Problem 28 - 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 \(+163.2 \mathrm{kJ} / \mathrm{mol}\) 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 \(\Delta G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}.\)

Step-by-Step Solution

Answer

The \(K_{\mathrm{p}}\) is approximately \(5.89 \times 10^{-29}\), indicating a non-spontaneous reaction with low \(\mathrm{O}_{3}\) concentration at equilibrium.

Step 1: Understand the Given Reaction and Parameters

We are given the reaction \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})\). We know that \(\Delta_{\mathrm{r}} G^{\circ} = +163.2 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\), which is \(298\,\mathrm{K}\). Our task is to calculate \(K_{\mathrm{p}}\) for this reaction.

Step 2: Use the Relation Between Free Energy and Equilibrium Constant

The relation between standard Gibbs free energy change and the equilibrium constant \(K_p\) is given by the equation: \[ \Delta_{\mathrm{r}} G^{\circ} = -RT \ln K_{\mathrm{p}} \]where \(R\) is the universal gas constant \(8.314 \mathrm{J/mol \cdot K}\) and \(T\) is the temperature in Kelvin \(298\,\mathrm{K}\).

Step 3: Solve for \(K_{\mathrm{p}}\)

Rearrange the equation to solve for \(K_{\mathrm{p}}\):\[ \ln K_{\mathrm{p}} = -\frac{\Delta_{\mathrm{r}} G^{\circ}}{RT} \]Substitute the values:\( \Delta_{\mathrm{r}} G^{\circ} = +163200 \mathrm{J/mol} \) (converted from kJ to J),\( R = 8.314 \mathrm{J/mol\cdot K} \), and \( T = 298 \, \mathrm{K} \):\[ \ln K_{\mathrm{p}} = -\frac{163200}{8.314 \times 298} \]Calculate \[ \ln K_{\mathrm{p}} = -65.59 \].

Step 4: Convert \(\ln K_{\mathrm{p}}\) to \(K_{\mathrm{p}}\)

To find \(K_{\mathrm{p}}\), take the exponential of both sides:\[ K_{\mathrm{p}} = e^{-65.59} \approx 5.89 \times 10^{-29} \].

Step 5: Interpret the Sign and Magnitude

Since \(\Delta_{\mathrm{r}} G^{\circ} > 0\), the formation of \(\mathrm{O}_{3}\) is non-spontaneous under standard conditions. The calculated \(K_{\mathrm{p}}\) is very small, indicating that at equilibrium, the concentration of \(\mathrm{O}_{3}\) is much lower than that of \(\mathrm{O}_{2}\).

Key Concepts

Gibbs free energy

Gibbs free energy, often represented as \( \Delta G \), is a critical concept in understanding chemical reactions and their spontaneity. It combines enthalpy and entropy to determine the direction of a chemical reaction. If \( \Delta G \) is negative, the reaction is spontaneous, meaning it can proceed without outside energy. If \( \Delta G \) is positive, the reaction is non-spontaneous, requiring energy to proceed. The standard free energy change, \( \Delta_{\text{r}} G^{\circ} \), is calculated under standard conditions, which we will explain later. The equation \[ \Delta_{\text{r}} G^{\circ} = -RT \ln K_{\text{p}} \] connects free energy with the equilibrium constant \( K_{\text{p}} \), highlighting that reactions with large negative \( \Delta_{\text{r}} G^{\circ} \) have large \( K_{\text{p}} \) values, indicating a spontaneous reaction.

non-spontaneous reactions

Non-spontaneous reactions are those that cannot occur on their own and require an input of energy to proceed. In these reactions, the Gibbs free energy change \( \Delta G \) is positive. For the reaction given in the exercise, \( 3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g}) \), \( \Delta_{\text{r}} G^{\circ} \) is +163.2 kJ/mol, indicating that it is non-spontaneous under standard conditions.
This means that without additional energy, the reaction will not favor the formation of \( \mathrm{O}_{3} \) and will remain primarily as \( \mathrm{O}_{2} \). Non-spontaneous processes are common in nature and often require catalysts or energy sources to occur.

• The small value of \( K_{\text{p}} \), calculated as approximately \( 5.89 \times 10^{-29} \), further underscores the non-spontaneity, arriving at a position where the formation of products is heavily disfavored.

gas constant

The gas constant, represented by \( R \), is a fundamental constant in chemistry and physics. It links various physical properties in formulas involving gases and is used frequently in calculations involving energy and reactions. Its value is approximately \( 8.314 \mathrm{J/mol\cdot K} \).
In the context of Gibbs free energy and the equilibrium constant, the gas constant connects temperature and energy quantities to help determine reaction spontaneity.
When using the formula \( \Delta_{\text{r}} G^{\circ} = -RT \ln K_{\text{p}} \), \( R \) plays a crucial role in converting energy units and aligning temperature and free energy relations.

standard conditions

Standard conditions, often abbreviated as STP (Standard Temperature and Pressure), are a set of conditions used for scientific experiments to allow for the comparability of different data sets and experiments. Specifically, in thermodynamics, standard conditions refer to a pressure of 1 atmosphere and a temperature of 298 Kelvin (25°C).
For the calculation of the standard Gibbs free energy \( \Delta_{\text{r}} G^{\circ} \), using standard conditions ensures that measured values are consistent and comparable across different studies.
In our exercise, these standard conditions help establish the baseline from which we calculate properties like the equilibrium constant \( K_{\text{p}} \). It highlights the fact that energy changes and reaction tendencies can vary significantly if conditions such as temperature or pressure are altered.