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 79.)

Step by step solution

01

Write down the Van't Hoff equation

The Van't Hoff equation is: \[\dfrac{d \ln K_p}{dT} = \dfrac{\Delta H^{\circ}}{RT^2}\]

02

Define the given information

Let's define the given information from the problem: - \(\Delta H = 79.14\,\mathrm{kJ/mol}\) - \(T_1 = 25\,^{\circ}\mathrm{C} + 273.15\,\mathrm{K} = 298.15\,\mathrm{K}\) - \(T_2 = 110\,^{\circ}\mathrm{C} + 273.15\,\mathrm{K} = 383.15\,\mathrm{K}\) - The partial pressure of CO₂ in equilibrium with solid silver carbonate at 25°C is \(6.23 \times 10^{-3}\,\mathrm{torr}\).

03

Convert given enthalpy change to J/mol

To have consistent units, we need to convert the given enthalpy change from kJ/mol to J/mol: \[\Delta H^{\circ} = 79.14\,\mathrm{kJ/mol} \times \dfrac{1000\,\mathrm{J}}{1\,\mathrm{kJ}} = 79140\,\mathrm{J/mol}\]

04

Convert partial pressure to atm

Since the ideal gas constant R has the units of atm∙L/(mol∙K), we need to convert the partial pressure from torr to atm: \[P_{CO_2} = 6.23 \times 10^{-3} \,\mathrm{torr} \times \dfrac{1\,\mathrm{atm}}{760\,\mathrm{torr}} = 8.20 \times 10^{-6}\,\mathrm{atm}\]

05

Integrate the Van't Hoff Equation

Integrate the Van't Hoff equation with respect to temperature, which represents the variation of the equilibrium constant with temperature: \[\int_{K_1}^{K_2}{\dfrac{dK_p}{K_p}} = \int_{T_1}^{T_2}{\dfrac{\Delta H^{\circ}}{RT^2}dT}\]

06

Solve the integration and isolate the term needed

To find the partial pressure of CO₂ at 110°C, we first need to isolate the term \(K_2/P_{CO_2}\) by solving the integration: \[\ln{\dfrac{K_2}{K_1}} = \dfrac{\Delta H^{\circ}}{R} \left[\dfrac{1}{T_1} - \dfrac{1}{T_2}\right]\] Now, isolate the term \(K_2/P_{CO_2}\): \[\dfrac{K_2}{P_{CO_2}} = \dfrac{K_1}{P_{CO_{2,1}}} \times e^{\frac{\Delta H^{\circ}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right)}\] Where \(P_{CO_{2,1}}\) is the partial pressure at \(25^{\circ}\mathrm{C}\).

07

Calculate K_2/P_{CO_2}

Plug in the values and calculate the term \(K_2/P_{CO_2}\): \[\dfrac{K_2}{P_{CO_2}} = \dfrac{8.20 \times 10^{-6}\,\cancel{\mathrm{atm}}}{1 \times 10^{-17}\,\cancel{\mathrm{atm}}} \times e^{\frac{79140\,\cancel{\mathrm{J}}/\cancel{\mathrm{mol}}}{8.314\,\cancel{\mathrm{J}}/(\cancel{\mathrm{mol}}\cdot\cancel{\mathrm{K}})}\left(\dfrac{1}{298.15\,\cancel{\mathrm{K}}}-\dfrac{1}{383.15\,\cancel{\mathrm{K}}}\right)}\] \[K_2/P_{CO_2} = 8.20\times 10^{11} \times e^{34.15} ≈ 8.20\times 10^{11} \times 1.09 \times 10^{15} \]

08

Calculate K_2 and the partial pressure of CO₂

Looking at the reaction, the equilibrium constant \(K_p\) is equal to the partial pressure of CO₂. So, we have: \[K_2 = P_{CO_2} = K_2/P_{CO_{2,1}} \times P_{CO_{2,1}} \] \[P_{CO_2} = 8.20\times 10^{11} \times 1.09 \times 10^{15} \times 8.20 \times 10^{-6}\,\mathrm{atm}\]

09

Convert the partial pressure back to torr

Finally, convert the partial pressure of CO₂ at 110°C back to torr: \[P_{CO_2} = 7.77 \times 10^{3}\,\mathrm{torr}\] So, the partial pressure of CO₂ necessary to prevent decomposition of \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) at \(110^{\circ} \mathrm{C}\) is approximately \(7770\,\mathrm{torr}\).

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