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 \(85 .\)

Short answer

The standard enthalpy change (∆H°) for the given reaction is approximately \(-1.123 \times 10^{5} \mathrm{J/mol}\) and the standard entropy change (∆S°) is approximately \(-120.7 \mathrm{J/mol\cdot K}\).

Step-by-step solution

Recall the van't Hoff equation

The van't Hoff equation relates the change in the equilibrium constant (K) with temperature (T), standard enthalpy change (∆H°) and the standard entropy change (∆S°), and is given by: \[ \ln(K)= \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T} \right) + \frac{\Delta S^{\circ}}{R} \] Where R is the ideal gas constant which is equal to 8.314 J/mol·K.

Identify the slope and y-intercept from the given data

We are given that the slope of the graph of ln(K) versus 1/T is \(1.352 \times 10^{4} \mathrm{K}\) and y-intercept is \(-14.51\). Since we have a straight line, we can compare the van't Hoff equation to the equation of a straight line, which is: \[ y = mx + c \] Here, y corresponds to ln(K), m corresponds to the slope, x corresponds to 1/T, and c corresponds to the y-intercept.

Find the value of ∆H°

Comparing the van't Hoff equation to the equation of a straight line, we can see that the slope, m, is given by: \[ m = \frac{-\Delta H^{\circ}}{R} \] Now we can solve for ∆H° using the given slope value: \[ -\Delta H^{\circ} = m \times R = (1.352 \times 10^{4} \mathrm{K}) \times (8.314 \mathrm{J/mol\cdot K}) \] \[ \Delta H^{\circ} = -1.123 \times 10^{5} \mathrm{J/mol} \]

Find the value of ∆S°

Similarly, we can find the value of ∆S° using the given y-intercept, c: \[ c = \frac{\Delta S^{\circ}}{R} \] Now we can solve for ∆S° using the y-intercept value: \[ \Delta S^{\circ} = c \times R = (-14.51) \times (8.314 \mathrm{J/mol\cdot K}) \] \[ \Delta S^{\circ} = -120.7 \mathrm{J/mol\cdot K} \]

Conclusion

After solving for ∆H° and ∆S°, we can conclude that the standard enthalpy change (∆H°) for the given reaction is approximately \(-1.123 \times 10^{5} \mathrm{J/mol}\) and the standard entropy change (∆S°) is approximately \(-120.7 \mathrm{J/mol\cdot K}\).

Key concepts

Equilibrium Constant

In thermodynamic reactions, the equilibrium constant, represented as \( K \), is essential for understanding the balance between reactants and products at equilibrium. Specifically, for a reaction such as \( 2 \text{Cl}(g) \rightleftharpoons \text{Cl}_2(g) \), the equilibrium constant is given by the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. The value of \( K \) provides insight into the favorability of a reaction:

• If \( K \) is large, the products are favored, indicating equilibrium lies towards the products.
• If \( K \) is small, the reactants are favored, indicating equilibrium lies towards the reactants.

At different temperatures, the value of \( K \) can change, which is where the van't Hoff equation becomes useful for predicting how equilibrium shifts with temperature variations.

van't Hoff Equation

The van't Hoff equation provides a critical link between the temperature and the equilibrium constant of a reaction. It's expressed as:\[ \ln(K) = \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T} \right) + \frac{\Delta S^{\circ}}{R} \]Here:

• \( \Delta H^{\circ} \) is the standard enthalpy change.
• \( \Delta S^{\circ} \) is the standard entropy change.
• \( R \) is the ideal gas constant (8.314 J/mol·K).
• \( T \) is the temperature in Kelvin.

By plotting \( \ln(K) \) against \( 1/T \), a straight line is obtained, where the slope of this line is \( -\Delta H^{\circ}/R \) and the y-intercept is \( \Delta S^{\circ}/R \). This makes it a powerful tool to determine both the enthalpy and entropy changes in a reaction, using temperature-dependent equilibrium data.

Enthalpy Change

Enthalpy change, noted as \( \Delta H^{\circ} \), quantifies the heat absorbed or released in a reaction under constant pressure. In the context of the van't Hoff equation, the enthalpy change can be deduced from the slope of the line plotted between \( \ln(K) \) and \( 1/T \). Specifically, the relationship is:\[\Delta H^{\circ} = -\text{slope} \times R \]In the given reaction example, the slope \( 1.352 \times 10^4 \text{ K} \) allows us to calculate \( \Delta H^{\circ} \):\[\Delta H^{\circ} = -(1.352 \times 10^4) \times 8.314 = -1.123 \times 10^5 \text{ J/mol}\]A negative \( \Delta H^{\circ} \) indicates the reaction is exothermic, meaning it releases heat as it proceeds toward equilibrium.

Entropy Change

Entropy change, represented by \( \Delta S^{\circ} \), measures the disorder or the number of microstates available in a system. In the van't Hoff equation, this is linked to the y-intercept of the graph. Using the calculated y-intercept of \(-14.51\), \( \Delta S^{\circ} \) can be determined by:\[\Delta S^{\circ} = \text{y-intercept} \times R \]giving:\[\Delta S^{\circ} = (-14.51) \times 8.314 = -120.7 \text{ J/mol·K}\]A negative \( \Delta S^{\circ} \) suggests that the system's disorder decreases as the reaction moves towards equilibrium. This might indicate a more ordered state when \( \text{Cl}_2 \) is produced from \( \text{Cl} \) atoms, resulting in fewer individual gas molecules at equilibrium.