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\(K_{\mathrm{p}}\) for the formation of phosgene, \(\mathrm{COCl}_{2},\) is \(6.5 \times 10^{11} \mathrm{at}\) \(25^{\circ} \mathrm{C}\) $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{COCl}_{2}(\mathrm{g}) $$ What is the value of \(K_{\mathrm{p}}\) for the dissociation of phosgene? $$ \operatorname{COCl}_{2}(g) \rightleftarrows \mathrm{CO}(g)+\mathrm{Cl}_{2}(\mathrm{g}) $$

Short Answer

Expert verified
\(K_\mathrm{p}\) for dissociation of phosgene is \(1.54 \times 10^{-12}\).

Step by step solution

01

Understand the Reaction Reversal

The given equilibrium constant, \(K_\mathrm{p}\), is for the forward reaction formation of phosgene: \( \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows \mathrm{COCl}_{2}(\mathrm{g}) \). We are asked to find \(K_\mathrm{p}\) for the reverse reaction: \( \operatorname{COCl}_{2}(g) \rightleftarrows \mathrm{CO}(g)+\mathrm{Cl}_{2}(\mathrm{g}) \). In a chemical equilibrium, the \(K_\mathrm{p}\) of the reverse reaction is the reciprocal of the \(K_\mathrm{p}\) of the forward reaction.
02

Calculate the Reciprocal

To find the \(K_\mathrm{p}\) for the dissociation of phosgene, we need to take the reciprocal of the given \(K_\mathrm{p}\) for the formation of phosgene. Given that \(K_\mathrm{p} = 6.5 \times 10^{11}\) for the forward reaction, the reciprocal is calculated as: \[ K_\mathrm{p,\: reverse} = \frac{1}{6.5 \times 10^{11}} \].
03

Solve for the Reverse \(K_\mathrm{p}\)

Perform the calculation for the reciprocal that represents the equilibrium constant of the reverse reaction. \[ K_\mathrm{p,\: reverse} = \frac{1}{6.5 \times 10^{11}} \approx 1.54 \times 10^{-12} \]. This is the \(K_\mathrm{p}\) value for the dissociation of phosgene.

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, typically represented as \( K \), is a critical value in chemical reactions that helps determine the balance between products and reactants. For gaseous reactions, it is often represented as \( K_p \), where "p" stands for pressure, indicating that the concentrations of gases are expressed in partial pressures. This constant is derived from the law of mass action and its value tells us about the extent of a reaction.

In the given exercise, the constant \( K_p \) for the formation of phosgene is \( 6.5 \times 10^{11} \). This large number implies that at equilibrium, phosgene (\( ext{COCl}_2 \)) is highly favored over its constituent gases, carbon monoxide (\( ext{CO} \)) and chlorine (\( ext{Cl}_2 \)).

The equilibrium constant provides insights by:
  • Quantifying the ratio of product concentrations to reactant concentrations at equilibrium.
  • Indicating the direction in which a reaction is favored; a large \( K \) suggests more products, while a small \( K \) indicates more reactants.
  • Helping predict whether a system will reach equilibrium quickly or if it might require external intervention."
Chemical Equilibrium
Chemical equilibrium is a dynamic state where the concentrations of all reactants and products remain constant over time. At this point, the rate of the forward reaction equals the rate of the reverse reaction. It is crucial to understand that equilibrium does not mean that the reactants and products are in equal concentrations, rather their ratios remain constant.

In the context of phosgene formation, equilibrium indicates that the forward reaction continuously converts carbon monoxide and chlorine into phosgene, while the reverse reaction converts some phosgene back into its constituents. However, due to a high \( K_p \) value for the forward reaction, there's much more phosgene than the reactants at equilibrium.

Key features of chemical equilibrium include:
  • Constant macroscopic properties, such as concentration and pressure.
  • The system is closed with no net change in concentrations or pressures over time.
  • Equilibrium can be shifted by changing conditions, like temperature or pressure, according to Le Chatelier's principle."
Reaction Reversal
Reaction reversal involves considering the opposite direction of a given chemical reaction. Understanding the reversal is essential because it allows chemists to predict outcomes when conditions change.

When you reverse a reaction, the equilibrium constant of the reverse reaction is the reciprocal of the forward reaction's equilibrium constant. For the dissociation of phosgene, the \( K_p \) is calculated by taking the reciprocal of its formation constant. This exercise illustrates the critical role of reaction reversals.The key points include:
  • The calculation for the reverse process involves finding \( \frac{1}{K_p} \) of the forward process.
  • A small \( K_p \) for the reverse reaction indicates that dissociation into \( \text{CO} \) and \( \text{Cl}_2 \) is not favored at equilibrium.
  • Understanding reaction reversals helps manage chemical processes, such as optimizing yield in industrial applications."

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Most popular questions from this chapter

Summary and Conceptual Questions The following questions may use concepts from preceding chapters. Decide whether each of the following statements is true or false. If false, change the wording to make it true. (a) The magnitude of the equilibrium constant is always independent of temperature. (b) When two chemical equations are added to give a net equation, the equilibrium constant for the net equation is the product of the equilibrium constants of the summed equations. (c) The equilibrium constant for a reaction has the same value as \(K\) for the reverse reaction. (d) Only the concentration of \(\mathrm{CO}_{2}\) appears in the equilibrium constant expression for the reaction \(\mathrm{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) (e) For the reaction \(\operatorname{CaCO}_{3}(\mathrm{s}) \rightleftarrows \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g})\) the value of \(K\) is numerically the same no matter whether the amount of \(\mathrm{CO}_{2}\) is expressed as moles/liter or as gas pressure.

The total pressure for a mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) is 1.5 atm. If \(K_{p}=6.75\) (at \(25^{\circ} \mathrm{C}\) ), calculate the partial pressure of each gas in the mixture. $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

A mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=5.0 \times 10^{-3}\) \(\mathrm{mol} / \mathrm{L},\left[\mathrm{O}_{2}\right]=1.9 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{SO}_{3}\right]=6.9 \times 10^{-3}\) mol/L. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) \quad K=279 $$

Sulfur dioxide is readily oxidized to sulfur trioxide. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) \quad K=279 $$ If we add \(3.00 \mathrm{g}\) of \(\mathrm{SO}_{2}\) and \(5.00 \mathrm{g}\) of \(\mathrm{O}_{2}\) to a 1.0 - \(\mathrm{L}\). flask, approximately what quantity of \(\mathrm{SO}_{3}\) will be in the flask once the reactants and the product reach equilibrium? (a) \(2.21 \mathrm{g}\) (c) \(3.61 \mathrm{g}\) (b) \(4.56 \mathrm{g}\) (d) \(8.00 \mathrm{g}\) (Note: The full solution to this problem results in a cubic equation. Do not try to solve it exactly. Decide only which of the answers is most reasonable.)

Write equilibrium constant expressions for the following reactions. For gases use either pressures or concentrations. (a) \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})\) (b) \(\mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{g})\) (c) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftarrows 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) (d) \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftarrows 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})\)

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