If \(K_{c}=0.042\) for \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) at \(500 \mathrm{~K}\) what is the value of \(K_{p}\) for this reaction at this temperature?

Short Answer

Expert verified
The value of \(K_p\) for the reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) at 500 K is approximately 0.977.

Step by step solution

01

Calculate Δn

To find Δn, you need to subtract the total moles of gas on the reactants' side from the total moles of gas on the products' side. For the given reaction, we have \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) Δn = moles of products - moles of reactants \[Δn = 1 - (1 + 1) = -1\]
02

Use the relationship between Kc and Kp

Now that we have Δn, we can use the equation for the relationship between Kc and Kp: \(K_p = K_c (RT)^{\Delta n}\) We have Kc = 0.042, R = 0.0821 L atm/mol K, T = 500 K, and Δn = -1.
03

Calculate Kp

Substitute all the values into the equation and solve for Kp. \(K_p = (0.042)((0.0821)(500))^(-1)\) \(K_p = (0.042)((41.05))^(-1)\) \(K_p = 1.023^{-1}\) \(K_p ≈ 0.977\) The value of Kp for this reaction at 500 K is approximately 0.977.

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Key Concepts

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

Kc and Kp Relationship
Understanding the relationship between the equilibrium constants Kc and Kp is crucial when dealing with gas-phase reactions. In chemical equilibrium, Kc is used when concentrations are expressed in molarity, whereas Kp is used when dealing with partial pressures. Their relationship is bridged by the ideal gas law and can be represented as:

\( K_p = K_c (RT)^{\Delta n} \)

In this equation, R represents the ideal gas constant (0.0821 L atm/mol K), T is the temperature in kelvins, and \(\Delta n\) is the change in moles of gas (moles of gaseous products minus moles of gaseous reactants).

This relationship is of particular importance because it accounts for the different units used in measuring concentrations (Kc) and the pressures (Kp). Learning to connect these two constants allows students to transition between different conditions and representations of chemical equilibria.
Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry where the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentration of reactants and products over time. At this point, a dynamic balance is achieved, although the reactions continue to occur.

The equilibrium constant (Kc for concentrations or Kp for pressures) is a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium, each raised to the power of their respective coefficients from the balanced equation.

A key thing to remember is that the equilibrium constant is only affected by temperature; changes in concentration or pressure do not affect its value. Therefore, understanding the exact conditions under which the constant was calculated, such as the temperature, is pivotal in accurately using it for predictions and calculations.
Gas-Phase Reactions
Gas-phase reactions are characterized by reactants and products that are in the gaseous state. Analysis of these reactions often involves understanding concepts such as partial pressure, molecular behavior in gases, and the use of the Kp equilibrium constant.

Partial pressure is an important factor as it represents the pressure a gas would exert if it occupied the entire volume alone. It allows chemists to understand the behavior of individual gases within a mixture, a perspective that's crucial when calculating equilibrium constants in terms of pressure (Kp).

For students tackling gas-phase equilibrium problems, a solid grasp on how temperature, volume, and pressure interact according to the Ideal Gas Law will greatly assist in accurately establishing the equilibrium constants and predicting the behavior of the gaseous system.

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Most popular questions from this chapter

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C},\) the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) . \quad\) If \(\quad K_{c}=0.802\) what are the concentrations of each substance in the equilibrium mixture?

Suppose that you worked at the U.S. Patent Office and a patent application came across your desk claiming that a newly developed catalyst was much superior to the Haber catalyst for ammonia synthesis because the catalyst led to much greater equilibrium conversion of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) into \(\mathrm{NH}_{3}\) than the Haber catalyst under the same conditions. What would be your response?

Bioremediation is the use of microorganisms to degrade environmental pollutants. Many pollutants contain only carbon and hydrogen (oil being one example). The chemical reactions are complicated, but in general the microorganisms react the pollutant hydrocarbon with \(\mathrm{O}_{2}\) to produce \(\mathrm{CO}_{2}\) and other carbon-containing compounds that are incorporated into the organism's biomass. How would increasing levels of \(\mathrm{CO}_{2}\) in the environment affect the bioremediation reaction?

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s)\) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) .\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

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