Problem 11
Question
The Gibbs energies of formation, \(\Delta G_{\mathrm{f}}^{\circ},\) for \(\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) are \(-379.09 \mathrm{kJ} \mathrm{mol}^{-1}\) and \(-449.63 \mathrm{kJ} \mathrm{mol}^{-1}\) respectively, at 298 K. Calculate the equilibrium constant for the reaction below at \(298 \mathrm{K} .\) Is \(\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s})\) thermodynamically stable with respect to \(\mathrm{Na}_{2} \mathrm{O}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g})\) at \(298 \mathrm{K} ?\) $$ \mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) $$
Step-by-Step Solution
Verified Answer
The equilibrium constant for the reaction at 298 K is \(4.811x10^{-13}\). \(\mathrm{Na}_{2}\mathrm{O}_{2}\) is thermodynamically stable with respect to \(\mathrm{Na}_{2}\mathrm{O}\) and \(\mathrm{O}_{2}\), but under standard conditions, it is not thermodynamically favorable for these substances to react to form \(\mathrm{Na}_{2}\mathrm{O}_{2}\).
1Step 1: Determine the reaction Gibbs free energy change
The Gibbs free energy change for a reaction \(\Delta G\) is calculated by subtracting the sum of the Gibbs energies of formation of reactants from sum of Gibbs energies of formation of the products. For the reaction: \[\mathrm{Na}_{2} \mathrm{O}_{2}(\mathrm{s}) \longrightarrow \mathrm{Na}_{2} \mathrm{O}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}),\] \(\Delta G_{rxn} = \Delta G_{f}^{\circ}(\mathrm{Na}_{2}\mathrm{O}(\mathrm{s})) + 0.5*\Delta G_{f}^{\circ}(\mathrm{O}_{2}(\mathrm{g})) - \Delta G_{f}^{\circ}(\mathrm{Na}_{2}\mathrm{O}_{2}(\mathrm{s}))\). Substituting the given values, we have \(\Delta G_{rxn} = -379.09 kJ/mol + 0.5*0 kJ/mol -(-449.63 kJ/mol) = 70.54 kJ/mol.\)
2Step 2: Calculate the equilibrium constant
The equilibrium constant K can be calculated from the Gibbs free energy change using the equation: \[K = e^{(-\Delta G_{rxn})/RT},\] where \(R\) is the gas constant (8.314 J/mol·K) and \(T\) is the temperature in Kelvin. Substituting the given temperature (298 K), we have: \[K = e^{(-70540 J/mol)/(8.314 J/mol·K*298 K)} = e^{-28.42} = 4.811x10^{-13}.\]
3Step 3: Discuss the thermodynamic stability
The value of \(\Delta G_{rxn}\) is positive, indicating that the reaction is non-spontaneous under standard conditions. However, this does not necessarily mean that \(\mathrm{Na}_{2}\mathrm{O}_{2}\) is not thermodynamically stable. It simply means that, under standard conditions (\(298 K\), 1 atm), it is more thermodynamically favorable for \(\mathrm{Na}_{2}\mathrm{O}\) and \(\mathrm{O}_{2}\) to remain as separate entities rather than react to form \(\mathrm{Na}_{2}\mathrm{O}_{2}\).
Key Concepts
Equilibrium ConstantThermodynamic StabilityReaction Gibbs Free Energy Change
Equilibrium Constant
The equilibrium constant (K) is a dimensionless number that provides a measure of the position of equilibrium for a reversible chemical reaction. In the context of Gibbs free energy, the equilibrium constant can be calculated using the expression
\[K = e^{(-\frac{\Delta G_{rxn}}{RT})}\]
where \(\Delta G_{rxn}\) is the change in Gibbs free energy for the reaction, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvins. A K value less than one indicates that reactants are favored at equilibrium, while a K value greater than one indicates that products are favored.
In the exercise, the equilibrium constant for the specified reaction was calculated to be approximately \(4.811 \times 10^{-13}\), which implies that the reactants are heavily favored at equilibrium. This low value suggests that under standard conditions, the conversion of \(\mathrm{Na}_2\mathrm{O}_2\) to the products is not favored, meaning that at equilibrium, there will be a significantly higher concentration of \(\mathrm{Na}_2\mathrm{O}_2\) compared to that of \(\mathrm{Na}_2\mathrm{O}\) and \(\mathrm{O}_2\).
\[K = e^{(-\frac{\Delta G_{rxn}}{RT})}\]
where \(\Delta G_{rxn}\) is the change in Gibbs free energy for the reaction, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvins. A K value less than one indicates that reactants are favored at equilibrium, while a K value greater than one indicates that products are favored.
