Question

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant and the free- energy change for these reactions at \(298 \mathrm{~K}:\) (a) \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)

Short answer

(a) For the reaction \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\), the equilibrium constant is \(K = 2.49 \times 10^{-3}\), and the free energy change is \(\Delta G = 16.6 \,\mathrm{kJ/mol}\). (b) For the reaction \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\), the equilibrium constant is \(K = 6.80 \times 10^{26}\), and the free energy change is \(\Delta G = -155.2 \,\mathrm{kJ/mol}\). (c) For the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), the equilibrium constant is \(K = 1.24 \times 10^{18}\), and the free energy change is \(\Delta G = -140.4 \,\mathrm{kJ/mol}\).

Step-by-step solution

a. Equilibrium constant expression

For this reaction, the equilibrium constant expression is: \(K = \frac{[\mathrm{CO}_{2}]}{[\mathrm{NaHCO}_{3}]}\)

b. Standard free energy change

Calculate the standard free energy change for the reaction by using the equation: \(\Delta G^\circ = \Delta G^\circ({\mathrm{products}}) - \Delta G^\circ({\mathrm{reactants}})\) We can look up the values in Appendix C: \(\Delta G^\circ_{\mathrm{NaHCO}_3} = -745.7 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{NaOH}} = -367.7 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{CO}_2} = -394.4 \,\mathrm{kJ/mol}\) \(\Delta G^\circ = (-367.7 - 394.4) \,\mathrm{kJ/mol} - (-745.7 \,\mathrm{kJ/mol}) = 16.6 \,\mathrm{kJ/mol}\)

c. Equilibrium constant

Calculate the equilibrium constant by using the equation: \(K = e^{-\frac{\Delta G^\circ}{RT}}\) Where \(R = 8.314 \,\mathrm{J/(mol\,K)}\) and \(T = 298 \,\mathrm{K}\): \(K = e^{-\frac{16600 \,\mathrm{J/mol}}{(8.314 \,\mathrm{J/(mol\,K)})(298 \,\mathrm{K})}} = 2.49 \times 10^{-3}\)

d. Free energy change

Since we have the equilibrium constant and standard free energy change, the free energy change for the reaction is: \(\Delta G = \Delta G^\circ = 16.6 \,\mathrm{kJ/mol}\) Now we repeat this analysis for the two other reactions. Reaction (b): \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) We follow the same steps as in reaction (a):

Equilibrium constant expression

Here the equilibrium constant expression is: \(K = \frac{[\mathrm{HCl}]^2[\mathrm{Br}_{2}]}{[\mathrm{HBr}]^2[\mathrm{Cl}_2]}\)

Standard free energy change

Calculate the standard free energy change for the reaction using the data from Appendix C: \(\Delta G^\circ_{\mathrm{HBr}} = -53.6 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{Cl}_2} = 0 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{HCl}} = -131.2 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{Br}_2} = 0 \,\mathrm{kJ/mol}\) \(\Delta G^\circ = 2(-131.2) + 0 - 2(-53.6) = -155.2 \,\mathrm{kJ/mol}\)

Equilibrium constant

Calculate the equilibrium constant: \(K = e^{-\frac{-155200 \,\mathrm{J/mol}}{(8.314 \,\mathrm{J/(mol\,K)})(298 \,\mathrm{K})}} = 6.80 \times 10^{26}\)

Free energy change

The free energy change for the reaction is: \(\Delta G = \Delta G^\circ = -155.2 \,\mathrm{kJ/mol}\) Reaction (c): \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) Again, we follow the same steps:

Equilibrium constant expression

The equilibrium constant expression is: \(K = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]}\)

Standard free energy change

Calculate the standard free energy change for the reaction: \(\Delta G^\circ_{\mathrm{SO}_2} = -300.2 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{O}_2} = 0 \,\mathrm{kJ/mol}\) \(\Delta G^\circ_{\mathrm{SO}_3} = -370.4 \,\mathrm{kJ/mol}\) \(\Delta G^\circ = 2(-370.4) - 2(-300.2) = -140.4 \,\mathrm{kJ/mol}\)

