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\(]\)

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: \[ \mathrm{N}_{2} (g) + 2\,\mathrm{O}_{2}(g) \rightarrow 2\,\mathrm{NO}_{2}(g) \]

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 (\(w = 0\)).

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.