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 85.)

Short answer

To prevent the decomposition of Ag₂CO₃ at 110°C, the partial pressure of CO₂ must be greater than or equal to \(5.36 \times 10^{-2}\, \mathrm{torr}\).

Step-by-step solution

Find the equilibrium constant (K) at 25°C

Given that the equilibrium partial pressure of CO₂ is \(6.23 \times 10^{-3}\) torr at 25°C. As there are no other gases except carbon dioxide involved in the reaction, we can directly use this partial pressure to represent the equilibrium constant. So, at 25°C, K₁ = \(6.23 \times 10^{-3}\).

Use Van't Hoff Equation

The Van't Hoff equation relates the change in the equilibrium constant (K) with temperature (T) and enthalpy change (∆H°) as: \[\frac{\mathrm{d} \ln K}{\mathrm{d} T}=\frac{\Delta H^{\circ}}{R T^{2}}\] where R is the gas constant. By integrating the equation, we find the relationship between the equilibrium constants at two different temperatures: \[\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\] We are given ∆H° = 79.14 kJ/mol. So, we convert it to J/mol: ∆H° = 79140 J/mol. Now, we have all the values to calculate the ratio of the equilibrium constants.

Calculate the equilibrium constant (K₂) at 110°C

First, we need to convert the given temperatures from Celsius to Kelvin: \(T_1 = 25^{\circ}C + 273.15 = 298.15K\) \(T_2 = 110^{\circ}C + 273.15 = 383.15K\) Now, we can use the Van't Hoff equation: \[\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{383.15}-\frac{1}{298.15}\right)\] Plugging in the given values and solving for K₂, we get: \[K_{2}=K_{1} \times \mathrm{e}^{-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{383.15}-\frac{1}{298.15}\right)}\] \[K_{2} = 6.23 \times 10^{-3} \times \mathrm{e}^{-\frac{79140}{8.314}\left(\frac{1}{383.15}-\frac{1}{298.15}\right)}\] After calculating the expression, we find that \[K_2 = 5.36 \times 10^{-2}\]

Calculate the partial pressure of CO₂ at 110°C

We have calculated the value of the equilibrium constant at 110°C, K₂ = 5.36 x 10⁻². In this case, as only CO₂ is the gas in the reaction, K₂ represents the equilibrium partial pressure of CO₂ at 110°C. To prevent decomposition, the partial pressure of CO₂ should be greater than or equal to the equilibrium partial pressure. Therefore, the partial pressure of CO₂ necessary to prevent decomposition of Ag₂CO₃ at 110°C is: \[P_{CO₂} \ge 5.36 \times 10^{-2}\, \mathrm{torr}\]

Key concepts

Van't Hoff Equation: Understanding the Relation Between Temperature and Equilibrium

In the world of chemical reactions, the Van't Hoff Equation is a powerful tool. It helps us see how the equilibrium position of a reaction changes with temperature. This formula is written as:\[\ln \frac{K_{2}}{K_{1}}=-\frac{\Delta H^{\circ}}{R}\left(\frac{1}{T_{2}}-\frac{1}{T_{1}}\right)\]Here, \(K\) is the equilibrium constant, \(T\) is the temperature, \(\Delta H^\circ\) is the change in enthalpy, and \(R\) is the gas constant. The equation shows us that if the enthalpy change is large, temperature will have a significant impact on \(K\), shifting the balance of the reaction.
Essentially, this equation connects thermodynamics with chemical equilibrium, revealing how energy changes in reactions.

Partial Pressure: Its Role in Chemical Equilibrium

Partial pressure is a term used to describe the pressure a single gas in a mixture exerts as if it were alone. In our exercise, carbon dioxide (CO₂) is the gas of interest, so its partial pressure is crucial. Knowing the partial pressure allows us to understand the contributions of CO₂ to the overall pressure in the system, which is key when considering chemical equilibrium.For a chemical reaction like the decomposition of silver carbonate, the partial pressure of CO₂ determines whether the reaction will proceed forward or remain stable. The partial pressure directly influences the equilibrium constant \(K\), as higher partial pressures of CO₂ can shift the equilibrium to favor the formation of reactants over products.
Thus, understanding partial pressure is vital to controlling and predicting the behaviors of gas-involving reactions.

Decomposition Reaction: Breaking Down Complex Compounds

A decomposition reaction involves a single complex compound breaking down into two or more simpler substances. The breakdown of silver carbonate into silver oxide and carbon dioxide is a classic example of such a reaction. This type of reaction is significant in both chemistry and everyday applications. Decomposition reactions are driven by the need to reach a state of lower energy or increased disorder, often influenced by temperature. In the given problem, we prevent the decomposition by ensuring that the partial pressure of carbon dioxide is sufficiently high, thus counteracting the formation of decomposition products.
Understanding decomposition is key in predicting how substances will react under various conditions, and how those conditions, like temperature or pressure, can be manipulated to maintain product stability.

Thermodynamics: The Energy Changes Behind Reactions

Thermodynamics is the study of energy changes within physical systems. In chemical reactions, it looks at how heat and work influence the direction and extent of reactions. When dealing with reactions, like the decomposition of silver carbonate, we use thermodynamic principles to understand and predict

• Reaction spontaneity
• Energy requirements

The enthalpy change \(\Delta H^\circ\) is a central thermodynamic quantity here. It tells us whether a reaction absorbs or releases heat. Positive \(\Delta H^\circ\) suggests that the reaction is endothermic, requiring heat from the surroundings to proceed.In our exercise, knowing the enthalpy change helps apply the Van't Hoff Equation to predict how equilibrium shifts with temperature changes, highlighting the profound link between thermodynamics and chemical behavior.