Problem 12 - General Chemistry: Principles and Modern Applications

Question

The Gibbs energies of formation, \(\Delta G_{f}^{Q}\), for \(\mathrm{KO}_{2}(\mathrm{s})\) and \(\mathrm{K}_{2} \mathrm{O}(\mathrm{s})\) are \(-240.59 \mathrm{kJmol}^{-1}\) and \(-322.09 \mathrm{kJmol}^{-1}\) respectively, at \(298 \mathrm{K}\). Calculate the equilibrium constant for the reaction below at \(298 \mathrm{K}\). Is \(\mathrm{KO}_{2}(\mathrm{s})\) thermodynamically stable with respect to \(\mathrm{K}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g})\) at \(298 \mathrm{K} ?\) $$ 2 \mathrm{KO}_{2}(\mathrm{s}) \longrightarrow \mathrm{K}_{2} \mathrm{O}(\mathrm{s})+\frac{3}{2} \mathrm{O}_{2}(\mathrm{g}) $$

Step-by-Step Solution

Answer

Substitute the values into the equations in Steps 1 and 2 to get the value for \(\Delta G\) and the equilibrium constant, K. Use the sign of the \(\Delta G\) to determine the thermodynamic stability in Step 3.

Step 1: Calculate the change in Gibbs free energy for the reaction

The change in Gibbs free energy for the reaction can be calculated using the Gibbs energies of formation of the products and reactants in the reaction. The formula is \(\Delta G_{reaction} = \Delta G_{f}^{products} - \Delta G_{f}^{reactants}\). Substituting the given values, we have, \(\Delta G = [(-322.09 kJ/mol) + (1.5*0)] - 2*(-240.59 kJ/mol)\) where 0 kJ/mol is the standard Gibbs free energy for \(O_2\).

Step 2: Calculate the equilibrium constant

The relationship between the Gibbs free energy and the equilibrium constant is given by the equation \(\Delta G = -R*T*log(K)\), where 'R' is the gas constant (8.314 J/K.mol) and 'T' is the temperature (in Kelvin). We can rearrange this equation to solve for the equilibrium constant, K: \(K = exp(-\Delta G / R*T)\). Here, T is given as 298 K. Substituting these values, we can find the equilibrium constant for the given reaction.

Step 3: Determine the thermodynamic stability of KO2

The sign of the Gibbs free energy change for the reaction determines whether a reaction is thermodynamically favourable. If \(\Delta G\) is negative, the reaction proceeds spontaneously, and the reverse reaction is not spontaneous. If \(\Delta G\) is positive, the reaction does not occur spontaneously, and the reverse reaction is favourable. Use this concept to determine the thermodynamic stability of KO2 with respect to K2O and \(O_2\).

Key Concepts

Gibbs free energy

Gibbs free energy, often denoted by the symbol \( \Delta G \), is a crucial concept in the field of chemistry and helps predict the spontaneity of a reaction.Understanding it requires a look at energy changes during a chemical reaction, alongside entropy changes.

The formula for Gibbs free energy is given by:

• \( \Delta G = \Delta H - T \Delta S \)

where:

• \( \Delta H \) is the change in enthalpy, or heat content,
• \( T \) is the temperature in Kelvin, and
• \( \Delta S \) is the change in entropy, or disorder.

Using this concept, the Gibbs free energy change for a chemical reaction can tell us whether the reaction is likely to be spontaneous.

Spontaneous reactions have a \( \Delta G \) that is negative, meaning they release free energy.This concept is essential when calculating equilibrium constants, as it relates to how far a reaction can go towards products under constant conditions. In our example, the Gibbs energy change \( \Delta G \) is calculated by finding the difference between the Gibbs energies of formation of the reactants and the products.

Thermodynamic stability

Thermodynamic stability essentially tells us how stable a substance is under given conditions.It is a measure of the tendency of a chemical species to remain in its current form without decomposing or reacting.

In the exercise, to determine the thermodynamic stability of \( \mathrm{KO}_{2} \) compared to \( \mathrm{K}_2\mathrm{O} \) and \( \mathrm{O}_2 \), we must consider the Gibbs free energy change \( \Delta G \) for the involved reaction.

• If \( \Delta G \) is negative, \( \mathrm{KO}_{2} \) is less stable, as it spontaneously converts into \( \mathrm{K}_2\mathrm{O} \) and \( \mathrm{O}_2 \).
• If \( \Delta G \) is positive, \( \mathrm{KO}_{2} \) is more stable, as the conversion does not proceed spontaneously.

The final value of \( \Delta G \) for the reaction can give us the insight needed to determine this stability.

Chemical equilibrium

Chemical equilibrium is when a chemical reaction reaches a state where the concentrations of reactants and products remain steady over time.This happens when the forward and reverse reactions occur at equal rates.

One key factor in understanding equilibrium is the equilibrium constant \( K \), which is directly related to \( \Delta G \) through the equation:

• \( \Delta G = -RT \ln K \)

Here, \( R \) is the gas constant and \( T \) is the temperature.The equilibrium constant \( K \) is a measure of how far a reaction will proceed before reaching equilibrium.

• If \( K \) is large, it indicates a greater concentration of products at equilibrium, suggesting the reaction favors product formation.
• If \( K \) is small, it means the reaction favors the reactants, and a substantial amount of reactants remain.

In the given exercise, calculating \( K \) will help us to understand the extent of the reaction and determine the equilibrium position.