because matter can neither be created nor destroyed, a principle known as the Law of Conservation of Mass. When a chemical equation is balanced, it means that there are equal numbers of each type of atom on both sides of the reaction. To balance the equation for the reaction between nitrogen ( ext{N}_2) and oxygen ( ext{O}_2) to form nitrogen dioxide ( ext{NO}_2), we first write the unbalanced equation: \[ ext{N}_2 + ext{O}_2 ightarrow ext{NO}_2\].Then, we balance it by accounting for the number of atoms of each element involved. In this case, we find:\[ ext{N}_2 + 2 ext{O}_2 ightarrow 2 ext{NO}_2\]. Here, each element has the same number of atoms on both sides of the equation. Nitrogen and oxygen are balanced in such a way that the mass is conserved. This balanced equation properly represents the stoichiometry of the reaction, giving us the correct proportions of reactants and products.

Constant Pressure

Reactions occurring at constant pressure are common in experimental chemistry. In these scenarios, the pressure remains unchanged while the reaction occurs. This condition is significant as it allows us to link pressure to other thermodynamic quantities such as work and enthalpy change. The term \(P\) involves pressure, and it plays a role in determining the work done in a reaction.Maintaining constant pressure is important in approximating real-world conditions, especially since many reactions in laboratories and industries are performed under such circumstances. For calculation purposes, this assumption simplifies the mathematics because changes in pressure often complicate energy calculations. The reaction between nitrogen and oxygen occurs at constant pressure, making it easier to analyze changes in volume and relate them to work.

Work Done by a System

In thermodynamics, the work done by a system can be assessed by understanding the changes in volume at constant pressure. This concept is vital for gas-phase reactions, like the reaction between nitrogen and oxygen to form nitrogen dioxide. When this reaction occurs, the system may do work on its surroundings if there is a change in volume.The work done by the system is calculated using the formula:\[w = -P(\Delta V)\], where \(w\) is the work done, \(P\) is the pressure, and \(\Delta V\) is the change in volume. For this specific reaction, since there are 3 moles of gas on the reactants' side and 2 moles on the products' side, the system's volume decreases. When the system's volume decreases, it does positive work on the surroundings. This can be understood as the contraction of gas exerting pressure outward, which in thermodynamic terms is considered as the system doing work on its surroundings. Thus, under these reaction conditions, the work \(w\) will be positive.