Problem 111 - Chemistry: The Central Science

Question

The Haber process is the principal industrial route for converting nitrogen into ammonia: $$ \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ (a) What is being oxidized, and what is being reduced? (b) Using the thermodynamic data in Appendix \(\mathrm{C}\), calculate the equilibrium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature.

Step-by-Step Solution

Answer

(a) Nitrogen is being reduced, and Hydrogen is being oxidized. (b) The equilibrium constant for the Haber process at room temperature is approximately 83.4. (c) The standard emf of the Haber process at room temperature is 0.094 V.

Step 1: (a) Identifying Oxidation and Reduction

In the Haber process, nitrogen gas (N2) reacts with hydrogen gas (H2) to form ammonia gas (NH3). To determine the molecules undergoing oxidation and reduction, we need to analyze the change in oxidation state of the elements. The oxidation state of: - Nitrogen in N2 is 0. - Hydrogen in H2 is 0. - Nitrogen in NH3 is -3. - Hydrogen in NH3 is +1. The nitrogen's oxidation state changed from 0 to -3, which means it gained electrons, so it was reduced. The hydrogen’s oxidation state changed from 0 to +1, which means it lost electrons and was oxidized. Answer: Nitrogen is being reduced, and Hydrogen is being oxidized.

Step 2: (b) Calculating the Equilibrium Constant at Room Temperature

To calculate the equilibrium constant at room temperature, we need to find the Gibbs free energy change (∆G) for the Haber process from Appendix C, and then use the equation: \(K = e^{(-\Delta G/RT)}\) where: - K is the equilibrium constant - ∆G is the Gibbs free energy change - R is the universal gas constant (8.314 J/mol·K) - T is the temperature in Kelvin (room temperature is approximately 298 K) From Appendix C: - ∆G(N2) = 0 J/mol - ∆G(H2) = 0 J/mol - ∆G(NH3) = -16.45 kJ/mol The balanced chemical equation for the Haber process is: \(N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)\) Applying the Hess law, we calculate ∆G for the process as follows: \(\Delta G = 2 \times \Delta G(NH_{3}) - (\Delta G(N_{2}) + 3 \times \Delta G(H_{2}))\) \(\Delta G = 2 \times (-16.45 \,\text{kJ/mol}) - (0 + 0) = -32.9 \,\text{kJ/mol}\) Now let's calculate the equilibrium constant K: \(K = e^{(-\Delta G/RT)}\) \(K = e^{(32900 \,\text{J/mol} / (8.314 \,\text{J/mol·K} \times 298 \,\text{K}))}\) \(K = e^{(4.425)}\) \(K \approx 83.4\) Answer: The equilibrium constant for the Haber process at room temperature is approximately 83.4.

Step 3: (c) Calculating the Standard Emf at Room Temperature

To calculate the standard emf (E°) of the Haber process, we need to use the relationship between Gibbs free energy change (∆G), the number of electrons transferred (n), and the standard emf (E°): \(\Delta G = -nFE^{\circ}\) where: - F is the Faraday constant (96,485 C/mol) From the balanced chemical equation and the change in oxidation states, we can see that 6 electrons are transferred in the process (3 H2 molecules lose two electrons each). Now, let's solve for E°: \(E^{\circ} = - \frac{\Delta G}{nF}\) \(E^{\circ} = - \frac{-32.9 \times 10^3 \,\text{J/mol}}{6 \times 96,485 \,\text{C/mol}}\) \(E^{\circ} = 0.094 \,\text{V}\) Answer: The standard emf of the Haber process at room temperature is 0.094 V.

Key Concepts

Oxidation and Reduction

Understanding oxidation and reduction is fundamental in chemistry, especially when analyzing reactions like the Haber process. The main takeaway here is that oxidation involves the loss of electrons and increase in oxidation state, whereas reduction involves the gain of electrons and decrease in oxidation state.

In the context of the Haber process, hydrogen atoms lose electrons and are therefore oxidized, while nitrogen gains electrons and is reduced. This transfer of electrons is essential for the chemical transformation that occurs, where diatomic nitrogen and hydrogen gases react to form ammonia. Remember that in determining what is oxidized and reduced, one should always look at the changes in oxidation states of the elements involved in the reaction.

Equilibrium Constant Calculation

Calculating the equilibrium constant, symbolized by K, is a way to understand the extent to which a reaction proceeds at a given temperature. To find the value of K, we often use the Gibbs free energy change, \(\Delta G\), as it relates to the energy available to do work in a reaction system.

For the Haber process, using the equation \(K = e^{(-\Delta G/RT)}\)—where R is the universal gas constant and T is temperature in Kelvin—allows us to compute the equilibrium constant based on the thermodynamic data provided. Keep in mind that this value indicates the ratio between the concentration of the products (ammonia) and the reactants (nitrogen and hydrogen) at equilibrium. A higher K value suggests a reaction heavily favoring product formation under standard conditions.

Standard EMF Calculation

The standard electromotive force (emf), \(E^\circ\), is indicative of a reaction’s capacity to do electrical work based on the transfer of electrons during a redox reaction. It is a crucial concept in electrochemistry but also, as the Haber process illustrates, relevant in other chemical reactions.

To calculate the standard emf, one uses the relationship between \(\Delta G\), the number of electrons involved in the reaction (n), and the Faraday constant (F). This process shows the direct connection between the chemical and electrical aspects of a reaction. The formula \(E^\circ = - \frac{\Delta G}{nF}\) gives you the standard emf in volts, allowing you to gauge the feasibility and spontaneity of the reaction. For the Haber process, this calculation reveals a positive emf, highlighting the reaction’s tendency to proceed as written under standard conditions.