Question

The Ostwald process for the commercial production of nitric acid from ammonia and oxygen involves the following steps: $$ \begin{aligned} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ a. Use the values of \(\Delta H_{\mathrm{f}}^{\circ}\) in Appendix 4 to calculate the value of \(\Delta H^{\circ}\) for each of the preceding reactions. b. Write the overall equation for the production of nitric acid by the Ostwald process by combining the preceding equations. (Water is also a product.) Is the overall reaction exothermic or endothermic?

Short answer

The overall equation for the Ostwald process is given by \(4 NH_3(g) + 5 O_2(g) \rightarrow 2 HNO_3(aq) + 3 H_2O(g) + H_2O(l)\). The reaction's enthalpy change, \(\Delta H^\circ_{overall}\), can be calculated as the sum of the individual enthalpy changes (\(\Delta H^{\circ}_{1}\), \(\Delta H^{\circ}_{2}\), and \(\Delta H^{\circ}_{3}\)). Once calculated, if \(\Delta H^\circ_{overall} < 0\), the reaction is exothermic; if \(\Delta H^\circ_{overall} > 0\), the reaction is endothermic.

Step-by-step solution

a. Calculate ∆H° for each reaction

To calculate the enthalpy change for each reaction, we will use the equation: \[ \Delta H^{\circ}_{reaction} = \sum \Delta H^{\circ}_{f,\,products} - \sum \Delta H^{\circ}_{f,\,reactants} \] For each reaction, we will find the enthalpy change using the enthalpy of formation values provided in Appendix 4. Reaction 1: \(4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO(g) + 6 H_2O(g)\) \[ \begin{aligned} \Delta H^{\circ}_{1} &= [4 \times \Delta H^{\circ}_{f,\,NO(g)} + 6 \times \Delta H^{\circ}_{f,\,H_{2}O(g)}] - [4 \times \Delta H^{\circ}_{f,\,NH_{3}(g)}]\\ \end{aligned} \] Reaction 2: \(2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)\) \[ \begin{aligned} \Delta H^{\circ}_{2} &= [2 \times \Delta H^{\circ}_{f,\,NO_{2}(g)}] - [2 \times \Delta H^{\circ}_{f,\,NO(g)}]\\ \end{aligned} \] Reaction 3: \(3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(aq) + NO(g)\) \[ \begin{aligned} \Delta H^{\circ}_{3} &= [2 \times \Delta H^{\circ}_{f,\,HNO_{3}(aq)} + \Delta H^{\circ}_{f,\,NO(g)}] - [3 \times \Delta H^{\circ}_{f,\,NO_{2}(g)} + \Delta H^{\circ}_{f,\,H_{2}O(l)}]\\ \end{aligned} \] Now, plug in the enthalpy of formation values from Appendix 4 and calculate the values of \(\Delta H^\circ_{1}\), \(\Delta H^\circ_{2}\) and \(\Delta H^\circ_{3}\).

b. Overall equation and determining exothermic or endothermic

In order to find the overall equation for the Ostwald process, we will add the three reactions together and eliminate any substance that appears on both sides of the equation. Reaction 1: \(4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO(g) + 6 H_2O(g)\) Reaction 2: \(2 NO(g) + O_2(g) \rightarrow 2 NO_2(g)\) Reaction 3: \(3 NO_2(g) + H_2O(l) \rightarrow 2 HNO_3(aq) + NO(g)\) Overall Equation: \(4 NH_3(g) + 5 O_2(g) \rightarrow 2 HNO_3(aq) + 3 H_2O(g) + H_2O(l)\) To determine whether the overall reaction is exothermic or endothermic, we will check the sum of the enthalpy changes of the individual reactions: \[ \Delta H^\circ_{overall} = \Delta H^{\circ}_{1} + \Delta H^{\circ}_{2} + \Delta H^{\circ}_{3} \] If \(\Delta H^\circ_{overall} < 0\), the overall reaction is exothermic. If \(\Delta H^\circ_{overall} > 0\), the overall reaction is endothermic. Calculate \(\Delta H^\circ_{overall}\) using the previously calculated values of \(\Delta H^\circ_{1}\), \(\Delta H^\circ_{2}\) and \(\Delta H^\circ_{3}\), and determine whether the reaction is exothermic or endothermic.