Question

The Haber process is the principal industrial route for converting nitrogen into ammonia: $${\mathrm{N}}_2(g)+3{\mathrm{H}}_2(g)⟶2{\mathrm{NH}}_3(g)$$ (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.

Short answer

(a) Nitrogen is being reduced, and hydrogen is neither oxidized nor reduced. (b) At room temperature, the equilibrium constant K is approximately $8.02\times {10}^4$. (c) The standard emf of the Haber process at room temperature is approximately 0.0901 V.

Step-by-step solution

(a) Identifying the oxidized and reduced elements

To identify the oxidized and reduced elements in the Haber process, we need to recall the definitions of oxidation and reduction. Oxidation refers to the loss of electrons, while reduction refers to the gain of electrons. Let's assign oxidation numbers to each element in the given reaction: $${\mathrm{N}}_2(g)+3{\mathrm{H}}_2(g)⟶2{\mathrm{NH}}_3(g)$$ In the $N_2$ molecule, each nitrogen atom has an oxidation number of 0 since they are in their elemental form. In the $H_2$ molecule, each hydrogen atom also has an oxidation number of +1 since they are in their elemental form. In the $N{H}_3$ molecule, nitrogen has an oxidation number of -3, and hydrogen has an oxidation number of +1. Comparing the oxidation numbers, nitrogen goes from an oxidation number of 0 in $N_2$ to -3 in $N{H}_3$. This is a decrease in the oxidation number, which means that nitrogen is being reduced. On the other hand, hydrogen's oxidation number remains the same (+1) throughout the reaction. It is neither oxidized nor reduced.

(b) Calculating the equilibrium constant at room temperature

Use the thermodynamic data provided in Appendix C to calculate the standard Gibbs free energy change, $\mathrm{\Delta }{G}^{\circ }$, for the reaction: $$\mathrm{\Delta }{G}^{\circ }=\sum \mathrm{\Delta }{G}_f^{\circ }(\text{products})-\sum \mathrm{\Delta }{G}_f^{\circ }(\text{reactants})$$ From Appendix C, we can find the standard Gibbs free energy of formation for each compound involved in the reaction: $\mathrm{\Delta }{G}_f^{\circ }(N_2)=0.0\text{kJ/mol}$ $\mathrm{\Delta }{G}_f^{\circ }(H_2)=0.0\text{kJ/mol}$ $\mathrm{\Delta }{G}_f^{\circ }(N{H}_3)=-16.45\text{kJ/mol}$ Now calculate the standard Gibbs free energy change for the reaction: $$\mathrm{\Delta }{G}^{\circ }=2\mathrm{\Delta }{G}_f^{\circ }({\text{NH}}_3)-(\mathrm{\Delta }{G}_f^{\circ }({\text{N}}_2)+3\mathrm{\Delta }{G}_f^{\circ }({\text{H}}_2))$$ $$\mathrm{\Delta }{G}^{\circ }=2(-16.45)-(0+3(0))=-32.9\text{kJ/mol}$$ Now, we can find the equilibrium constant, K, for the reaction at room temperature (25°C or 298K) using the relation: $$K=e^{-\frac{\mathrm{\Delta }{G}^{\circ }}{RT}}$$ where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. Plugging in the values, we get: $$K=e^{-\frac{-32.9\times {10}^3\text{J/mol}}{(8.314\text{J/mol·K})(298\text{K})}}\approx 8.02\times {10}^4$$

(c) Calculating the standard emf of the Haber process at room temperature

The standard emf (E°) of the Haber process at room temperature can be calculated using the relationship between the standard Gibbs free energy change, ΔG°, and the standard emf: $$\mathrm{\Delta }{G}^{\circ }=-nF{E}^{\circ }$$ Where n is the number of electrons transferred in the reaction (in this case, n = 6, as each nitrogen gains 3 electrons), F is Faraday's constant (96,485 C/mol), and E° is the standard emf we want to calculate. Rearrange the equation to find E°: $$E^{\circ }=-\frac{\mathrm{\Delta }{G}^{\circ }}{nF}$$ Now, plug in the values: $$E^{\circ }=-\frac{-32.9\times {10}^3\text{J/mol}}{6(96,485\text{C/mol})}\approx 0.0901\text{V}$$ The standard emf of the Haber process at room temperature is approximately 0.0901 V.

Key concepts

Oxidation and Reduction

Understanding the fundamental concepts of oxidation and reduction is crucial when studying chemical reactions such as the Haber process. In a chemical reaction, oxidation is the process where an element loses electrons, while reduction involves the gain of electrons.

Oxidation and reduction always occur together; they are two halves of a whole, often referred to as redox reactions. One species gets oxidized by losing electrons, and another gets reduced by gaining those electrons. To identify the changes in oxidation states, we look at the elements before and after the reaction.

In the Haber process, ${\mathrm{N}}_2(g)+3{\mathrm{H}}_2(g)⟶2{\mathrm{NH}}_3(g)$, nitrogen is reduced as its oxidation state goes from 0 to -3. This negative shift indicates a gain of electrons. Hydrogen's oxidation number stays constant; hence, it isn't oxidized or reduced in this process. Awareness of these changes helps us track electron flow and understand the mechanisms of chemical reactions.

Equilibrium Constant

The equilibrium constant (K) expresses the ratio of concentrations of products to reactants at equilibrium. It is an essential indicator of the extent to which a reaction will proceed under a given set of conditions.

In chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in a constant concentration of the reaction components. The equilibrium constant is unique for every reaction and depends only on the temperature, not on the initial concentrations of reactants and products.

For the Haber process, once we evaluate the standard Gibbs free energy change, we can determine K using the formula $K=e^{-\frac{\mathrm{\Delta }{G}^{\circ }}{RT}}$, where R is the gas constant and T is the temperature in Kelvin. K tells us how far the reaction moves towards completion; a large K indicates the products are favored, while a small K indicates reactants are favored at equilibrium.

Standard Gibbs Free Energy

The standard Gibbs free energy ($\mathrm{\Delta }{G}^{\circ }$) is crucial in predicting the direction of a chemical reaction and the point at which the reaction comes to equilibrium. It represents the maximum reversible work that can be done by a chemical reaction at constant temperature and pressure.

This thermodynamic quantity combines the concepts of enthalpy and entropy to predict the spontaneity of a process. A negative $\mathrm{\Delta }{G}^{\circ }$ indicates that a reaction is spontaneous in the forward direction under standard conditions. In contrast, a positive value suggests a non-spontaneous process, requiring input of energy to proceed.

The calculation for the Haber process involves subtracting the Gibbs free energy of the reactants from that of the products. This evaluation informs us not just whether ammonia formation is spontaneous, but also how much energy changes in the process, contributing to our understanding of the reaction's efficiency and potential for work.

Standard EMF

Electromotive force (emf) is a measurement of the energy provided by a battery or galvanic cell per coulomb of charge passing through. In the context of the Haber process, the standard emf (E°) is linked to the chemical potential energy change when nitrogen and hydrogen react to form ammonia.

The standard emf can be directly related to the standard Gibbs free energy change through the equation $\mathrm{\Delta }{G}^{\circ }=-nF{E}^{\circ }$, where F is Faraday's constant and n is the number of moles of electrons exchanged in the reaction. In this equation, a negative $\mathrm{\Delta }{G}^{\circ }$ results in a positive E°, suggesting the reaction can operate as a galvanic cell, providing electrical current.

For the Haber process, the calculation of the standard emf gives us insight into the electric potential of the reaction. This value helps us understand the electrical work that could be extracted from the chemical reaction under standard conditions, bridging the gap between chemistry and electricity in energy conversion processes.