Equilibrium constant

Calculate the equilibrium constant: \(K = e^{-\frac{-140400 \,\mathrm{J/mol}}{(8.314 \,\mathrm{J/(mol\,K)})(298 \,\mathrm{K})}} = 1.24 \times 10^{18}\)

Free energy change

The free energy change for the reaction is: \(\Delta G = \Delta G^\circ = -140.4 \,\mathrm{kJ/mol}\) To sum up, we found the following results for the three reactions: (a) K = \(2.49 \times 10^{-3}\), \(\Delta G = 16.6 \,\mathrm{kJ/mol}\) (b) K = \(6.80 \times 10^{26}\), \(\Delta G = -155.2 \,\mathrm{kJ/mol}\) (c) K = \(1.24 \times 10^{18}\), \(\Delta G = -140.4 \,\mathrm{kJ/mol}\)

Key concepts

Free Energy Change

The concept of free energy change, often denoted as \( \Delta G \), is crucial in understanding chemical reactions. It tells us whether a reaction will occur spontaneously. In simple terms, if \( \Delta G \) is negative, the reaction is spontaneous, meaning it can happen without input from outside energy.
The standard free energy change \( \Delta G^\circ \) for a reaction is calculated using the formula:
\[ \Delta G^\circ = \Delta G^\circ(\text{products}) - \Delta G^\circ(\text{reactants}) \]
If \( \Delta G^\circ \) is positive, the reaction isn't likely to happen on its own. In our examples from the exercise, the reaction involving \( \mathrm{NaHCO}_3 \) has a positive \( \Delta G^\circ \) of \( 16.6 \ kJ/mol \), indicating it is non-spontaneous under standard conditions. But, the reaction of \( 2 \mathrm{HBr}(g) + \mathrm{Cl}_2(g) \) with a negative \( \Delta G^\circ \) of \( -155.2 \ kJ/mol \) is spontaneous.
Understanding free energy change helps us predict if and how quickly a reaction might proceed, focusing on energy dynamics rather than just reactants and products configuration.

Equilibrium Reactions

Equilibrium reactions are reversible processes where the forward and reverse reactions occur at the same rate. At equilibrium, there's no net change in the concentration of reactants and products. To express this state mathematically, we use the equilibrium constant \( K \). Each reaction has a unique expression for \( K \), depending on its balanced equation.
For example, for the decomposition of \( \mathrm{NaHCO}_3 \), the expression is:

• \( K = \frac{[\mathrm{CO}_2]}{[\mathrm{NaHCO}_3]} \)

The value of \( K \) indicates the position of equilibrium. A large \( K \) value means the reaction favors products; a small \( K \) value means it favors reactants. For instance, the equilibrium constant for \( 2 \mathrm{HBr}(g) + \mathrm{Cl}_2(g) \) is very large (\( 6.80 \times 10^{26} \)), suggesting a strong favor towards products under standard conditions.
This understanding of equilibrium helps recognize how different conditions affect reaction balance and helps in designing reactions favoring desired products.

Thermodynamics in Chemistry

Thermodynamics offers tools to measure and predict the direction and extent of chemical processes. It involves energy changes and principles, particularly concerning heat and work.
In thermodynamics, there are key concepts like enthalpy, entropy, and free energy. These determine how energy is transferred and how systems change over time. For chemical reactions, \( \Delta G \) is derived from these concepts, illustrating how spontaneous a reaction is.
Particularly, the equation:

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

links enthalpy \( \Delta H \) (heat absorbed or released) and entropy \( \Delta S \) (disorder or randomness change) to \( \Delta G \), using temperature \( T \).
Thermodynamics not only helps chemists predict reaction behaviors but also aids industries in optimizing processes for energy efficiency, linking chemistry with practical applications. Understanding these principles is critical for innovations in pharmaceuticals, materials science, and environmental solutions.