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Phosgene, \(\mathrm{COCl}_{2}\), is a toxic gas used in the manufacture of urethane plastics. The gas dissociates at high temperature. $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ At \(400^{\circ} \mathrm{C}\), the equilibrium constant \(K_{c}\) is \(8.05 \times 10^{-4}\). Find the percentage of phosgene that dissociates at this temperature when \(1.00\) mol of phosgene is placed in a \(25.0\) - \(\mathrm{L}\) vessel.

Short Answer

Expert verified
Approximately 14.18% of phosgene dissociates at 400°C.

Step by step solution

01

Identify Initial Conditions

Start by noting that we have initially 1 mol of \( \text{COCl}_{2} \) in a 25.0 L vessel. The initial concentration \( [\text{COCl}_{2}]_0 \) is calculated by dividing moles by volume: \[ [\text{COCl}_{2}]_0 = \frac{1.00}{25.0} = 0.040\, \text{mol/L}. \]
02

Set up the ICE Table

Create an ICE (Initial, Change, Equilibrium) table to track concentrations:- Initial: \( [\text{COCl}_{2}] = 0.040 \), \([\text{CO}] = 0\), \([\text{Cl}_2] = 0\).- Change: \([\text{COCl}_{2}] = -x\), \([\text{CO}] = +x\), \([\text{Cl}_2] = +x\).- Equilibrium: \([\text{COCl}_{2}] = 0.040 - x\), \([\text{CO}] = x\), \([\text{Cl}_2] = x\).
03

Write the Expression for Equilibrium Constant

The equilibrium constant \( K_c \) for the balanced reaction is given by:\[ K_c = \frac{[\text{CO}][\text{Cl}_2]}{[\text{COCl}_{2}]} = \frac{x^2}{0.040 - x}. \]Given that \( K_c = 8.05 \times 10^{-4} \).
04

Solve for x

Solve the equation:\[ \frac{x^2}{0.040 - x} = 8.05 \times 10^{-4}. \]Assume \( x \) is small compared to 0.040, so \( 0.040 \approx 0.040 - x \).This simplifies to:\[ x^2 \approx 8.05 \times 10^{-4} \times 0.040. \]\[ x^2 \approx 3.22 \times 10^{-5}. \]\[ x \approx \sqrt{3.22 \times 10^{-5}} \approx 0.00567. \]
05

