Equilibrium Constant and the Gibbs Free Energy

Oct 13, 2025 by ludoplayers7.com 37 views

Introduction

The reaction $I_2(s) \rightleftharpoons I_2(g)$ is a classic example of a phase transition, where iodine in its solid state (s) reacts to form iodine gas (g). This reaction is characterized by a specific enthalpy change ( $\Delta H$ ) and entropy change ( $\Delta S$ ). In this article, we will use the given values of $\Delta H$ and $\Delta S$ to determine the temperature at which the reaction is at equilibrium.

Equilibrium Constant and the Gibbs Free Energy

At equilibrium, the Gibbs free energy change ( $\Delta G$ ) is equal to zero. The Gibbs free energy change is related to the equilibrium constant ( $K$ ) by the following equation:

$\Delta G = -RT \ln K$

where $R$ is the gas constant, $T$ is the temperature in Kelvin, and $\ln K$ is the natural logarithm of the equilibrium constant.

The equilibrium constant ( $K$ ) is related to the enthalpy change ( $\Delta H$ ) and entropy change ( $\Delta S$ ) by the following equation:

$\ln K = \frac{\Delta S}{R} - \frac{\Delta H}{RT}$

Substituting Values and Solving for Temperature

We are given the following values:

$\Delta H = 62.4 \, \text{kJ/mol}$
$\Delta S = 0.145 \, \text{kJ/(mol} \cdot \text{K)}$
$R = 8.314 \, \text{J/(mol} \cdot \text{K)}$

We can substitute these values into the equation for $\ln K$ :

$\ln K = \frac{0.145 \, \text{kJ/(mol} \cdot \text{K)}}{8.314 \, \text{J/(mol} \cdot \text{K)}} - \frac{62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T}$

Simplifying the equation, we get:

$\ln K = 0.0175 - \frac{62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T}$

Equating $\Delta G$ to Zero and Solving for Temperature

At equilibrium, $\Delta G$ is equal to zero. We can set up the following equation:

$-RT \ln K = 0$

Substituting the expression for $\ln K$ from the previous equation, we get:

$-RT \left(0.0175 - \frac{62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T}\right) = 0$

Simplifying the equation, we get:

$-RT \cdot 0.0175 + \frac{RT \cdot 62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T} = 0$

Solving for Temperature

We can simplify the equation further by canceling out the $T$ terms:

$-RT \cdot 0.0175 + \frac{RT \cdot 62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T} = 0$

Q: What is the equilibrium temperature of the iodine reaction?

A: The equilibrium temperature of the iodine reaction can be calculated using the given values of $\Delta H$ and $\Delta S$ . By substituting these values into the equation for $\ln K$ , we can solve for the temperature at which the reaction is at equilibrium.

Q: What is the significance of the Gibbs free energy change ( $\Delta G$ ) in this reaction?

A: At equilibrium, the Gibbs free energy change ( $\Delta G$ ) is equal to zero. This means that the reaction is at a state of balance, where the forward and reverse reactions occur at the same rate.

Q: How is the equilibrium constant ( $K$ ) related to the enthalpy change ( $\Delta H$ ) and entropy change ( $\Delta S$ )?

A: The equilibrium constant ( $K$ ) is related to the enthalpy change ( $\Delta H$ ) and entropy change ( $\Delta S$ ) by the following equation:

$\ln K = \frac{\Delta S}{R} - \frac{\Delta H}{RT}$

Q: What is the value of the gas constant ( $R$ ) used in this calculation?

A: The value of the gas constant ( $R$ ) used in this calculation is $8.314 \, \text{J/(mol} \cdot \text{K)}$ .

Q: How do we solve for the temperature at which the reaction is at equilibrium?

A: To solve for the temperature at which the reaction is at equilibrium, we can set up the following equation:

$-RT \ln K = 0$

Substituting the expression for $\ln K$ from the previous equation, we get:

$-RT \left(0.0175 - \frac{62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T}\right) = 0$

Simplifying the equation, we get:

$-RT \cdot 0.0175 + \frac{RT \cdot 62.4 \, \text{kJ/mol}}{8.314 \, \text{J/(mol} \cdot \text{K)} \cdot T} = 0$

Q: What is the final answer for the temperature at which the reaction is at equilibrium?

A: After simplifying the equation, we get:

$T = \frac{62.4 \, \text{kJ/mol}}{0.0175 \cdot 8.314 \, \text{J/(mol} \cdot \text{K)} \cdot 1000}$

$T = 373.2 \, \text{K}$

Therefore, the equilibrium temperature of the iodine reaction is $373.2 \, \text{K}$ .

Q: What is the significance of this result?

A: This result indicates that the iodine reaction is at equilibrium at a temperature of $373.2 \, \text{K}$ . This means that the forward and reverse reactions occur at the same rate, and the reaction is in a state of balance.

Q: How can this result be applied in real-world scenarios?

A: This result can be applied in real-world scenarios where the iodine reaction is used as a model system to study phase transitions and chemical equilibria. It can also be used to design experiments and predict the behavior of the reaction under different conditions.