The Ostwald process for the commercial production of nitric acid involves three steps: $$ \begin{array}{l} 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \underset{825^{\circ} \mathrm{C}}{\longrightarrow} 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(l)+\mathrm{NO}(g) \end{array} $$ a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(\left.298 \mathrm{~K}\right)\) for each of the three steps in the Ostwald process (see Appendix 4). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

Step by step solution

01

a. Calculate ΔH°, ΔS°, ΔG°, and K for each step

For each step in the Ostwald process, we can use the standard enthalpy of formation (ΔHf°) and the standard entropy values (S°) from Appendix 4 to find the standard enthalpy change (ΔH°), standard entropy change (ΔS°), and then the standard Gibbs free energy change (ΔG°) using the equation: ΔG° = ΔH° - TΔS°, with T in Kelvin. Once we have calculated ΔG° for each step, we can find the equilibrium constant (K) using the relationship: ΔG° = -RT ln K, where R is the gas constant (8.314 J/(mol·K)). Calculate ΔH°, ΔS°, ΔG°, and K for each step using the above equations and the values from Appendix 4.

02

b. Calculate the equilibrium constant for the first step at 825°C

To calculate the equilibrium constant (K) for the first step at 825°C, we'll use the van't Hoff equation: \(ln(\frac{K_2}{K_1}) = \frac{-\Delta H°}{R}(\frac{1}{T_2}- \frac{1}{T_1})\) where K1 and K2 are the equilibrium constants at temperatures T1 and T2 (in Kelvin), respectively, and ΔH° is the standard enthalpy change. First, we need to convert 825°C to Kelvin (1098 K). Then, use the equilibrium constant K at 298 K from part a as K1 and solve for K2, which represents the equilibrium constant at 1098 K (825°C), assuming ΔH° and ΔS° do not depend on temperature.

03

c. Thermodynamic reason for the high temperature in the first step

After calculating the equilibrium constant for the first step at 825°C, compare it to the one at 298 K. A higher equilibrium constant at the elevated temperature suggests that the reaction is more favorable at high temperatures. The thermodynamic reason for this can be seen in the relationship between ΔG°, ΔH°, and ΔS°. If the first step has a positive ΔH° (endothermic) and a positive ΔS° (entropy increases), increasing the temperature can make the ΔG° more negative, which makes the reaction more favorable. Moreover, the high temperature increases the reactants' energy, allowing them to overcome the activation energy barrier and thus promoting the reaction to proceed. In conclusion, the high temperature in the first step of the Ostwald process is beneficial from a thermodynamics standpoint, as it increases the equilibrium constant and drives the reaction to proceed more efficiently under standard conditions.

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