Problem 34 - Chemistry and Chemical Reactivity

Question

The formation of \(\mathbf{O}_{3}(\mathrm{g})\) from \(\mathbf{O}_{2}(\mathrm{g})\) has a standard free energy change, \(\Delta G^{\circ},\) of \(+163.2 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{p}}\) at this temperature. Comment on the connection between the sign of \(\Delta G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}\)

Step-by-Step Solution

Answer

The equilibrium constant \( K_{\mathrm{p}} \approx 5.85 \times 10^{-29} \), indicating a reaction favoring reactants due to positive \( \Delta G^{\circ} \).

Step 1: Understand the Relation Formula

To find the relation between the standard free energy change, \( \Delta G^{\circ} \), and the equilibrium constant, \( K_{\mathrm{p}} \), we use the formula: \[ \Delta G^{\circ} = -RT \ln K_{\mathrm{p}} \] where \( R \) is the universal gas constant \( 8.314 \: \mathrm{J/(mol \cdot K)} \) and \( T \) is the temperature in Kelvin.

Step 2: Convert Temperature to Kelvin

Convert the temperature from Celsius to Kelvin using the formula: \[ T(K) = T(^{\circ}\mathrm{C}) + 273.15 \]. Therefore, \[ T = 25 + 273.15 = 298.15 \: \mathrm{K} \].

Step 3: Calculate \( K_{\mathrm{p}} \)

Rearrange the formula to solve for \( K_{\mathrm{p}} \): \[ \ln K_{\mathrm{p}} = \frac{-\Delta G^{\circ}}{RT} \]. Plug in the known values \( \Delta G^{\circ} = 163.2 \: \mathrm{kJ/mol} = 163200 \: \mathrm{J/mol} \), \( R = 8.314 \: \mathrm{J/(mol \cdot K)} \), and \( T = 298.15 \: \mathrm{K} \). Compute: \[ \ln K_{\mathrm{p}} = \frac{-163200}{8.314 \times 298.15} \approx -65.48 \].

Step 4: Determine \( K_{\mathrm{p}} \) using Exponentiation

To find \( K_{\mathrm{p}} \), take the exponent of each side of the equation: \( \ln K_{\mathrm{p}} = -65.48 \). Thus, \[ K_{\mathrm{p}} = e^{-65.48} \approx 5.85 \times 10^{-29} \].

Step 5: Comment on \( \Delta G^{\circ} \) and \( K_{\mathrm{p}} \) Relation

Since \( \Delta G^{\circ} \) is positive, the reaction is non-spontaneous under standard conditions. This results in a very small \( K_{\mathrm{p}} \), indicating that the reactants are favored at equilibrium rather than the products (\( O_3 \)).

Key Concepts

Free Energy Change

The concept of free energy change, denoted as \( \Delta G^{\circ} \), is pivotal in understanding the spontaneity of a chemical reaction. It refers to the difference in energy between reactants and products at standard conditions (298 K or 25°C, and 1 atm pressure). When \( \Delta G^{\circ} \) is negative, the reaction occurs spontaneously, promoting conversion of reactants to products.
However, when \( \Delta G^{\circ} \) is positive, the reaction tends to be non-spontaneous, meaning it requires external energy input to proceed. In the case of \( O_3 \) formation, \( \Delta G^{\circ} = +163.2 \, \mathrm{kJ/mol} \) suggests that forming ozone from oxygen is not favored without energy input. This insight helps predict whether a chemical process will favor one set of substances over another under given conditions. By understanding free energy change, scientists can assess the feasibility of industrial processes and environmental interventions.

Thermodynamics

Thermodynamics is the study of energy and its transformations. In chemical reactions, it focuses on understanding energy changes, like heat and work, and how they affect the system – in this case, a reaction forming \( O_3 \) from \( O_2 \).
In the realm of thermodynamics for this exercise, one of the crucial equations used is \( \Delta G^{\circ} = -RT \ln K_{\mathrm{p}} \). This formula links free energy change with equilibrium constants, showcasing how thermodynamic principles quantify chemical equilibria.

• \( R \) is the universal gas constant \( 8.314 \, \mathrm{J/(mol \cdot K)} \)
• \( T \) represents temperature (Kelvin) converted from Celsius using the formula \( T(K) = T(°C) + 273.15 \)
• \( K_{\mathrm{p}} \) is the equilibrium constant for pressure-based equilibria

Students exploring thermodynamics through this lens understand how thermal energy influences reactions, bridging theoretical concepts with real-world phenomena. Calculations relating to these principles reveal why some reactions need added energy to reach equilibrium.

Chemical Equilibrium

Chemical equilibrium refers to the state in a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction, resulting in no net change in the concentration of reactants and products over time. It is represented by the equilibrium constant \( K_{\mathrm{p}} \) for reactions involving gases.
In analyzing the formation of ozone, \( K_{\mathrm{p}} = e^{-65.48} \approx 5.85 \times 10^{-29} \), calculated through the free energy relation. A tiny \( K_{\mathrm{p}} \) value indicates an equilibrium heavily favoring reactants, which aligns with the positive \( \Delta G^{\circ} \).
This relationship highlights:

• The extent to which reactants are converted to products
• Conditions under which equilibrium favors a particular direction
• The practical meaning of very small or large \( K_{\mathrm{p}} \) values

Understanding chemical equilibrium is vital for predicting how systems behave when subjected to external changes, such as temperature or pressure shifts. This helps in fields like chemical engineering, environmental science, and pharmacology.