Calculate the Percentage Dissociation

The dissociation of \( \text{COCl}_{2} \) is given by \( \frac{x}{0.040} \times 100\% \).Substituting the value of \( x \):\[ \text{Percentage} = \frac{0.00567}{0.040} \times 100\% \approx 14.18\%. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phosgene Dissociation
Phosgene, denoted as \( \text{COCl}_2 \), is a chemical compound that can undergo dissociation at elevated temperatures. In this dissociation process, phosgene breaks down into carbon monoxide \( \text{CO} \) and chlorine gas \( \text{Cl}_2 \). This reaction is an example of a reversible chemical reaction and can be represented as: \[ \text{COCl}_2 (g) \rightleftharpoons \text{CO} (g) + \text{Cl}_2 (g) \]Being a toxic gas used in industries, phosgene's behavior in chemical reactions is significant. Understanding the conditions under which it dissociates helps in managing its chemical stability and safety during its use in manufacturing processes. The dissociation process influences how phosgene is handled in industrial settings, especially relevant in high-temperature environments where the forward reaction is more favored.
Equilibrium Constant
The equilibrium constant \( K_c \) is a crucial aspect of chemical reactions that reach equilibrium. It gives insight into the proportions of products and reactants at equilibrium, indicating whether the reactants or products are favored. For the dissociation of phosgene:\[ K_c = \frac{[\text{CO}][\text{Cl}_2]}{[\text{COCl}_{2}]} \]In this reaction, \( K_c \) is given as \( 8.05 \times 10^{-4} \) at \( 400^{\circ} \text{C} \). This relatively small value of \( K_c \) suggests that, at equilibrium, the concentration of phosgene remains significantly higher than that of its dissociated products.Here's why it matters:
  • A smaller \( K_c \) value indicates the reaction does not go to completion and remains mostly as reactants.
  • It is a measure that assists chemists in predicting the extent of dissociation under given conditions.
  • Understanding \( K_c \) is vital for controlling the conditions to enhance or inhibit the dissociation based on the desired chemical process outcome.
ICE Table
The ICE table is a fundamental tool used in chemistry to solve equilibrium problems. ICE stands for Initial, Change, and Equilibrium, and helps to visualize how concentrations of reactants and products change as the reaction proceeds to equilibrium.For phosgene dissociation, we start with the initial amount of moles of each species:
  • Initial: \([\text{COCl}_2] = 0.040\,\text{mol/L}, \,[\text{CO}] = 0, \,[\text{Cl}_2] = 0\)
Then account for changes as the reaction progresses:
  • Change: \([\text{COCl}_2] = -x, \,[\text{CO}] = +x, \,[\text{Cl}_2] = +x\)
At equilibrium, all concentrations can be expressed in terms of \( x \), making it easier to apply mathematical expressions to find unknowns:
  • Equilibrium: \([\text{COCl}_2] = 0.040 - x, \,[\text{CO}] = x, \,[\text{Cl}_2] = x\)
Using an ICE table helps ensure accurate calculation of equilibrium concentrations and is often the first step in tackling equilibrium problems in chemistry.
Percentage Dissociation
Percentage dissociation is a term used to express how much of a compound has dissociated in relation to the total amount initially present. It is calculated using the expression:\[ \text{Percentage Dissociation} = \frac{x}{\text{Initial concentration}} \times 100\% \]Where \( x \) is the change in concentration of the dissociating species (in this case, phosgene). For our phosgene problem:
  • Change in concentration, \( x = 0.00567 \).
  • Initial concentration = \( 0.040 \text{ mol/L} \).
  • Percentage dissociation is \( \frac{0.00567}{0.040} \times 100\% \approx 14.18\% \).
This value of 14.18% signifies that a modest portion of the phosgene has dissociated at \( 400^{\circ} \text{C} \), which again correlates with the relatively low \( K_c \) value indicating that equilibrium lies towards the undissociated molecules. Such calculations are useful in predicting how much of a gas will undergo reaction under specific conditions.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C}\) in a closed system, ammonium hydrogen sulfide exists as the following equilibrium: $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ a. When a sample of pure \(\mathrm{NH}_{4} \mathrm{HS}(s)\) is placed in an evacuated reaction vessel and allowed to come to equilibrium at \(25^{\circ} \mathrm{C}\), total pressure is \(0.660 \mathrm{~atm} .\) What is the value of \(K_{p} ?\) b. To this system, sufficient \(\mathrm{H}_{2} \mathrm{~S}(g)\) is injected until the pressure of \(\mathrm{H}_{2} \mathrm{~S}\) is three times that of the ammonia at equilibrium. What are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{~S}\) ? c. In a different experiment, \(0.750\) atm of \(\mathrm{NH}_{3}\) and \(0.500 \mathrm{~atm}\) of \(\mathrm{H}_{2} \mathrm{~S}\) are introduced into a \(1.00-\mathrm{L}\) vessel at \(25^{\circ} \mathrm{C}\). How many moles of \(\mathrm{NH}_{4} \mathrm{HS}\) are present when equilibrium is established?

Which of the following reactions involve homogeneous equilibria and which involve heterogeneous equilibria? Explain the difference. a. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) b. \(2 \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(s) \rightleftharpoons 2 \mathrm{CuO}(s)+4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) c. \(2 \mathrm{~N}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) d. \(2 \mathrm{NH}_{3}(g)+3 \mathrm{CuO}(s) \rightleftharpoons 3 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \operatorname{Cu}(s)\)

(a) Predict the direction of reaction when chlorine gas is added to an equilibrium mixture of \(\mathrm{PCl}_{3}, \mathrm{PCl}_{5}\), and \(\mathrm{Cl}_{2}\). The reaction is $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ (b) What is the direction of reaction when chlorine gas is removed from an equilibrium mixture of these gases?

Iodine and bromine react to give iodine monobromide, IBr. $$ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) $$ What is the equilibrium composition of a mixture at \(150^{\circ} \mathrm{C}\) that initially contained \(0.0015\) mol each of iodine and bromine in a 5.0-L vessel? The equilibrium constant \(K_{c}\) for this reaction at \(150^{\circ} \mathrm{C}\) is \(1.2 \times 10^{2}\)

When \(1.0\) mol each of \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) are mixed at a certain high temperature, they react to give a final mixture consisting of \(0.5 \mathrm{~mol}\) each of \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) and \(1.0 \mathrm{~mol} \mathrm{HI}(g)\). Why do you obtain the same final mixture when you bring \(2.0 \mathrm{~mol} \mathrm{HI}(g)\) to the same temperature?

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