Question

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) 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?

Short answer

In the Ostwald process, the three steps for commercial production of nitric acid involve the following reactions: 1. \(4 \mathrm{NH_{3}(g)} + 5 \mathrm{O_{2}(g)} \rightarrow 4 \mathrm{NO(g)} + 6 \mathrm{H_{2}O(g)}\) 2. \(2 \mathrm{NO(g)} + \mathrm{O_{2}(g)} \rightarrow 2 \mathrm{NO_{2}(g)}\) 3. \(\mathrm{NO_{2}(g)} + \mathrm{H_{2}O(l)} \rightarrow \mathrm{HNO_{3}(aq)} + \mathrm{NO(g)}\) After calculating the \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) values for each step, we can calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\) using the relation \(K = e^{[-\Delta G^{\circ}/(RT)]}\). For the first step at \(825^{\circ} \mathrm{C}\), we can use the Van't Hoff equation to calculate the equilibrium constant \(K\): \(K = e^{[\Delta S^{\circ}/R - \Delta H^{\circ}/(RT)]}\). The high temperature in the first step ensures that the reaction moves forward because the rate of the reaction depends on temperature. Moreover, at higher temperatures, the entropy becomes more significant, and since the reaction has a positive entropy change, it suggests a more spontaneous reaction at high temperatures. Thus, the high temperature in the first step helps the Ostwald process run efficiently and ensures the desired product formation.

Step-by-step solution

Calculate thermodynamic properties for each step

Refer to Appendix 4 of the book and use the given thermodynamic data to find the \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for each step: Step 1: \(\Delta H^{\circ} = 4\Delta H_{f}^{\circ}(\mathrm{NO}) - (4\Delta H_{f}^{\circ}(\mathrm{NH_{3})} + 5\Delta H_{f}^{\circ}(\mathrm{O_{2})})\) \(\Delta S^{\circ} = (4S^{\circ}(\mathrm{NO}) + 6S^{\circ}(\mathrm{H_{2}O}) - (4S^{\circ}(\mathrm{NH_{3})} + 5S^{\circ}(\mathrm{O_{2})})\) \(\Delta G^{\circ} = 4\Delta G_{f}^{\circ}(\mathrm{NO}) - (4\Delta G_{f}^{\circ}(\mathrm{NH_{3})} + 5\Delta G_{f}^{\circ}(\mathrm{O_{2})})\) Step 2: Similar calculations as in step 1, for the second reaction. Step 3: Similar calculations as in step 1, for the third reaction. After calculating the \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) values for each step, we can calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\) using the following relation: \(K = e^{[-\Delta G^{\circ}/(RT)]}\)

Calculate equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\)

To find 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, we can use the Van't Hoff equation: \(\frac{\Delta H^{\circ}}{T} = \frac{\Delta S^{\circ}}{R} \ln(K)\) Rearrange to solve for K: \(K = e^{[\Delta S^{\circ}/R - \Delta H^{\circ}/(RT)]}\) Now, plug in the values of \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), R, and temperature (in Kelvin, \(825^{\circ} \mathrm{C} = 1098 \mathrm{K}\)) and calculate the equilibrium constant \(K\).

Discuss the thermodynamic reason for the high temperature in the first step

The high temperature in the first step ensures that the reaction moves forward because the rate of the reaction depends on temperature. Moreover, at higher temperatures, the entropy becomes more significant, and since the reaction has a positive entropy change, it suggests a more spontaneous reaction at high temperatures. In conclusion, the high temperature in the first step helps the Ostwald process run efficiently and ensures the desired product formation.

Key concepts

Thermodynamic Properties

When studying chemical reactions, thermodynamic properties such as enthalpy (\( \Delta H^{\circ} \)), entropy (\( \Delta S^{\circ} \)), and Gibbs free energy (\( \Delta G^{\circ} \)) are crucial in determining the feasibility and direction of a reaction. These properties offer insights into the energy changes and disorder associated with a reaction.

- **Enthalpy (\( \Delta H^{\circ} \)):** Represents the heat absorbed or released under constant pressure during a reaction. Negative values indicate exothermic reactions, where heat is released, while positive values suggest endothermic reactions.

- **Entropy (\( \Delta S^{\circ} \)):** Measures the disorder or randomness in a system. A positive change in entropy indicates an increase in disorder, often favorable for the spontaneity of a reaction.

- **Gibbs Free Energy (\( \Delta G^{\circ} \)):** Combines enthalpy and entropy to determine if a reaction is spontaneous. Calculated using the equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), where \( T \) is the temperature in Kelvin. A negative \(\Delta G^{\circ}\) signifies a spontaneous process at given conditions.

Understanding these properties helps in calculating the equilibrium constant, performance at different temperatures, and overall reaction behavior.

Equilibrium Constant

The equilibrium constant (\( K \)) quantifies the ratio of product and reactant concentrations when a reaction reaches equilibrium, which means no further net change in concentration. It's crucial for assessing the position of equilibrium and predicting the composition of a reaction mixture.

- **Relation to Gibbs Free Energy:** The equilibrium constant is linked to Gibbs free energy by the equation: \( K = e^{[-\Delta G^{\circ}/(RT)]} \), where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

- **Interpreting \( K \):** A large value of \( K \) (>1) suggests that the reaction favors the formation of products, while a small \( K \) (<1) implies that reactants are favored at equilibrium.

- **Temperature Dependency:** Since \( \Delta G^{\circ} \) and \( K \) depend on \( T \), changes in temperature can shift the equilibrium position, impacting \( K \).
Knowing the equilibrium constant is essential for predicting the yields and efficiencies of industrial processes like the Ostwald process.

Van't Hoff Equation

The Van’t Hoff equation describes how the equilibrium constant (\( K \)) changes with temperature, providing valuable insights into the effects of temperature on reaction equilibria. This equation is vital in understanding how industrial processes, like the Ostwald process, are affected by temperature changes.

- **Equation Form:** The Van't Hoff equation is given by \[ \ln \left( \frac{K_2}{K_1} \right) = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \( K_1 \) and \( K_2 \) are the equilibrium constants at temperatures \( T_1 \) and \( T_2 \) respectively. It shows that if \( \Delta H^{\circ} \) is known, we can predict how \( K \) changes as temperature varies.

- **Temperature Effects:** For exothermic reactions (\( \Delta H^{\circ} < 0 \)), an increase in temperature generally reduces \( K \), driving the reaction towards reactants. Conversely, for endothermic reactions, \( K \) increases with temperature, moving towards products.

Employing the Van't Hoff equation allows chemists to optimize reaction conditions and comprehend processes like the Ostwald process, guiding decisions on optimal operational temperatures.