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\(]\)

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.

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.

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 \(33.18\, \text{kJ/mol}\).