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Chapter 17: Problem 77

At \(1200 \mathrm{~K}\), an equilibrium mixture of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) gases contains 98.31 mol percent CO and some solid carbon. The total pressure of the mixture is \(1.00 \mathrm{~atm}\). For the system $$ \mathrm{CO}_{2}(g)+\mathrm{C}(s) \rightleftharpoons 2 \mathrm{CO}(g) $$ calculate (a) \(P_{\mathrm{Co}}\) and \(P_{\mathrm{CO}_{2}}\) (b) \(K\) (c) \(\Delta G^{\circ}\) at \(1200 \mathrm{~K}\)

Short Answer

Expert verified
The partial pressures of CO and CO2 are 0.9831 atm and 0.0169 atm, respectively. The equilibrium constant, K, is 56.93, and the change in Gibbs free energy at 1200 K is approximately -70.0 kJ/mol.

Step by step solution

01

Calculate the partial pressures of CO and CO2

Given that the CO mixture contains 98.31 mol % CO, that means CO2 is the remaining 1.69 mol % of the gas mixture. The total pressure is 1.00 atm, so to determine the partial pressures we can multiply the mol percentages by the total pressure: \(P_{\mathrm{CO}} = 0.9831 \times 1.00 \mathrm{~atm} = 0.9831 \mathrm{~atm}\) \(P_{\mathrm{CO_2}} = 0.0169 \times 1.00 \mathrm{~atm} = 0.0169 \mathrm{~atm}\)
02

Write the equilibrium expression for the reaction

The equilibrium expression for the given reaction is: \(K = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}} \times P_{\mathrm{C}}} = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}}}\) Since carbon is in a solid state, its pressure can be considered as 1, so we have: \(K = \frac{(P_{\mathrm{CO}})^2}{P_{\mathrm{CO_2}}}\)
03

Calculate the equilibrium constant (\(K\))

Now we will use the partial pressures from step 1 to calculate the equilibrium constant: \(K = \frac{(0.9831)^2}{0.0169} = 56.93\)
04

Calculate the change in Gibbs free energy (\(\Delta G^{\circ}\)) at 1200 K

The relationship between \(\Delta G^{\circ}\) and \(K\) is given by the equation: \(\Delta G^{\circ} = -R \times T \times \ln{K}\) Where \(R\) is the gas constant (8.314 J/mol K) and \(T\) is the temperature (1200 K). Now we will substitute the values into the equation: \(\Delta G^{\circ} = -8.314 \times 1200 \times \ln{56.93} = -70046.72 \mathrm{~J/mol}\) So, the change in Gibbs free energy at 1200 K is approximately -70.0 kJ/mol.

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Most popular questions from this chapter

Predict the sign of \(\Delta S\) for the following. (a) ice cream melting (b) boiling water (c) dissolving instant coffee in hot water (d) sugar, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), decomposing to carbon and steam

Some bacteria use light energy to convert carbon dioxide and water to glucose and oxygen: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \quad \Delta G^{\circ}=2870 \mathrm{~kJ}\) at \(25^{\circ} \mathrm{C}\) Other bacteria, those that do not have light available to them, couple the reaction $$ \mathrm{H}_{2} \mathrm{~S}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{S}(s) $$ to the glucose synthesis above. Coupling the two reactions, the overall reaction is \(24 \mathrm{H}_{2} \mathrm{~S}(g)+6 \mathrm{CO}_{2}(g)+6 \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+18 \mathrm{H}_{2} \mathrm{O}(l)+24 \mathrm{~S}(s)\) Show that the reaction is spontaneous at \(25^{\circ} \mathrm{C}\).

Use Table \(17.1\) to calculate \(\Delta S^{\circ}\) for each of the following reactions. (a) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (b) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)\) (c) \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (d) \(2 \mathrm{NaCl}(s)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{NaF}(s)+\mathrm{Cl}_{2}(g)\)

The normal boiling point for ethyl alcohol is \(78.4^{\circ} \mathrm{C} .5^{\circ}\) for \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) is \(282.7 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\). At what temperature is the vapor pressure of ethyl alcohol \(357 \mathrm{~mm} \mathrm{Hg} ?\)

Pencil "lead" is almost pure graphite. Graphite is the stable elemental form of carbon at \(25^{\circ} \mathrm{C}\) and 1 atm. Diamond is an allotrope of graphite. Given $$ \text { diamond: } \Delta H_{\mathrm{f}}^{\circ}=1.9 \mathrm{~kJ} / \mathrm{mol} ; S^{\circ}=2.4 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K} $$ at what temperature are the two forms in equilibrium at 1 atm? \(\mathrm{C}\) (graphite) \(\rightleftharpoons \mathrm{C}\) (diamond)
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