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

Short Answer

Expert verified
The equilibrium pressure of CO2 at the given temperatures can be calculated using the Gibbs free energy data and the equilibrium constant expression. The equilibrium pressures at the two temperatures are: a) \( 400^{\circ} \mathrm{C} (673.15 \mathrm{K}) \): Equilibrium pressure of CO2 = [Value] (appropriate units) b) \( 180^{\circ} \mathrm{C} (453.15 \mathrm{K}) \): Equilibrium pressure of CO2 = [Value] (appropriate units)

Step by step solution

01

In this exercise, temperatures are given in Celsius. We need to convert them to Kelvin for further calculations. Use the following formula: \( T_{K} = T_{C} + 273.15 \) Temperatures in Kelvin: a) \( 400^{\circ} \mathrm{C} = 400 + 273.15 = 673.15 \mathrm{K} \) b) \( 180^{\circ} \mathrm{C} = 180 + 273.15 = 453.15 \mathrm{K} \) #Step 2: Calculate ΔG and K for each temperature#

Use the Gibbs energies of formation (∆Gf°) from Appendix C to find the Gibbs energy change (∆G°) for the reaction at the given temperatures. ΔG° = ΔGf° (PbO) + ΔGf° (CO2) - ΔGf°(PbCO3) Using the equation ΔG° = -RT ln(K), we can find the equilibrium constant K at each temperature. #Step 3: Solve for equilibrium pressure at each temperature#
02

Now that we have the equilibrium constants (K) for the reactions at both temperatures, we can use the equilibrium constant expression to find the equilibrium pressure of CO2. K = \[ \frac{[CO_2]}{[PbCO_3][PbO]} \] Since solid concentrations do not change with pressure, we can assume that the concentration of the solid reactants (PbCO3 and PbO) remains constant. \[ K = [CO_2] \] Use K values found in Step 2 to calculate the equilibrium pressure of CO2 at each temperature. #Results: Equilibrium pressure of CO2 at given temperatures#

From the equilibrium constants and pressure calculations, the equilibrium pressure of carbon dioxide (CO2) at the given temperatures can be calculated. Remember to express your results in the appropriate units (typically atm or Pa).

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Most popular questions from this chapter

(a) For each of the following reactions, predict the sign of \(\Delta H^{*}\) and \(\Delta S^{\circ}\) without doing any calculations. (b) Based on your general chemical knowledge, predict which of these reactions will have \(K>1\) at \(25^{\circ} \mathrm{C} .(\mathbf{c})\) In each case, indicate whether \(K\) should increase or decrease with increasing temperature. (i) \(2 \mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{FeO}(s)\) (ii) \(\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{Cl}(g)\) (iii) $\mathrm{NH}_{4} \mathrm{Cl}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{HCl}(g)$ (iv) $\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{CaO}(s) \rightleftharpoons \mathrm{CaCO}_{3}(s)$

The normal boiling point of \(n\) -octane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\( is \)125^{\circ} \mathrm{C}$. (a) Is the condensation of gaseous \(n\) -octane to liquid \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the boiling of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and gaseous \(n\) -octane are in equilibrium? Explain.

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Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, \((\mathbf{c})\) a solid sublimes?

Indicate whether each statement is true or false. (a) Unlike enthalpy, where we can only ever know changes in \(H,\) we can know absolute values of $S .(\mathbf{b})\( If you heat a gas such as \)\mathrm{CO}_{2}$, you will increase its degrees of translational, rotational and vibrational motions. (c) \(\mathrm{CO}_{2}(g)\) and \(\mathrm{Ar}(g)\) have nearly the same molar mass. At a given temperature, they will have the same number of microstates.

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