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 71 .

Short answer

The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction are approximately -112400 J/mol and -120.6 J/K·mol, respectively.

Step-by-step solution

Recall the Van 't Hoff equation

Recall the Van 't Hoff equation, which relates the equilibrium constant K, temperature (T), and the Gibbs free energy change ∆G°: \( \ln K = -\frac{\Delta G^\circ}{R \cdot T} \) where R is the gas constant (approximately 8.314 J/K·mol) and T is the temperature in Kelvin.

Relate ∆G° to ∆H° and ∆S°

Now we need to establish a connection between ∆G°, ∆H°, and ∆S°. The relevant relationship is given by the Gibbs-Helmholtz equation: \( \Delta G^\circ = \Delta H^\circ - T \cdot \Delta S^\circ \) Now we can substitute this expression for ∆G° into the Van 't Hoff equation: \( \ln K = -\frac{\Delta H^\circ - T \cdot \Delta S^\circ}{R \cdot T} \)

Rearrange the equation for ln(K) vs. 1/T

We need to rewrite this equation to match the given information about the graph: \( \ln K = -\frac{\Delta H^\circ}{R} \cdot \frac{1}{T} + \frac{\Delta S^\circ}{R} \) This equation is in the form of a linear equation y = mx + b, where y = ln(K), x = 1/T, m = -∆H°/R, and b = ∆S°/R. We are given the slope (m = -∆H°/R) and y-intercept (b = ∆S°/R) as 13520 K and -14.51, respectively.

Calculate ∆H° and ∆S°

Now that we have the slope and y-intercept values, we can calculate ∆H° and ∆S°: For the slope, m = -∆H°/R, we have: \( -\Delta H^\circ = m \cdot R \) \( \Delta H^\circ = -m \cdot R = -(1.352 \times 10^4 \,\text{K}) \cdot (8.314 \,\text{J/K·mol}) \) \( \Delta H^\circ \approx -112400 \, \text{J/mol} \) For the y-intercept, b = ∆S°/R, we have: \( \Delta S^\circ = b \cdot R = (-14.51) \cdot (8.314 \, \text{J/K·mol}) \) \( \Delta S^\circ \approx -120.6 \, \text{J/K·mol} \)

Final Answer

The values of ∆H° and ∆S° for this reaction are approximately -112400 J/mol and -120.6 J/K·mol, respectively.

Key concepts

Van 't Hoff equation

The Van 't Hoff equation comes into play when exploring the relationship between equilibrium constants and temperature. It sets the stage for understanding how chemical equilibria respond to temperature variations.

Mathematically, it is expressed as:- \(\ln K = -\frac{\Delta G^\circ}{R \cdot T}\)- Here, \(K\) represents the equilibrium constant.- \(\Delta G^\circ\) is the Gibbs free energy change.- \(R\) is the universal gas constant, approximately 8.314 J/K·mol.- \(T\) is the temperature in Kelvin.

This equation suggests that as temperature changes, the equilibrium constant will adjust to balance the reaction. The Van 't Hoff equation also indicates a linear relationship between \(\ln K\) and \(1/T\), which can be graphically represented by a straight line. This insight is crucial for predicting the behavior of reactions under different thermal conditions.

Gibbs-Helmholtz equation

The Gibbs-Helmholtz equation introduces a connection between the Gibbs free energy change (\(\Delta G^\circ\)), enthalpy change (\(\Delta H^\circ\)), and entropy change (\(\Delta S^\circ\)) for a reaction. It's a cornerstone for understanding how energy changes drive chemical processes.

Expressed as:- \(\Delta G^\circ = \Delta H^\circ - T \cdot \Delta S^\circ\)- It reflects how enthalpy (heat absorbed or released) and entropy (disorder) changes influence the spontaneity of reactions.

By inserting this equation into the Van 't Hoff equation, we derive a form that links these thermodynamic quantities to the equilibrium constant:- \(\ln K = -\frac{\Delta H^\circ - T \cdot \Delta S^\circ}{R \cdot T}\)- This rearrangement highlights how the balance of enthalpic and entropic contributions affects equilibrium at varying temperatures.

Thermodynamics

Thermodynamics is the realm of physics concerned with heat, energy, and how they influence matter. In the context of chemical reactions, it helps decipher how reactions either absorb or release energy, as well as their efficiency.

The main laws of thermodynamics that apply are: - The First Law: Energy cannot be created or destroyed, only transformed. It ensures energy conservation. - The Second Law: Entropy, a measure of disorder, tends to increase over time, dictating the spontaneous direction of processes.

These principles guide the calculations involving enthalpy and entropy changes that accompany reactions. A solid grasp of thermodynamics enables chemists to anticipate how reactions will progress and adjust conditions to control them.

Gibbs free energy

Gibbs free energy (\(\Delta G^\circ\)) is a crucial concept for predicting the spontaneity of chemical reactions. It integrates the effects of enthalpy, entropy, and temperature on a reaction's feasbility.

When \(\Delta G^\circ\) is:- Negative: The reaction is spontaneous and will proceed under the given conditions.- Positive: The reaction is non-spontaneous, requiring energy input to proceed.- Zero: The system is at equilibrium, with no net change occurring.

This makes Gibbs free energy an invaluable tool in designing chemical processes and understanding the energy landscapes of reactions.

Enthalpy and entropy changes

Enthalpy (\(\Delta H^\circ\)) and entropy (\(\Delta S^\circ\)) are two fundamental components of thermodynamics that describe energy changes in reactions.

• Enthalpy change (\(\Delta H^\circ\)) measures the total heat change in a system. It tells whether a reaction absorbs heat (endothermic) or releases heat (exothermic).
• Entropy change (\(\Delta S^\circ\)) provides a measure of disorder or randomness. An increase in entropy means more disorder.

In the context of reactions:- The balance between \(\Delta H^\circ\) and \(\Delta S^\circ\) influences equilibrium.- A reaction with a negative \(\Delta H^\circ\) and positive \(\Delta S^\circ\) is usually spontaneous at all temperatures.

Understanding these changes helps chemists fine-tune reaction conditions to favor desired outcomes.