Question

Using data from Appendix \(\mathrm{C}\), write the equilibrium-constant expression and calculate the value of the equilibrium constant 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

The equilibrium constants for the given reactions at \(298 \mathrm{~K}\) are: (a) \(K_p \approx 2.67 \times 10^{-5}\) (b) \(K_p \approx 2.51 \times 10^4\) (c) \(K_p \approx 6.63 \times 10^{12}\)

Step-by-step solution

(a) Writing the equilibrium constant expression for the first reaction

For the given reaction: \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\) The equilibrium constant expression can be written using the concentrations of the products and reactants: \(K_p = \frac{[\mathrm{CO}_{2}]^{1}}{[\mathrm{NaHCO}_{3}]^{1} [\mathrm{NaOH}]^{1}}\)

(a) Finding the equilibrium constant using Appendix C data

In order to find the equilibrium constant value, we need to use the standard Gibbs free energy changes \(\Delta G^\circ\) of each species involved, which can be found in Appendix C. Then, we can calculate the standard Gibbs free energy change for the reaction \(\Delta G_r^\circ\) by using the formula: \(\Delta G_r^\circ = \sum v_i \Delta G_i^\circ\) where \(v_i\) are the stoichiometric coefficients and \(\Delta G_i^\circ\) the standard Gibbs free energy changes for the species involved. Using the data from Appendix C, the standard Gibbs free energy changes for the species involved are: \(\Delta G^\circ_{\mathrm{NaHCO}_{3}}=-851.2 \mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{NaOH}}=-380.6\mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{CO}_{2}}=-394.4\mathrm{~kJ/mol} \) Now, we can determine the standard Gibbs free energy change for the reaction: \(\Delta G_r^\circ = \Delta G^\circ_{\mathrm{NaOH}} + \Delta G^\circ_{\mathrm{CO}_{2}} - (\Delta G^\circ_{\mathrm{NaHCO}_{3}})\) \(\Delta G_r^\circ = (-380.6 - 394.4) - (-851.2)\) \(\Delta G_r^\circ = 76.2 \mathrm{~kJ/mol} \) Finally, we can find the equilibrium constant using the relation between \(\Delta G\) and \(K\): \(\Delta G_r^\circ = -RT \ln K_p\) \(K_p = e^{-\frac{\Delta G_r^\circ}{RT}}\) Rearrange and substituting the known values: \(K_p = e^{-\frac{76.2 \mathrm{~kJ/mol}}{(8.314 \mathrm{~J/molK})(298 \mathrm{~K})}}\) \(K_p \approx 2.67 \times 10^{-5}\) So the equilibrium constant for reaction (a) is approximately \(2.67 \times 10^{-5}\) at 298 K.

(b) Writing the equilibrium constant expression for the second reaction

For the given reaction: \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\) The equilibrium constant expression can be written as: \(K_p = \frac{[\mathrm{HCl}]^{2} [\mathrm{Br}_{2}]^{1}}{[\mathrm{HBr}]^{2} [\mathrm{Cl}_{2}]^{1}} \) Repeat the same steps as in (a) using the data from Appendix C, the standard Gibbs free energy changes for the species involved are: \(\Delta G^\circ_{\mathrm{HBr}}=-53.3 \mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{HCl}}=-131.2\mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{Cl}_{2}}=0\mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{Br}_{2}}=0\mathrm{~kJ/mol} \) By calculating \(\Delta G_r^\circ\) and subsequently finding \(K_p\), we get the equilibrium constant for reaction (b) to be approximately \(2.51\times 10^4\) at 298 K.

(c) Writing the equilibrium constant expression for the third reaction

For the given reaction: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) The equilibrium constant expression can be written as: \(K_p = \frac{[\mathrm{SO}_{3}]^{2}}{[\mathrm{SO}_{2}]^{2} [\mathrm{O}_{2}]^{1}} \) Repeat the same steps as in (a) using the data from Appendix C, the standard Gibbs free energy changes for the species involved are: \(\Delta G^\circ_{\mathrm{SO}_{2}}=-300.4 \mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{O}_{2}}=0\mathrm{~kJ/mol} \) \(\Delta G^\circ_{\mathrm{SO}_{3}}=-371.1\mathrm{~kJ/mol} \) By calculating \(\Delta G_r^\circ\) and subsequently finding \(K_p\), we get the equilibrium constant for reaction (c) to be approximately \(6.63\times 10^{12}\) at 298 K. In conclusion, the equilibrium constants for the given reactions at 298 K are: (a) \(2.67 \times 10^{-5}\), (b) \(2.51 \times 10^4\), and (c) \(6.63 \times 10^{12}\).

