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Chapter 17: Problem 111

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.)

Short Answer

Expert verified
The partial pressure of CO2 necessary to prevent decomposition of Ag2CO3 at 110°C can be calculated using the Van't Hoff equation. Based on the given data, we find the partial pressure at 25°C, convert it to atmospheres, and then use the Van't Hoff equation to find the partial pressure at 110°C. Finally, convert the pressure back to Torr. The required partial pressure at 110°C is approximately \(3.29 \times 10^{-5}\) Torr.

Step by step solution

01

Understand and list known variables

The information provided in the problem can be listed as the following variables: - Enthalpy change (\( \Delta H \)) = 79.14 kJ/mol. Reminder that 1 kJ = 1000 J, thus \( \Delta H \) = 79140 J/mol - Initial temperature (T1) = 25°C = 298.15 K (Converted from Celsius to Kelvin) - Final temperature (T2) = 110°C = 383.15 K - Partial pressure of CO2 at T1 (P1) = 6.23 x 10^-3 Torr. - R is the gas constant = 8.314 J/(mol*K)
02

Convert the pressure to atmospheric pressure

Conversion is required because the gas constant 'R' we are using is in terms of the J/(mol*K), and pressure needs to be in atmospheres. Torr can be converted to atmospheres using the conversion factor 1 atm = 760 Torr: Pressure, P1_atm = \( \frac{6.23 \times 10^{-3} \, \text{Torr}}{760} \) = 8.197 \times 10^-6 atm.
03

Use the Van't Hoff equation

The Van't Hoff equation \[ \ln \left(\frac{P_2}{P_1}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \] is usually used to solve this kind of problem. Substituting the known values into the equation gives: \[\ln \left(\frac{P_2}{8.197 \times 10^{-6}}\right) = -\frac{79140}{8.314} \left(\frac{1}{383.15} - \frac{1}{298.15}\right)\] where \(P_2\) is the partial pressure of carbon dioxide at \(110^{\circ} C\), which is what we're trying to find.
04

Solve for P2

By calculating the right side of the equation first and then rearranging the formula we solve for \(P_2\): \(P_2 = 8.197 \times 10^{-6} \cdot e^{-\frac{79140}{8.314} \left(\frac{1}{383.15} - \frac{1}{298.15}\right)}\)
05

Convert the answer back to Torr

The final step is to convert the \(P_2\) from atm to Torr for which a simple multiplication operation will suffice. \(P_2\) in Torr = \(P_2 \times 760\). By following these steps and using the given data, a calculation of the partial pressure necessary to prevent decomposition of silver carbonate at a temperature of \(110^{\circ} C\) can be made.

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Most popular questions from this chapter

For the reaction at \(298 \mathrm{~K}\), $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\), respectively. What is the value of \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) ? Assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

Ethanethiol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\right.\) also called ethyl mercaptan \()\) is commonly added to natural gas to provide the "rotten egg" smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is \(27.5 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy of vaporization for this substance?

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right) .\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g)\) $$ \begin{aligned} \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\ \mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\ \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K} \end{aligned} $$
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