Problem 82 - Chemistry: The Central Science

Question

Using data from Appendix \(C\), 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) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\)

Step-by-Step Solution

Answer

(a) For \(\mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g)\), the equilibrium-constant expression is \(K_p = P_{CO_2}\). The value of the equilibrium constant \(K_p\) is approximately \(3.16 \times 10^{-10}\), and the free-energy change \(\Delta G^\circ\) is \(56.6\, kJ/mol\). (b) For \(2 \mathrm{HBr}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)+\mathrm{Br}_{2}(g)\), the equilibrium-constant expression is \(K_p = \frac{P_{HCl}^2 \cdot P_{Br_2}}{P_{HBr}^2 \cdot P_{Cl_2}}\). The value of the equilibrium constant \(K_p\) is approximately \(1.77 \times 10^{8}\), and the free-energy change \(\Delta G^\circ\) is \(-71.4\, kJ/mol\). (c) For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), the equilibrium-constant expression is \(K_p = \frac{P_{SO_3}^2}{P_{SO_2}^2 \cdot P_{O_2}}\). The value of the equilibrium constant \(K_p\) is approximately \(2.37 \times 10^{13}\), and the free-energy change \(\Delta G^\circ\) is \(-142.1\, kJ/mol\).

Step 1: Write the Equilibrium-Constant Expression

The equilibrium-constant expression for this reaction is: \(K_p = P_{CO_2}\) Here, \(K_p\) is the equilibrium constant, and \(P_{CO_2}\) is the partial pressure of CO2.

Step 2: Calculate the Value of the Equilibrium Constant

At equilibrium, we know: \(\Delta G^\circ = Standard\, Free\, Energy\, Change\) and \(K_p = e^{-\frac{\Delta G^\circ}{RT}}\) where R is the gas constant (8.314 J / (mol K)) and T is the temperature (298 K in this case). According to Appendix C, the Standard Free Energy Change for this reaction is: \(\Delta G^\circ = 56.6\, kJ/mol\) Now, we can calculate \(K_p\): \(K_p = e^{-\frac{(56.6 \times 10^3)\, J/mol}{(8.314\, J / (mol\, K)) (298\, K)}}\) \(K_p \approx 3.16 \times 10^{-10}\)

Step 3: Calculate the Free-Energy Change

We have already found the Standard Free Energy Change in Step 2, which is: \(\Delta G^\circ = 56.6\, kJ/mol\) For (b) 2 HBr(g) + Cl2(g) ⇌ 2 HCl(g) + Br2(g):

Step 1: Write the Equilibrium-Constant Expression

The equilibrium-constant expression for this reaction is: \(K_p = \frac{P_{HCl}^2 \cdot P_{Br_2}}{P_{HBr}^2 \cdot P_{Cl_2}}\)

Step 2: Calculate the Value of the Equilibrium Constant

As before, we have: \(\Delta G^\circ = Standard\, Free\, Energy\, Change\) and \(K_p = e^{-\frac{\Delta G^\circ}{RT}}\) According to Appendix C, the Standard Free Energy Change for this reaction is: \(\Delta G^\circ = -71.4\, kJ/mol\) Now, we can calculate \(K_p\): \(K_p = e^{-\frac{(-71.4 \times 10^3)\, J/mol}{(8.314\, J / (mol\, K)) (298\, K)}}\) \(K_p \approx 1.77 \times 10^{8}\)

Step 3: Calculate the Free-Energy Change

We have already found the Standard Free Energy Change in Step 2, which is: \(\Delta G^\circ = -71.4\, kJ/mol\) For (c) 2 SO2(g) + O2(g) ⇌ 2 SO3(g):

Step 1: Write the Equilibrium-Constant Expression

The equilibrium-constant expression for this reaction is: \(K_p = \frac{P_{SO_3}^2}{P_{SO_2}^2 \cdot P_{O_2}}\)

Step 2: Calculate the Value of the Equilibrium Constant

As before, we have: \(\Delta G^\circ = Standard\, Free\, Energy\, Change\) and \(K_p = e^{-\frac{\Delta G^\circ}{RT}}\) According to Appendix C, the Standard Free Energy Change for this reaction is: \(\Delta G^\circ = -142.1\, kJ/mol\) Now, we can calculate \(K_p\): \(K_p = e^{-\frac{(-142.1 \times 10^3)\, J/mol}{(8.314\, J / (mol\, K)) (298\, K)}}\) \(K_p \approx 2.37 \times 10^{13}\)

Step 3: Calculate the Free-Energy Change

We have already found the Standard Free Energy Change in Step 2, which is: \(\Delta G^\circ = -142.1\, kJ/mol\)

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentrations of the reactants and products do not change over time. This balance does not imply that the reactants and products are equal in concentration, but rather that their ratios do not change.

To represent this state, an equilibrium expression is written using the concentration (in mol/L) of the reactants and products, raised to the power of their stoichiometric coefficients from the balanced chemical equation. For gas-phase reactions, the expression typically uses partial pressures instead of concentrations.

Understanding chemical equilibrium is crucial in many fields such as chemistry, engineering, biology, and environmental science, as it determines the composition of the mixture at equilibrium and helps predict the direction and extent of chemical reactions under various conditions.

Gibbs Free Energy

Gibbs Free Energy (G) is a thermodynamic quantity that represents the amount of energy capable of doing work during a chemical process at constant temperature and pressure. The change in Gibbs Free Energy, denoted as \(\Delta G\), helps determine whether a reaction is spontaneous or non-spontaneous.

A negative \(\Delta G\) indicates that a reaction will occur spontaneously, whereas a positive \(\Delta G\) suggests that the reaction is non-spontaneous and requires additional energy to proceed. Reactions with a \(\Delta G\) of zero are at equilibrium. By linking \(\Delta G\) to the equilibrium constant (\(K\)), it's possible to calculate the set point where the system has reached balance—neither favoring the reactants nor the products excessively.

Equilibrium Constant Expression

The equilibrium constant expression quantifies the position of the equilibrium and is derived from the law of mass action. For a gas-phase reaction, the equilibrium constant is expressed in terms of partial pressures (\(K_p\)) and is defined as the ratio of the equilibrium partial pressures of the products to the reactants, each raised to the power of their respective stoichiometric coefficients in the balanced equation.

In general, the equilibrium constant expression for a reaction such as \(aA + bB \rightleftharpoons cC + dD\) is written as:\[K_p = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b}\]
Understanding and calculating \(K_p\) is essential for predicting the extent of a reaction and for calculating the Gibbs Free Energy change associated with reaching equilibrium.

Partial Pressure

Partial pressure is the pressure that a gas in a mixture of gases would exert if it alone occupied the entire volume of the mixture at the same temperature. It's an important concept in chemical equilibria involving gaseous reactants and products because the equilibrium position is determined by the partial pressures of the gases involved.

For example, in the reaction \(2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)\), each gas has its own partial pressure represented as \(P_{SO_2}\), \(P_{O_2}\), and \(P_{SO_3}\). The equilibrium constant expression in terms of partial pressure (\(K_p\)) for this reaction would involve these individual pressures. It's useful to remember that the total pressure of the system is the sum of all partial pressures of the gases present.