Key concepts

Gibbs Free Energy

Gibbs Free Energy, symbolized as \( G \), is a thermodynamic quantity that measures the maximum amount of work a system can perform at constant temperature and pressure. In chemical reactions, it serves as a powerful tool to predict the spontaneity and feasibility of a reaction. The change in Gibbs Free Energy, denoted by \( \Delta G \), helps us understand if a reaction can occur on its own (\( \Delta G < 0 \) for spontaneous reactions) or if it requires additional energy to proceed (\( \Delta G > 0 \) for non-spontaneous reactions).

When a system is at equilibrium, \( \Delta G \) is zero because the system is in its most stable state, and no net work can be done. The relationship between Gibbs Free Energy and the equilibrium constant \( K \) is given by the equation \( \Delta G_r^\circ = -RT \ln K_p \), where \( R \) is the universal gas constant, \( T \) is the temperature in Kelvin, and \( K_p \) is the equilibrium constant for gaseous reactions under constant pressure. This equation can be rearranged to find \( K_p \) when \( \Delta G_r^\circ \) is known, assuming that the reaction is conducted at standard conditions (denoted by the superscript \( \circ \) on \( \Delta G \) ).

To enhance comprehension, it's important to remember that Gibbs Free Energy involves energy changes at molecular levels, with decreases indicating the release of energy (exergonic processes) and increases corresponding to energy absorption (endergonic processes). This concept applies universally to physical, chemical, and biological systems.

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to constant concentrations of reactants and products, even though individual molecules continue to react. At this point, the rate of the forward reaction equals the rate of the backward reaction. Because there is no net change in the system, the system is said to be in a state of dynamic equilibrium.

It’s essential to understand that equilibrium does not imply equal concentrations of reactants and products but rather stable concentrations. The point of equilibrium is governed by the equilibrium constant (\( K \) ), which provides a ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their stoichiometric coefficients. Conditions such as temperature and pressure can influence the position of equilibrium; however, the value of the equilibrium constant only changes with temperature.

Underlying the state of equilibrium is the concept of free energy. When a system reaches equilibrium, its free energy is at a minimum, and no further work can be extracted. The system is stable, and any deviation from equilibrium will shift the reactions to re-establish it. This principle is known as Le Chatelier's Principle and is critical for predicting how changes in conditions affect the system.

Equilibrium Constant Expression

The equilibrium constant expression quantifies the state of chemical equilibrium. It is derived from the law of mass action, which states that for a chemical equilibrium \( aA + bB \rightleftharpoons cC + dD \) the equilibrium constant \( K \) is given by \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \) for concentrations (\( K_c \) ) or \( K_p = \frac{P_C^cP_D^d}{P_A^aP_B^b} \) for partial pressures (\( K_p \) ).

Each concentration or pressure is raised to the power of its coefficient in the balanced equation. For gases, \( K_p \) is most commonly used, whereas \( K_c \) is used for species in solution. Importantly, pure solids and liquids do not appear in these expressions because their concentrations do not change.

Factors Affecting Equilibrium Constant Values

While the equilibrium constant is indeed a constant at a given temperature, it is temperature-dependent. It does not change with changes in concentration, pressure, or volume, as the system will adjust itself (according to Le Chatelier’s Principle) to maintain the constant value of \( K \).

To foster a deeper understanding, we should note that the magnitude of \( K \) gives insight into the reaction's favorability: a high \( K \) value indicates a reaction favoring products, while a low \( K \) value shows a reaction favoring reactants. By manipulating the reaction conditions, chemists can shift the equilibrium to obtain more products or reactants as needed, which is of vital importance in industrial chemical processes.