Question

The equilibrium constant and standard Gibbs free energy change for ammonia synthesis at \(400^{\circ} \mathrm{C}\) or \(673^{\circ} \mathrm{K}\) are \(1.64 \times\) \(10^{-4}\) and \(11,657 \mathrm{cal} /\) mole, respectively. The equation for this reaction is (1) \(\quad \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) Calculate the equilibrium constants and standard free energy changes for (2) \((1 / 2) \mathrm{N}_{2}(\mathrm{~g})+3 / 2 \mathrm{H}_{2}(\mathrm{~g})=\mathrm{NH}_{3}(\mathrm{~g})\) and (3) \(2 \mathrm{NH}_{3}(\mathrm{~g}) \rightarrow \mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g})\)

Short answer

The equilibrium constants and standard Gibbs free energy changes for the given reactions are: Reaction (2): \(K_2 = 0.0128\), \(\Delta G_2 = 5828.5 \text{cal/mol}\) Reaction (3): \(K_3 = 6097.56\), \(\Delta G_3 = -11,657 \text{cal/mol}\)

Step-by-step solution

Find the equilibrium constants for reactions (2) and (3)

We can use the relationships between the reactions to find the equilibrium constants. The equilibrium constant of reaction (2) can be derived from reaction (1) since reaction (2) is just half of reaction (1). Let's denote the equilibrium constants for reactions (1), (2), and (3) as \(K_1\), \(K_2\), and \(K_3\), respectively. We are given that: \(K_1 = 1.64 \times 10^{-4}\) Now, Since reaction (2) is half of reaction (1), we can calculate the equilibrium constant of reaction (2) as the square root of the equilibrium constant of reaction (1), i.e., \(K_2 = \sqrt{K_1}\) For reaction (3), it's the reverse of reaction (1), so the equilibrium constant will be the inverse of the equilibrium constant of reaction (1), i.e., \(K_3 = \frac{1}{K_1}\)

Calculate the equilibrium constants for reactions (2) and (3)

Now we can calculate the equilibrium constants for both reactions (2) and (3) using the relationships found in Step 1: \(K_2 = \sqrt{K_1} = \sqrt{1.64 \times 10^{-4}} = 0.0128\) \(K_3 = \frac{1}{K_1} = \frac{1}{1.64 \times 10^{-4}} = 6097.56\)

Find the relationships between standard Gibbs free energy changes of the reactions

Let's denote the standard Gibbs free energy changes for reactions (1), (2), and (3) as \(\Delta G_1\), \(\Delta G_2\), and \(\Delta G_3\), respectively. We are given that: \(\Delta G_1 = 11,657 \text{cal/mol}\) The relationship between the standard Gibbs free energy change and the equilibrium constant is given by the following equation: \(\Delta G = -RT \ln K\), where \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. For our given problem, the temperature \(T\) is \(673 K\). Now we can find the relationships between the standard Gibbs free energy changes for the reactions: Since reaction (2) is half of reaction (1), \(\Delta G_2 = \frac{1}{2} \Delta G_1\) For reaction (3), it's the reverse of reaction (1), so \(\Delta G_3 = -\Delta G_1\)

Calculate the standard Gibbs free energy changes for reactions (2) and (3)

Now we can calculate the standard Gibbs free energy changes for reactions (2) and (3) using the relationships found in Step 3: \(\Delta G_2 = \frac{1}{2} \Delta G_1 = \frac{1}{2} \times 11,657 \text{cal/mol} = 5828.5 \text{cal/mol}\) \(\Delta G_3 = -\Delta G_1 = -11,657 \text{cal/mol}\) The equilibrium constants and standard Gibbs free energy changes for the reactions are as follows: Reaction (2): \(K_2 = 0.0128\), \(\Delta G_2 = 5828.5 \text{cal/mol}\) Reaction (3): \(K_3 = 6097.56\), \(\Delta G_3 = -11,657 \text{cal/mol}\)

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. It's represented by an equilibrium constant \(K\), which is a dimensionless value giving the ratio of product concentrations to reactant concentrations, each raised to the power of their respective coefficients in the balanced equation.

For the synthesis of ammonia, the equilibrium expression derived from the reaction \(\mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \leftrightarrow 2\mathrm{NH}_{3}(g)\) is written as \(K = \frac{[\mathrm{NH}_{3}]^2}{[\mathrm{N}_{2}][\mathrm{H}_{2}]^3}\). Understanding how to manipulate this expression is key when dealing with different stoichiometric versions of the reaction or when the reaction is reversed, as seen in the example provided.

Gibbs Free Energy Change

Gibbs free energy change \(\Delta G\) is a thermodynamic quantity that predicts the direction of chemical reactions and the feasibility of processes at constant temperature and pressure. If \(\Delta G\) is negative, the process is spontaneous, whereas if it's positive, the process is nonspontaneous. At equilibrium, \(\Delta G = 0\).

The equation \(\Delta G = -RT \ln K\) where \(R\) is the gas constant and \(T\) is the absolute temperature in Kelvins, relates \(\Delta G\) to the equilibrium constant \(K\). By knowing the Gibbs free energy change for a reaction, we can calculate \(K\) and vice versa, as demonstrated in the solution steps for the ammonia synthesis at different stoichiometries.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy. In the context of chemical reactions, it involves studying the energy changes that accompany chemical processes.

In addition to Gibbs free energy, two other essential thermodynamic functions are enthalpy \(\Delta H\) and entropy \(\Delta S\). The interplay between these quantities determines the spontaneity and equilibrium position of chemical reactions. It's crucial for students to understand that \(\Delta G\), \(\Delta H\), and \(\Delta S\) are state functions, depending only on the initial and final states of a system, not on the path taken to get there.

Ammonia Synthesis

Ammonia synthesis, typically referred to as the Haber-Bosch process, is an industrial method for producing ammonia (\(\mathrm{NH}_3\)) by reacting nitrogen gas (\(\mathrm{N}_2\)) with hydrogen gas (\(\mathrm{H}_2\)) under high pressure and temperature, using an iron catalyst. The balanced equation \(\mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g)\) shows a synthesis reaction where the equilibrium constant and the Gibbs free energy change provide information about the reaction's tendency towards product or reactant formation.

Students must realize that manipulating the stoichiometry or reversing the equation influences the value of the equilibrium constant and Gibbs free energy, as exemplified in the calculated constants and energy changes for the different reactions in the provided exercise.