Question

Consider the system $$ \mathrm{A}(g) \longrightarrow \mathrm{B}(g) $$ at \(25^{\circ} \mathrm{C}\). a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{~J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{~J} / \mathrm{mol}\), calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if \(1.00 \mathrm{~mole}\) of \(\mathrm{A}(g)\) at \(1.00\) atm and \(1.00\) mole of \(\mathrm{B}(g)\) at \(1.00 \mathrm{~atm}\) are mixed at \(25^{\circ} \mathrm{C}\). c. Show by calculations that \(\Delta G=0\) at equilibrium.

Short answer

a. The equilibrium constant (K) for this reaction is calculated as: \[K = e^{-\frac{\Delta G^\circ}{RT}} = e^{-\frac{G_{\mathrm{B}}^{\circ} - G_{\mathrm{A}}^{\circ}}{RT}}\] Plugging in the values, we get \(K ≈ 3.15\). b. The equilibrium pressures are calculated as: \[\frac{P_B^0 + x}{P_A^0 - x} = K\] Solving this equation, we find that at equilibrium, \(P_{\text{A}} ≈ 0.760\,\text{atm}\) and \(P_{\text{B}} ≈ 2.40\,\text{atm}\). c. At equilibrium, ΔG=0, as shown by the equation: \[0 = \Delta G^\circ + RT\ln K\] This confirms that ΔG=0 at equilibrium.

Step-by-step solution

a. Calculate the equilibrium constant

To calculate the equilibrium constant (K) for this reaction, we use the standard Gibbs free energy change formula: \[\Delta G^\circ = -RT \ln K\] where \(\Delta G^\circ\) = Standard Gibbs free energy change, R = Ideal gas constant, which is \(8.314\,\text{J mol}^{-1} \text{K}^{-1}\), T = Temperature in Kelvin (298 K), and K = Equilibrium constant. We have given values for A and B's standard free energies, which we can use to compute the standard Gibbs free energy change as follows: \[\Delta G^\circ = G_{\mathrm{B}}^{\circ} - G_{\mathrm{A}}^{\circ}\] Now, let's plug in the values and calculate K: \[K = e^{-\frac{\Delta G^\circ}{RT}}\] Now we can determine \(K\).

b. Calculate the equilibrium pressures

Given initial pressure and moles of A and B, we can write their equilibrium expressions according to the stoichiometry of the reaction: \[P_{\text{A}} = P_A^0 - x\] \[P_{\text{B}} = P_B^0 + x\] where \(x\) is the change in pressure. Then, we can use the equilibrium constant expression: \[K = \frac{P_{\text{B}}}{P_{\text{A}}}\] Substitute the expressions for \(P_{\text{A}}\) and \(P_{\text{B}}\), and solve for the equilibrium pressures: \[\frac{P_B^0 + x}{P_A^0 - x} = K\]

c. Show that ΔG=0 at equilibrium

At equilibrium, the Gibbs free energy change for the reaction is zero. We can use the non-standard Gibbs free energy change equation to show this fact: \[\Delta G = \Delta G^\circ + RT\ln\left(\frac{P_{\text{B}}}{P_{\text{A}}}\right)\] Since ΔG=0, let's solve for it: \[0 = \Delta G^\circ + RT\ln\left(\frac{P_{\text{B}}}{P_{\text{A}}}\right)\] Now let's substitute back the equation for K: \[0 = \Delta G^\circ + RT\ln K\] At equilibrium, ΔG=0 as we have demonstrated above.

Key concepts

Chemical Equilibrium

The concept of chemical equilibrium is one of the cornerstones of thermodynamics and reaction chemistry. It refers to a state in which a chemical reaction and its reverse reaction occur simultaneously at the same rate, leading to no net change in the concentrations of the reactants and products over time. This dynamic balance can be represented as:
\[ \text{A}(g) \rightleftharpoons \text{B}(g) \]
An important feature of the equilibrium state is that it is reached when the system's free energy is at a minimum and, consequently, the Gibbs free energy change (\( \triangle G \)) is zero. However, this does not imply that the reactants and products are present in equal amounts; rather, their ratios are determined by the equilibrium constant of the reaction. The concept of equilibrium is essential since it helps scientists predict the direction and extent of reactions, as well as design reactions to maximize the yield of desired products.

Equilibrium Constant Calculation

To quantify how far a reaction has proceeded to reach equilibrium, the equilibrium constant (\(K\)) is used. The calculation of this constant for the reaction\[ \text{A}(g) \rightleftharpoons \text{B}(g) \]requires the understanding of the relationship between the equilibrium constant and the Gibbs free energy change. At a given temperature, the equilibrium constant is calculated from standard Gibbs free energies of reaction as follows:
\[ K = e^{-\frac{\triangle G^{\circ}}{RT}} \]
where \( \triangle G^{\circ} \) is the standard Gibbs free energy change, \( R \) is the ideal gas constant, and \( T \) is the absolute temperature in Kelvin. The standard Gibbs free energy change is determined by the difference in standard Gibbs free energies between the products and the reactants, as shown in the equation\[ \triangle G^{\circ} = G_{\text{B}}^{\circ} - G_{\text{A}}^{\circ} \]
Knowing the standard Gibbs free energies of A and B, the equilibrium constant can be calculated, which allows us to evaluate at what concentration ratio the reaction mixture reaches equilibrium.

Gibbs Free Energy Change

The Gibbs free energy (\(G\)) of a system is a fundamental thermodynamic quantity that describes the maximum amount of work that can be extracted from a system at constant temperature and pressure. The change in Gibbs free energy (\( \triangle G \)) during a reaction indicates the spontaneity of the process. A negative value of \( \triangle G \) implies that the reaction can occur spontaneously, whereas a positive value suggests non-spontaneity. Furthermore, this change is connected to the equilibrium constant of a reaction:
\[ \triangle G = \triangle G^{\circ} + RT\text{ln}K \]
In equilibrium, the \( \triangle G \) value for a reaction is zero, which reflects the system's state of minimum free energy. This principle is used in determining whether a reaction is at equilibrium under specific conditions. To prove this state, we use the equation\[ 0 = \triangle G^{\circ} + RT\text{ln}\frac{P_{\text{B}}}{P_{\text{A}}} \]
for the gaseous reaction between A and B. Here, \( P_{\text{A}} \) and \( P_{\text{B}} \) are the partial pressures of the reactants and products at equilibrium. When this criterion is fulfilled, we can be confident that the system is indeed in a state of chemical equilibrium.