Question

If wet silver carbonate is dried in a stream of hot air, the air must have a certain concentration level of carbon dioxide to prevent silver carbonate from decomposing by the reaction $$\mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g)$$ \(\Delta H^{\circ}\) for this reaction is \(79.14 \mathrm{~kJ} / \mathrm{mol}\) in the temperature range of 25 to \(125^{\circ} \mathrm{C}\). Given that the partial pressure of carbon dioxide in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), calculate the partial pressure of \(\mathrm{CO}_{2}\) necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110 .{ }^{\circ} \mathrm{C}\). (Hint: Manipulate the equation in Exercise 71 .)

Short answer

In order to prevent the decomposition of Ag₂CO₃ at \(110^{\circ} \mathrm{C}\), the partial pressure of CO₂ required is approximately \(1.976 \times 10^{-3}\ \mathrm{torr}\).

Step-by-step solution

Set up the equilibrium constant expression and Clausius-Clapeyron equation

We have the Gibbs free energy equation for this reaction: \[\Delta G^{\circ} = -RT \ln K\] And we know, \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] So, \[\Delta H^{\circ} - T\Delta S^{\circ} = -RT \ln K\] We can rearrange this equation to be more useful in this problem: \[K = e^{\frac{\Delta H^{\circ}}{RT} - \frac{\Delta S^{\circ}}{R}}\] This is a modified form of the Clausius-Clapeyron equation that relates the equilibrium constant of a reaction to temperature, enthalpy change, and entropy change.

Calculate the initial equilibrium constant at \(25^{\circ} \mathrm{C}\)

Given that the partial pressure of CO₂ in equilibrium with pure solid silver carbonate is \(6.23 \times 10^{-3}\) torr at \(25^{\circ} \mathrm{C}\), we can write the equilibrium constant expression as: \[K_{25} = \frac{P_{CO_{2}}} {(1- P_{CO_{2}})}\] Now plug in the given value of the partial pressure of CO₂ and solve for K. \[K_{25} = \frac{6.23 \times 10^{-3}} {(1- 6.23 \times 10^{-3})} = 6.26 \times 10^{-3}\]

Calculate the final equilibrium constant at \(110^{\circ} \mathrm{C}\)

Now we use the modified Clausius-Clapeyron equation to calculate the equilibrium constant at \(110^{\circ} \mathrm{C}\). \[K_{110} = K_{25} \times e^{-\frac{\Delta H^{\circ}(T_{110} - T_{25})}{RT_{110}T_{25}}}\] \[K_{110} = 6.26 \times 10^{-3} \times e^{-\frac{(79.14 \times 10^3)[(110+273.15) - (25+273.15)]}{8.314(110+273.15)(25+273.15)}}\] \[K_{110} = 6.26 \times 10^{-3} \times e^{-7.80} = 1.976 \times 10^{-6}\]

Calculate the partial pressure of CO₂ required at \(110^{\circ} \mathrm{C}\)

Finally, use the calculated equilibrium constant expression to find the partial pressure of CO₂ required to prevent the decomposition of Ag₂CO₃ at \(110^{\circ} \mathrm{C}\). \[K_{110} = \frac{P_{CO_{2}}} {(1- P_{CO_{2}})}\] Replacing K with the calculated \(K_{110}\) value and solving for \(P_{CO_{2}}\): \[1.976 \times 10^{-6} = \frac{P_{CO_{2}}} {(1- P_{CO_{2}})}\] \[P_{CO_{2}} = 1.976 \times 10^{-3} \ \mathrm{torr}\] Therefore, the partial pressure of CO₂ necessary to prevent the decomposition of Ag₂CO₃ at \(110^{\circ} \mathrm{C}\) is \(1.976 \times 10^{-3}\ \mathrm{torr}\).

Key concepts

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation is a vital tool in physical chemistry, especially when we want to understand how the equilibrium state of a phase transition, such as between a liquid and a gas, changes with temperature. It relates the pressure of a system in equilibrium to its temperature, allowing us to predict the conditions under which a substance will change phase.

Mathematically, the equation commonly takes the form:
\[\frac{dP}{dT} = \frac{\Delta H_{vap}}{T(V_{g} - V_{l})} \]
Here, \(\frac{dP}{dT}\) represents the rate of change of pressure with temperature, \(\Delta H_{vap}\) is the enthalpy of vaporization, and \(V_{g} - V_{l}\) is the change in volume between gaseous and liquid states.

In exercises like the one provided, a version of this equation allows us to determine how changes in temperature affect the equilibrium of a chemical reaction involving gas and solid phases. It shows that as the temperature increases, more of the solid will decompose to maintain equilibrium if the partial pressure of the gas phase decreases.

Equilibrium Constant

The equilibrium constant, often denoted as \(K\), quantifies the ratio of concentration of products to reactants at chemical equilibrium given a balanced chemical equation. It provides a numerical value that helps us to grasp the extent of a reaction and the concentrations of different species at equilibrium.

The equilibrium constant expression for a general reaction:
\[a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}\]
can be written as:
\[ K = \frac{[\text{C}]^{c}[\text{D}]^{d}}{[\text{A}]^{a}[\text{B}]^{b}} \]
where the square brackets denote the concentration of the respective species. The equilibrium constant is a reflection of the reaction conditions such as temperature and can be used to predict which way a reaction will proceed under a given set of circumstances.

Understanding and calculating the equilibrium constant is crucial in exercises like ours, where changes in conditions such as temperature require us to calculate changes in the equilibrium state of a reaction.

Partial Pressure

Partial pressure is a critical concept in the context of gas-phase reactions and equilibria. It refers to the pressure that an individual gas component in a mixture would exert if it alone occupied the entire volume of the mixture at the same temperature. Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is the sum of the partial pressures of each individual gas.

For a gas \(A\), the partial pressure can be denoted as \(P_A\) and can be calculated using the ideal gas law provided the number of moles, temperature, and volume are known:
\[ P_A = \frac{n_A RT}{V} \]
In the context of chemical equilibrium, particularly in reactions where gases are involved, the partial pressure of a gas is used in place of its concentration when writing the equilibrium constant expression. This is crucial for solving problems regarding the equilibria of gas-phase reactions, like determining the necessary partial pressure of CO₂ to prevent the decomposition of silver carbonate at higher temperatures as shown in our exercise.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relationships between heat and other forms of energy, and by extension, it covers the ways in which energy transformations govern the direction and extent of chemical reactions.

Within the realm of thermodynamics, the behavior of systems in equilibrium is especially relevant. Equilibrium in thermodynamics is a condition where the macroscopic properties of a system such as pressure and temperature remain constant over time.

In thermal equilibrium, two systems in contact with each other exchange no heat energy, because they are at the same temperature. In chemical equilibrium, which our exercise explores, the forward and reverse rates of a chemical reaction are equal, and the concentrations of the reactants and products remain stable over time. This state can be described quantitatively by the equilibrium constant.

The laws of thermodynamics constrain all chemical reactions, defining limits such as the direction in which reactions can occur spontaneously, and enabling us to predict such behaviors using parameters like Gibbs free energy, enthalpy, and entropy.