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Chapter 13: Problem 60

Lexan is a plastic used to make compact discs, eyeglass lenses, and bullet- proof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right)\), an extremely poisonous gas. Phosgene decomposes by the reaction $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an jnitial pressure of \(1.0\) atm decomposes, calculate the equilibrium pressures of all species.

Short Answer

Expert verified
The equilibrium pressures of all species are \(\mathrm{COCl}_{2}: 1.0\; \mathrm{atm}\), \(\mathrm{CO}: 2.6\times10^{-5}\; \mathrm{atm}\), and \(\mathrm{Cl}_{2}: 2.6\times10^{-5}\; \mathrm{atm}\).

Step by step solution

01

Write the equilibrium expression

The equilibrium expression for the given reaction is: $$ K_{\mathrm{p}} = \frac{[\mathrm{CO}] [\mathrm{Cl}_{2}]}{[\mathrm{COCl}_{2}]} $$ where \(K_{\mathrm{p}} = 6.8 \times 10^{-9}\), \([\mathrm{CO}]\) is the partial pressure of \(\mathrm{CO}\), \([\mathrm{Cl}_{2}]\) is the partial pressure of \(\mathrm{Cl}_{2}\), and \([\mathrm{COCl}_{2}]\) is the partial pressure of \(\mathrm{COCl}_{2}\).
02

Set up an ICE table

An ICE (Initial, Change, and Equilibrium) table is a way to organize information about concentrations or pressures for a reaction at equilibrium. In this case, we will organize the partial pressures of the three species. It looks like this: ``` Initial Change Equilibrium COCl2 1.0 atm -x 1.0-x CO 0 atm +x x Cl2 0 atm +x x ``` Here, "Initial" refers to the initial pressures, "Change" refers to the changes in pressures, and "Equilibrium" refers to the equilibrium pressures.
03

Use the equilibrium expression with the ICE table values

Now, we can plug in the equilibrium pressures from the ICE table into our equilibrium expression: $$ 6.8 \times 10^{-9} = \frac{x \cdot x}{1.0-x} $$
04

Solve for x

To simplify the expression, we can rewrite it as: $$ x^2 = 6.8 \times 10^{-9}(1.0-x) $$ Since the value of \(K_{\mathrm{p}}\) is quite small, we can make an approximation that x will be much smaller than 1.0. Therefore, we can write the equation as: $$ x^2 = 6.8 \times 10^{-9} $$ Taking the square root of both sides, we get: $$ x = \sqrt{6.8 \times 10^{-9}} = 2.6\times10^{-5} $$
05

Calculate the equilibrium pressures of all species

Based on our ICE table, the equilibrium pressures of the three species are as follows: $$ [\mathrm{COCl}_{2}]_{\text{eq}} = 1.0 - x = 1.0 - 2.6\times10^{-5} \approx 1.0\; \mathrm{atm} $$ $$ [\mathrm{CO}]_{\text{eq}} = x = 2.6\times10^{-5}\; \mathrm{atm} $$ $$ [\mathrm{Cl}_{2}]_{\text{eq}} = x = 2.6\times10^{-5}\; \mathrm{atm} $$ So, the equilibrium pressures of all species are \(\mathrm{COCl}_{2}: 1.0\; \mathrm{atm}\), \(\mathrm{CO}: 2.6\times10^{-5}\; \mathrm{atm}\), and \(\mathrm{Cl}_{2}: 2.6\times10^{-5}\; \mathrm{atm}\).

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Most popular questions from this chapter

Write expressions for \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) b. \(\mathrm{CO}_{2}(g)+\mathrm{MgO}(s) \rightleftharpoons \mathrm{MgCO}_{3}(s)\) c. \(\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2}(g)\) d. \(4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g)\)

A 1.00-L flask was filled with \(2.00\) moles of gaseous \(\mathrm{SO}_{2}\) and \(2.00\) moles of gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) moles of gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ If \(2.0\) moles of \(\mathrm{NO}\) and \(1.0\) mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.

Le Châtelier's principle is stated (Section \(13.7)\) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

An important reaction in the commercial production of hydrogen is $$ \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) $$ How will this system at equilibrium shift in each of the five following cases? a. Gaseous carbon dioxide is removed. b. Water vapor is added. c. In a rigid reaction container, the pressure is increased by adding helium gas. d. The temperature is increased (the reaction is exothermic). e. The pressure is increased by decreasing the volume of the reaction container.
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