Problem 80 - Chemistry An Atoms First Approach

Question

The equilibrium constant \(K\) for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{K}\) and a \(y\) -intercept of \(-14.51 .\) Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79.

Step-by-Step Solution

Answer

The values of ΔH⁰ and ΔS⁰ for the given reaction are \(-1.124 \times 10^5 \textrm{ J mol}^{-1}\) and \(-120.6 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), respectively.

Step 1: Write down the Van't Hoff equation

The Van't Hoff equation is given by the following formula: \[\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}\] Here, \(K\) is the equilibrium constant, \(R\) is the ideal gas constant, which is equal to \(8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), \(T\) is the temperature in Kelvin, ΔH⁰ is the change in enthalpy, and ΔS⁰ is the change in entropy.

Step 2: Express the equation in the form of y = mx + b

We can rewrite the Van't Hoff equation as: \[\ln K = \left(-\frac{\Delta H^{\circ}}{R}\right) \times \frac{1}{T} +\frac{\Delta S^{\circ}}{R}\] Now, comparing it with the linear equation form, \(y = mx + b\), we can identify the following: \(y = \ln K\) \(m = -\frac{\Delta H^{\circ}}{R}\) \(x = \frac{1}{T}\) \(b = \frac{\Delta S^{\circ}}{R}\)

Step 3: Use the given slope and y-intercept to calculate ΔH⁰ and ΔS⁰

We are given the slope, m, as \(1.352 \times 10^4 \textrm{K}\), and the y-intercept, b, as \(-14.51\). Therefore, we can find ΔH⁰ and ΔS⁰ as follows: First, find ΔH⁰ using the slope: \[m = -\frac{\Delta H^{\circ}}{R} \implies \Delta H^{\circ} = -m \times R\] Using the given slope, \(m = 1.352 \times 10^4 \textrm{K}\) and the gas constant, \(R = 8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), we get: \[\Delta H^{\circ} = -(1.352 \times 10^4 \textrm{K}) \times (8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}) = -1.124 \times 10^5 \textrm{ J mol}^{-1}\] Next, find ΔS⁰ using the y-intercept: \[b = \frac{\Delta S^{\circ}}{R} \implies \Delta S^{\circ} = b \times R\] Using the given y-intercept, \(b = -14.51\) and the gas constant, \(R = 8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), we get: \[\Delta S^{\circ} = (-14.51) \times (8.314 \textrm{ J K}^{-1}\textrm{mol}^{-1}) = -120.6 \textrm{ J K}^{-1}\textrm{mol}^{-1}\] Thus, the values of ΔH⁰ and ΔS⁰ for this reaction are \(-1.124 \times 10^5 \textrm{ J mol}^{-1}\) and \(-120.6 \textrm{ J K}^{-1}\textrm{mol}^{-1}\), respectively.

Key Concepts

Equilibrium Constant

For any chemical reaction, the equilibrium constant, denoted as \( K \), plays a crucial role in understanding the reaction's dynamics. It quantifies the ratio of product concentrations to reactant concentrations when a reaction reaches a state of equilibrium. This means no further net change in the concentrations of the reactants or products occurs.
Various factors can affect \( K \), including the nature of the reactants, the concentration, and the temperature. In a simple sense:

• A higher \( K \) value indicates that the reaction favors the formation of products.
• A lower \( K \) value suggests that reactants are favored.

In the context of Van't Hoff equation, \( K \) can vary with temperature, which is essential for deriving thermodynamic parameters like enthalpy and entropy changes.

Enthalpy Change (ΔH°)

Enthalpy change, expressed as \( \Delta H^{\circ} \), is a measure of the total energy change during a reaction. It indicates whether a reaction is exothermic (releases heat) or endothermic (absorbs heat).
When calculated, a negative \( \Delta H^{\circ} \) suggests an exothermic process, with the system releasing heat to the surroundings. Conversely, a positive \( \Delta H^{\circ} \) indicates an endothermic process, requiring heat from the surroundings.

• The Van't Hoff equation uses the slope from a graph of \( \ln(K) \) vs. \( 1/T \) to calculate \( \Delta H^{\circ} \).
• The slope is related to \( -\Delta H^{\circ}/R \), showing that the enthalpy change is directly proportional to this slope.

Understanding \( \Delta H^{\circ} \) is critical for predicting how temperature changes influence the reaction's equilibrium.

Entropy Change (ΔS°)

Entropy change, denoted as \( \Delta S^{\circ} \), reflects the disorder or randomness in a system. It's a measure of the amount of disorder introduced during a reaction.
When a reaction results in a higher disorder, \( \Delta S^{\circ} \) is positive, indicating a spontaneous increase in entropy. Conversely, a negative \( \Delta S^{\circ} \) means a decrease in randomness.

• In the Van't Hoff equation, we find \( \Delta S^{\circ} \) using the y-intercept, where \( b = \Delta S^{\circ} / R \).
• A positive \( \Delta S^{\circ} \) hints at higher spontaneity, often favoring the forward reaction.

Entropy is central to determining the direction and feasibility of chemical processes at varying temperatures.

Ideal Gas Constant

The ideal gas constant, often represented as \( R \), is a fundamental constant in thermodynamics. Its value is approximately \( 8.314 \,\mathrm{J \, K^{-1} \, mol^{-1}} \).
The purpose of \( R \) is to relate thermal energy to mechanical work in ideal gas laws and thermodynamic equations.

• In the context of the Van't Hoff equation, \( R \) acts as a proportionality constant connecting changes in enthalpy and entropy to temperature variations.
• It's a bridge between macroscopic and molecular-level descriptions of gases.

Understanding \( R \) is essential for applying thermodynamic calculations involving gases, linking energy transfer to physical phenomena like pressure and volume.