In the exercise, the equilibrium constant for the specified reaction was calculated to be approximately \(4.811 \times 10^{-13}\), which implies that the reactants are heavily favored at equilibrium. This low value suggests that under standard conditions, the conversion of \(\mathrm{Na}_2\mathrm{O}_2\) to the products is not favored, meaning that at equilibrium, there will be a significantly higher concentration of \(\mathrm{Na}_2\mathrm{O}_2\) compared to that of \(\mathrm{Na}_2\mathrm{O}\) and \(\mathrm{O}_2\).
Thermodynamic Stability
Thermodynamic stability relates to the Gibbs free energy of a substance and provides insight into its relative stability under a set of conditions like temperature and pressure.
A substance is considered thermodynamically stable if its Gibbs free energy is low compared to the Gibbs free energy of its potential decomposition products. The smaller (or more negative) the Gibbs free energy, the more stable the substance.
Therefore, assessing the thermodynamic stability of \(\mathrm{Na}_2\mathrm{O}_2\) involves comparing its Gibbs free energy of formation with those of its decomposition products (\(\mathrm{Na}_2\mathrm{O}(s)\) and \(\mathrm{O}_2(g)\)). In the given exercise, despite the positive change in Gibbs free energy for the reaction under standard conditions, indicating a non-spontaneous reaction, it does not directly equate to the instability of \(\mathrm{Na}_2\mathrm{O}_2\). Instead, it suggests that the compound would not spontaneously decompose into \(\mathrm{Na}_2\mathrm{O}\) and \(\mathrm{O}_2\) at 298 K and 1 atm pressure; thus, under these conditions, it is thermodynamically stable.
A substance is considered thermodynamically stable if its Gibbs free energy is low compared to the Gibbs free energy of its potential decomposition products. The smaller (or more negative) the Gibbs free energy, the more stable the substance.
Therefore, assessing the thermodynamic stability of \(\mathrm{Na}_2\mathrm{O}_2\) involves comparing its Gibbs free energy of formation with those of its decomposition products (\(\mathrm{Na}_2\mathrm{O}(s)\) and \(\mathrm{O}_2(g)\)). In the given exercise, despite the positive change in Gibbs free energy for the reaction under standard conditions, indicating a non-spontaneous reaction, it does not directly equate to the instability of \(\mathrm{Na}_2\mathrm{O}_2\). Instead, it suggests that the compound would not spontaneously decompose into \(\mathrm{Na}_2\mathrm{O}\) and \(\mathrm{O}_2\) at 298 K and 1 atm pressure; thus, under these conditions, it is thermodynamically stable.
Reaction Gibbs Free Energy Change
The reaction Gibbs free energy change \(\Delta G_{rxn}\) indicates whether a process is thermodynamically favorable under constant pressure and temperature. A negative \(\Delta G_{rxn}\) implies that the reaction proceeds spontaneously in the direction written, while a positive value indicates that the reaction is non-spontaneous and requires an input of energy to proceed.
The calculation of \(\Delta G_{rxn}\) is key to predicting the direction of a spontaneous reaction. It incorporates the Gibbs energies of formation for both the reactants and the products, with the general formula being:
\[\Delta G_{rxn} = \sum (\Delta G^{\circ}_f \text{ of products}) - \sum (\Delta G^{\circ}_f \text{ of reactants})\]
In our exercise, the computed \(\Delta G_{rxn}\) was 70.54 kJ/mol, meaning that the reaction, as written, is not spontaneous under standard conditions. This information combined with the equilibrium constant gives a complete picture regarding the favorability of the reaction. It's important to note that the term 'non-spontaneous' does not mean the reaction will not happen; instead, it might proceed under different conditions or require activation energy to initiate the process.
The calculation of \(\Delta G_{rxn}\) is key to predicting the direction of a spontaneous reaction. It incorporates the Gibbs energies of formation for both the reactants and the products, with the general formula being:
\[\Delta G_{rxn} = \sum (\Delta G^{\circ}_f \text{ of products}) - \sum (\Delta G^{\circ}_f \text{ of reactants})\]
In our exercise, the computed \(\Delta G_{rxn}\) was 70.54 kJ/mol, meaning that the reaction, as written, is not spontaneous under standard conditions. This information combined with the equilibrium constant gives a complete picture regarding the favorability of the reaction. It's important to note that the term 'non-spontaneous' does not mean the reaction will not happen; instead, it might proceed under different conditions or require activation energy to initiate the process.
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