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Chapter 15: Problem 89

At \(700 \mathrm{K},\) the equilibrium constant for the reaction $$\operatorname{CCI}_{4}(g) \Longrightarrow \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=0.76 .\) A flask is charged with 2.00 atm of \(\mathrm{CCl}_{4}\) ,which then reaches equilibrium at 700 \(\mathrm{K}\) . (a) What fraction of the CCl\(_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?(\mathbf{b})\) what are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

Short Answer

Expert verified
At 700 K, the equilibrium constant for the reaction is \(K_p = 0.76\). When a flask containing 2.00 atm of CCl\(_4\) reaches equilibrium, 50.7% of the CCl\(_4\) is converted into C and Cl\(_2\). At equilibrium, the partial pressures are 1.493 atm for CCl\(_4\) and 1.014 atm for Cl\(_2\).

Step by step solution

01

Write down the balanced chemical equation

The given reaction is: \[CCl_4(g) ⇄ C(s) + 2Cl_2(g)\]
02

Express the equilibrium constant (Kp) in terms of partial pressures

For the given reaction, the equilibrium constant expression is: \[K_p = \frac{P_{Cl_2}^2}{P_{CCl_4}}\]
03

Set up an equilibrium pressure table (ICE table)

Let x represent the fraction of CCl\(_4\) that dissociates. We set up the following table: | | Initial | Change | Equilibrium | |---------|---------|---------|-----------------| | CCl\(_4\) | 2.00 atm| -x | 2.00 - x | | Cl\(_2\) | 0 atm | 2x | 2x |
04

Substitute equilibrium pressures in Kp expression

Using the information from the table, we can write the Kp expression as: \[K_p = \frac{(2x)^2}{2.00 - x}\] We are given that \(K_p = 0.76\). So, we have: \[0.76 = \frac{(2x)^2}{2.00 - x}\]
05

Solve for x (fraction of CCl4 converted)

We can now solve this equation for x: \[0.76 (2.00 - x) = (2x)^2\] \[1.52 - 0.76x = 4x^2\] Rearrange the equation: \[4x^2 + 0.76x - 1.52 = 0\] Now solve the quadratic equation for x either by factoring, completing the square, or using the quadratic formula. Using the quadratic formula, we find: \[x = 0.507 (rounded)\] This means 50.7% of the CCl\(_4\) is converted to C and Cl\(_2\) at equilibrium.
06

Calculate the partial pressures at equilibrium

Now we can find the equilibrium partial pressures using the equilibrium pressure values from the table: Partial pressure of CCl\(_4\): \[P_{CCl_4} = 2.00 - x = 2.00 - 0.507 = 1.493 \, atm\] Partial pressure of Cl\(_2\): \[P_{Cl_2} = 2x = 2(0.507) = 1.014 \, atm\] Therefore, at equilibrium, the partial pressures are 1.493 atm for CCl\(_4\) and 1.014 atm for Cl\(_2\).

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Most popular questions from this chapter

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at 1000 \(\mathrm{K}\) . (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2},\) or does it favor NOBr? (b) Calculate \(K_{c}\) for \(2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) Calculate \(K_{c}\) for \(\operatorname{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\)

The following equilibria were measured at 823 K: \begin{equation} \begin{aligned} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67 \\\ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) & \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=0.14 \end{aligned} \end{equation} (a) Use these equilibria to calculate the equilibrium constant, \(K_{c},\) for the reaction \(\operatorname{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)\) \(+\mathrm{CO}_{2}(g)\) at 823 \(\mathrm{K}\) . (b) Based on your answer to part (a), would you say that carbon monoxide is a stronger or weaker reducing agent than \(\mathrm{H}_{2}\) at \(T=823 \mathrm{K} ?\) (c) If you were to place 5.00 \(\mathrm{g}\) of \(\mathrm{CoO}(s)\) in a sealed tube with a volume of 250 \(\mathrm{mL}\) that contains \(\mathrm{CO}(g)\) at a pressure of 1.00 atm and a temperature of \(298 \mathrm{K},\) what is the concentration of the CO gas? Assume there is no reaction at this temperature and that the CO behaves as an ideal gas (you can neglect the volume of the solid). (d) If the reaction vessel from part (c) is heated to 823 \(\mathrm{K}\) and allowed to come to equilibrium, how much \(\mathrm{CoO}(s)\) remains?

A mixture of 0.2000 mol of \(\mathrm{CO}_{2}, 0.1000 \mathrm{mol}\) of \(\mathrm{H}_{2},\) and 0.1600 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.000 -L vessel. The following equilibrium is established at \(500 \mathrm{K} :\) $$\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ (a) Calculate the initial partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2,}\) and \(\mathrm{H}_{2} \mathrm{O}\) (b) At equilibrium \(P_{\mathrm{H}_{2} \mathrm{O}}\) \(=3.51\) atm. Calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}, \mathrm{H}_{2},\) and \(\mathrm{CO} .(\mathbf{c})\) Calculate \(K_{p}\) for the reaction. (d) Calculate \(K_{c}\) for the reaction.

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C} .\) The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \Longrightarrow \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g)$$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S}\) taken together) is 0.614 atm. What is \(K_{p}\) for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the pharmaceutical industry. This reaction is catalyzed by strong acid (usually \(\mathrm{H}_{2} \mathrm{SO}_{4} ) .\) A simple example is the reaction of acetic acid with ethyl alcohol to produce ethyl acetate and water: $$\begin{aligned} \mathrm{CH}_{3} \mathrm{COOH}(s o l v)+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(s o l v) & \rightleftharpoons \\ \mathrm{CH}_{3} \mathrm{COOCH}_{2} \mathrm{CH}_{3}(\mathrm{solv}) &+\mathrm{H}_{2} \mathrm{O}(\text {solv}) \end{aligned}$$ where \(^{a}(s o l v)^{\prime \prime}\) indicates that all reactants and products are in solution but not an aqueous solution. The equilibrium constant for this reaction at \(55^{\circ} \mathrm{C}\) is 6.68 . A pharmaceutical chemist makes up 15.0 \(\mathrm{L}\) of a solution that is initially 0.275 \(\mathrm{M}\) in acetic acid and 3.85\(M\) in ethanol. At equilibrium, how many grams of ethyl acetate are formed?
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