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.)

Step by step solution

01

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 = \(6.23 \times 10^{-3}\).

02

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.

03

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 \[K_2 = 5.36 \times 10^{-2}\]

04

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: \[P_{CO₂} \ge 5.36 \times 10^{-2}\, \mathrm{torr}\]

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