Consider the reaction $$ \mathrm{PbCO}_{3}(s) \rightleftharpoons \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) $$ Using data in Appendix C, calculate the equilibrium pressure of \(\mathrm{CO}_{2}\) in the system at (a) \(400^{\circ} \mathrm{C}\) and (b) \(180^{\circ} \mathrm{C}\).

Short Answer

Expert verified
Using the given data and the Van't Hoff equation, we can calculate the equilibrium constant at each temperature and subsequently determine the equilibrium pressure of CO2. At \( 400 ^\circ C \), we find \( P_{CO_2}^{(400)} = K_{400} \) and at \( 180 ^\circ C \), we find \( P_{CO_2}^{(180)} = K_{180} \). Thus, we can determine the equilibrium pressure of CO2 at both temperatures.

Step by step solution

01

Gather Gibbs free energies of formation

Using data from Appendix C, we find the Gibbs free energies of formation (∆Gf) at 298 K for the following substances: - \( PbCO_3(s): ∆Gf(PbCO_3) = -699.9\,\text{k} \text{J/mol} \) - \( PbO(s): ∆Gf(PbO) = -217.3\,\text{k} \text{J/mol} \) - \( CO_2 (g): ∆Gf(CO_2) = -394.4\,\text{k} \text{J/mol} \)
02

Calculate the equilibrium constant at each temperature

Using the Gibbs free energies of formation, we can calculate the Gibbs free energy change (∆G) for the reaction at 298 K by the following equation: \( ∆G = ∆G_{PbO} + ∆G_{CO_2} - ∆G_{PbCO_3} \) Now, we can calculate the equilibrium constant (K) at each temperature by using the Van't Hoff equation: \( K = e^{-\frac{∆G}{R \times T}} \), where \( R = 8.314 \,\text{J/mol K} \) is the gas constant and T is the temperature. (a) For \( 400^\circ C = 673 \,\text{K} \): \( K_{400} = e^{-\frac{∆G}{R \times 673}} \) (b) For \( 180^\circ C = 453 \,\text{K} \): \( K_{180} = e^{-\frac{∆G}{R \times 453}} \)
03

Use the equilibrium constant expression to solve for the equilibrium pressure of CO2

The equilibrium constant expression for the reaction is: \( K = \frac{P_{CO_2}}{1} \) because the stoichiometric coefficients are 1 and the reactants are in solid form. Hence, we can calculate the equilibrium pressure of CO2: (a) At \( 400^\circ C \): \( P_{CO_2}^{(400)} = K_{400} \) (b) At \( 180^\circ C \): \( P_{CO_2}^{(180)} = K_{180} \) By calculating these values, we will find the equilibrium pressure of CO2 at both 400 ° C and 180 ° C.

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Most popular questions from this chapter

A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to that for going from state 2 back to state \(1 ?\) (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?(\mathrm{c})\) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\)

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