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Chapter 14: Problem 31

The following equilibrium constants were determined at \(1123 \mathrm{~K}\) $$\begin{array}{ll}\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) & K_{P}^{\prime}=1.3 \times 10^{14} \\\\\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) & K_{P}^{\prime \prime}=6.0 \times 10^{-3}\end{array}$$ Write the equilibrium constant expression \(K_{P}\), and calculate the equilibrium constant at \(1123 \mathrm{~K}\) for $$\mathrm{C}(s)+\mathrm{CO}_{2}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{COCl}_{2}(g)$$

Short Answer

Expert verified
The equilibrium constant \(K_P\) at \(1123 \mathrm{~K}\) for the reaction \(C(s) + CO_2(g) + 2 Cl_2(g) \rightleftharpoons 2 COCl_2(g)\) is \(7.8 \times 10^{11}\).

Step by step solution

01

Analyze the reactions and the respective equilibrium constants

The first reaction is the formation of CO from C and CO2 with an equilibrium constant of \(1.3 \times 10^{14}\). The second reaction shows the formation of COCl2 from CO and Cl2 with an equilibrium constant of \(6.0 \times 10^{-3}\).
02

Write the given reaction juxtaposed with the old reactions

The target reaction is: \(C(s) + CO_2(g) + 2 Cl_2(g) \rightleftharpoons 2 COCl_2(g)\). We observe that the two given reactions can add up to give the target reaction. Thus, the equilibrium constant of target reaction would be the product of equilibrium constants of the reactions that make up the target reaction.
03

Determine the equilibrium constant of the final reaction

The product of the two given equilibrium constants will be the equilibrium constant of the target reaction. Therefore, \(K_P = K_P' \times K_P'' = 1.3 \times 10^{14} \times 6.0 \times 10^{-3} = 7.8 \times 10^{11}\).

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Most popular questions from this chapter

Industrially, sodium metal is obtained by electrolyzing molten sodium chloride. The reaction at the cathode is \(\mathrm{Na}^{+}+e^{-} \longrightarrow \mathrm{Na}\). We might expect that potassium metal would also be prepared by electrolyzing molten potassium chloride. However, potassium metal is soluble in molten potassium chloride and therefore is hard to recover. Furthermore, potassium vaporizes readily at the operating temperature, creating hazardous conditions. Instead, potassium is prepared by the distillation of molten potassium chloride in the presence of sodium vapor at \(892^{\circ} \mathrm{C}\) : $$\mathrm{Na}(g)+\mathrm{KCl}(l) \rightleftharpoons \mathrm{NaCl}(l)+\mathrm{K}(g)$$ In view of the fact that potassium is a stronger reducing agent than sodium, explain why this approach works. (The boiling points of sodium and potassium are \(892^{\circ} \mathrm{C}\) and \(770^{\circ} \mathrm{C},\) respectively. \()\)

Consider the reaction $$\begin{array}{r}2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) \\\\\Delta H^{\circ}=-198.2 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Comment on the changes in the concentrations of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at equilibrium if we were to (a) increase the temperature; (b) increase the pressure; (c) increase \(\mathrm{SO}_{2}\); (d) add a catalyst; (e) add helium at constant volume.

A reaction vessel contains \(\mathrm{NH}_{3}, \mathrm{~N}_{2},\) and \(\mathrm{H}_{2}\) at equilibrium at a certain temperature. The equilibrium concentrations are \(\left[\mathrm{NH}_{3}\right]=0.25 M,\left[\mathrm{~N}_{2}\right]=0.11 M\) and \(\left[\mathrm{H}_{2}\right]=1.91 \mathrm{M} .\) Calculate the equilibrium constant \(K_{\mathrm{c}}\) for the synthesis of ammonia if the reaction is represented as (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (b) \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g)\)

The equilibrium constant \(K_{\mathrm{c}}\) for the following reaction is 1.2 at \(375^{\circ} \mathrm{C}\). $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ (a) What is the value of \(K_{P}\) for this reaction? (b) What is the value of the equilibrium constant \(K_{\mathrm{c}}\) for \(2 \mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) ?\) (c) What is the value of \(K_{c}\) for \(\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)\) \(\rightleftharpoons \mathrm{NH}_{3}(g) ?\) (d) What are the values of \(K_{P}\) for the reactions described in (b) and (c)?

At \(25^{\circ} \mathrm{C}\), a mixture of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) gases are in equilibrium in a cylinder fitted with a movable piston. The concentrations are \(\left[\mathrm{NO}_{2}\right]=0.0475 \mathrm{M}\) and \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.487 \mathrm{M} .\) The volume of the gas mixture is halved by pushing down on the piston at constant temperature. Calculate the concentrations of the gases when equilibrium is reestablished. Will the color become darker or lighter after the change? [Hint: \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) to \(\mathrm{NO}_{2}\) is \(4.63 \times 10^{-3} . \mathrm{N}_{2} \mathrm{O}_{4}(g)\) is colorless and \(\mathrm{NO}_{2}(g)\) has a brown color. \(]\)
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