Problem 46 - Chemistry and Chemical Reactivity

Question

The standard free energy change, \(\Delta_{\mathrm{r}} G^{\circ},\) for the formation of \(\mathrm{O}_{3}(\mathrm{g})\) from \(\mathrm{O}_{2}(\mathrm{g})\) is \(+163.2 \mathrm{kJ} / \mathrm{mol}\) -ren at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{p}}\) at this temperature for the equilibrium $$3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})$$ Comment on the sign of \(\Delta_{r} G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}.\)

Step-by-Step Solution

Answer

The equilibrium constant \( K_{\mathrm{p}} \) at 25°C is approximately \( 4.73 \times 10^{-29} \), indicating a strong favorability towards the reactants. The positive \( \Delta_{r} G^{\circ} \) reflects a non-spontaneous formation of \( \mathrm{O}_{3}(g) \).

Step 1: Understand the Relationship between ΔrG° and Kp

The standard free energy change, \( \Delta_{r} G^{\circ} \), is related to the equilibrium constant \( K_{\mathrm{p}} \) by the equation:\[\Delta_{r} G^{\circ} = -RT \ln K_{\mathrm{p}}\]where \( R \) is the universal gas constant \( 8.314 \ \text{J/mol} \cdot \text{K} \), and \( T \) is the temperature in Kelvin. Convert the temperature from Celsius to Kelvin: \( 25^{\circ} \text{C} = 298 \text{K} \).

Step 2: Convert Energy Units

The given \( \Delta_{r} G^{\circ} = +163.2 \ \text{kJ/mol} \) must be converted into joules for calculation purposes:\[\Delta_{r} G^{\circ} = 163.2 \times 1000 \ \text{J/mol} = 163200 \ \text{J/mol} \].

Step 3: Plug Values into the Equation

Use \( \Delta_{r} G^{\circ} = -RT \ln K_{\mathrm{p}} \) to solve for \( K_{\mathrm{p}} \):\[163200 \ \text{J/mol} = -8.314 \ \text{J/mol} \cdot \text{K} \times 298 \ \text{K} \times \ln K_{\mathrm{p}}\].

Step 4: Solve for ln(Kp)

Rearrange the equation to solve for \( \ln K_{\mathrm{p}} \):\[\ln K_{\mathrm{p}} = -\frac{163200}{8.314 \times 298}\]Calculate the value:\[\ln K_{\mathrm{p}} = -65.61\].

Step 5: Calculate Kp

Exponentiate both sides to solve for \( K_{\mathrm{p}} \):\[K_{\mathrm{p}} = e^{-65.61} \approx 4.73 \times 10^{-29}\].

Step 6: Analyze Results

The positive \( \Delta_{r} G^{\circ} \) indicates that the formation of \( \mathrm{O}_{3}(g) \) is non-spontaneous under standard conditions. The very small magnitude of \( K_{\mathrm{p}} \) implies the reaction strongly favors the reactants \( \mathrm{O}_{2}(g) \) at equilibrium.

Key Concepts

free_energy_change

Free energy change, denoted as \( \Delta_r G^\circ \), is a significant concept in thermodynamics and chemistry. It represents the maximum amount of work that can be performed by a system at constant temperature and pressure.
In the context of a chemical reaction, it indicates whether the reaction can occur spontaneously. A negative \( \Delta_r G^\circ \) implies that the reaction can proceed without external aid, which means it is spontaneous. Conversely, a positive \( \Delta_r G^\circ \), like +163.2 kJ/mol for the formation of ozone from oxygen, suggests that the reaction is non-spontaneous under standard conditions.
To bring this idea into calculations, \( \Delta_r G^\circ \) is intricately linked to the equilibrium constant, \( K_p \), through the equation \( \Delta_r G^\circ = -RT \ln K_p \). This relation helps us understand not just the spontaneity but also the extent of a reaction's completion at equilibrium.

chemical_equilibrium

Chemical equilibrium is a state in which the forward and reverse reactions occur at equal rates, resulting in constant concentrations of reactants and products. In this condition, while reactions are still occurring, there is no net change in the concentration of the substances involved.
The equilibrium constant, \( K_p \), is a vital parameter in equilibrium scenarios. It quantitatively expresses the ratio of product concentrations to reactant concentrations at equilibrium, taking into account their respective coefficients in the balanced chemical equation.
A small \( K_p \) value, such as \( 4.73 \times 10^{-29} \), indicates that under equilibrium conditions, the concentration of reactants far exceeds that of products. Hence, in the context of the ozone formation reaction, the equilibrium strongly favors the reactants, \( O_2(g) \), over the formation of \( O_3(g) \).

reaction_spontaneity

Reaction spontaneity refers to the ability of a reaction to proceed on its own without the input of external energy. This characteristic is greatly influenced by the free energy change, \( \Delta_r G^\circ \). Whether a reaction is spontaneous or not is primarily indicated by the sign of \( \Delta_r G^\circ \).
If \( \Delta_r G^\circ \) is negative, the reaction can progress on its own, which implies spontaneous behavior. However, a positive \( \Delta_r G^\circ \), as seen in the conversion of \( O_2 \) to \( O_3 \), indicates that the process is not spontaneous under standard conditions, meaning energy input is required for the reaction to occur.
It's important to understand that spontaneous reactions do not always proceed at a noticeable rate. Factors like temperature, catalyst presence, and reaction kinetics can influence the actual occurrence of a reaction.

thermodynamics_in_chemistry

Thermodynamics is the branch of chemistry that deals with the relationships between heat, work, temperature, and energy in systems. It provides valuable insights into whether a chemical process can occur and how energetically favorable it is.
This field of study uses fundamental laws, such as the first and second laws of thermodynamics, to predict the feasibility and direction of a reaction. The first law emphasizes energy conservation, while the second law introduces entropy and the notion of increasing disorder.
When applied to chemical reactions, these principles allow us to evaluate essential parameters like \( \Delta_r G^\circ \), internal energy changes, and entropy alterations. For instance, in evaluating the ozone formation reaction, thermodynamics shows us the reaction's non-spontaneity under standard conditions and its significant favorability towards the reactants at equilibrium.