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Chapter 12: Problem 60

Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof 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,$$\operatorname{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$,for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

Short Answer

Expert verified
The equilibrium pressures of the species involved in the decomposition of phosgene gas (\(\mathrm{COCl}_{2}\)) are approximately 0.99992 atm for COCl2, 8.2 × 10^{-5} atm for CO, and 8.2 × 10^{-5} atm for Cl2.

Step by step solution

01

Set up ICE table

An ICE table keeps track of the initial pressures of the species involved, the change in pressure, and the equilibrium pressures. The equilibrium constant expression for the given reaction is: \(K_\mathrm{p} = \frac{[\mathrm{CO}][\mathrm{Cl}_{2}]}{[\mathrm{COCl}_{2}]}\) The table is filled with the following information: - The initial pressure of phosgene is 1.0 atm, and the initial pressures of CO and Cl2 are 0 atm, as they are not initially present. - We will set the change in pressure for CO and Cl2 as x and phosgene as -x, since they are being formed and phosgene is decomposing. - The equilibrium pressures of the species are: COCl2 = 1.0 - x, CO = x, and Cl2 = x.
02

Write the expression for \(K_{\mathrm{p}}\)

Substituting the equilibrium pressures from the ICE table into the equilibrium constant expression, we have: \(K_{\mathrm{p}} = \frac{x^{2}}{(1.0 - x)}\) We are given that \(K_{\mathrm{p}} = 6.8 \times 10^{-9}\). So, we can solve for x: \(6.8 \times 10^{-9} = \frac{x^{2}}{(1.0 - x)}\)
03

Solve for x

Since \(K_{\mathrm{p}}\) is very small, we can use the approximation that x is much smaller than 1. Therefore, the denominator 1.0 - x can be approximated as 1.0. This simplifies the equation to: \(6.8 \times 10^{-9} = \frac{x^{2}}{1.0}\) Rearranging and taking the square root of both sides, we get: \(x = \sqrt{6.8 \times 10^{-9}}\) Now, calculate the value of x: \(x \approx 8.2 \times 10^{-5}\)
04

Find the equilibrium pressures

Using the value of x, we can now calculate the equilibrium pressures of each species from the equilibrium values in the ICE table: For COCl2: 1.0 - x ≈ 1.0 - 8.2 × 10^{-5} = 0.99992 atm For CO: x = 8.2 × 10^{-5} atm For Cl2: x = 8.2 × 10^{-5} atm The equilibrium pressures of the species are approximately 0.99992 atm for COCl2, 8.2 × 10^{-5} atm for CO, and 8.2 × 10^{-5} atm for Cl2.

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Most popular questions from this chapter

A sample of \(S_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{K}\) at an initial pressure of 1.00 atm, where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$\mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{S}_{2}(g)$$. At equilibrium, the partial pressure of \(S_{8}\) is 0.25 atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{K}\)

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?

At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$. At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

Predict the shift in the equilibrium position that will occur for each of the following reactions when the volume of the reaction container is increased. a. \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) b. \(\mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g)\) d. \(\operatorname{COCl}_{2}(g) \rightleftharpoons \operatorname{CO}(g)+\mathrm{Cl}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\)

For the reaction $$2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,it is determined that, at equilibrium at a particular temperature, the concentrations are as follows: \([\mathrm{NO}(g)]=8.1 \times 10^{-3} \mathrm{M}\) \(\left[\mathrm{H}_{2}(g)\right]=4.1 \times 10^{-5} \mathrm{M},\left[\mathrm{N}_{2}(g)\right]=5.3 \times 10^{-2} \mathrm{M},\) and \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=\) \(2.9 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.
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