/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 A sample of pure \(\mathrm{PCl}_... [FREE SOLUTION] | 91Ó°ÊÓ

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A sample of pure \(\mathrm{PCl}_{5}(\mathrm{g})\) is introduced into an evacuated flask and allowed to dissociate. $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If the fraction of \(\mathrm{PCl}_{5}\) molecules that dissociate is denoted by \(\alpha,\) and if the total gas pressure is \(P\) show that $$ K_{\mathrm{p}}=\frac{\alpha^{2} P}{1-\alpha^{2}} $$

Short Answer

Expert verified
The derivation for the given equation of Kp is found by assigning the pressures based on the fraction of PCl5 that dissociates, α, and applying the equation for Kp.

Step by step solution

01

Understanding the reaction

First, understand the given chemical reaction. A sample of PCl5 dissociates into PCl3 and Cl2. This can be represented as: PCl5(g) ⇌ PCl3(g) + Cl2(g). This reaction is at equilibrium, meaning that the rate of the reaction in the forward direction is equal to the rate of the reverse reaction, and the concentrations of the reactants and products are not changing over time.
02

Assigning partial pressure

Since PCl5 dissociates into PCl3 and Cl2, if α is the fraction of PCl5 dissociated then αP is the partial pressure for both PCl3 and Cl2 and (1-α)P is the partial pressure for PCl5, at equilibrium.
03

Determining Kp

The equilibrium constant Kp for the reaction is given by the product of the partial pressures of the products divided by the partial pressure of the reactant. Kp = (PCl3 * Cl2) / PCl5 = [(αP)^2]/[(1-α)P]
04

Simplify the expression for Kp

Simplify the expression for Kp which leads to the final results. The P cancels out and we are left with Kp = α²/(1- α²)

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Key Concepts

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

Le Chatelier's Principle
Le Chatelier's Principle is a foundational concept in chemical equilibrium that helps predict how a system at equilibrium will respond to changes in concentration, temperature, or pressure. If a change is imposed on a system at equilibrium, the system will adjust to minimize that change and restore a new equilibrium state.
For the dissociation reaction of \[\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\]let's consider what happens if we alter the pressure. According to Le Chatelier's Principle:
  • Increasing the pressure will favor the side of the reaction with fewer moles of gas. Here, the reactant side has one mole of gas, compared to two moles on the product side. Therefore, increasing pressure will shift the equilibrium towards the formation of more \( \text{PCl}_5 \).
  • Conversely, decreasing pressure would shift the equilibrium towards the products due to having more moles.
Understanding this principle aids in manipulating a chemical reaction to achieve desired concentrations of reactants or products under given conditions while maintaining equilibrium.
Equilibrium Constant
The equilibrium constant \(K_p\) is a vital concept in chemical equilibrium that quantifies the ratio of the concentrations of products to reactants at equilibrium. In our dissociation reaction, \(PCl_5\) dissociates to form \(PCl_3\) and \(Cl_2\), and this can be expressed under pressure terms in the equation for \(K_p\):
\[K_p = \frac{(\text{PCl}_3)\cdot(\text{Cl}_2)}{\text{PCl}_5}\]
This equation uses partial pressures for gases, making it essential to know how those pressures relate with fractions of dissociation. Given that when a fraction \(\alpha\) of \(\text{PCl}_5\) dissociates:
  • Partial pressures for both \(\text{PCl}_3\) and \(\text{Cl}_2\) become \(\alpha P\).
  • Remaining pressure for \(\text{PCl}_5\) becomes \((1-\alpha)P\).
Plugging these into our equilibrium expression gives:\[K_p = \frac{(\alpha P)^2}{(1-\alpha)P} = \frac{\alpha^2 P}{1 - \alpha^2}\]This formula provides a clear quantitative relationship between dissociation, pressure, and equilibrium, making \(K_p\) a crucial factor in determining the state of equilibrium.
Dissociation Reaction
A dissociation reaction involves the breaking down of a compound into simpler substances or elements. For example, the reaction \[\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)\]is a typical dissociation reaction where a single compound \(\text{PCl}_5\) splits into two different gases, \(\text{PCl}_3\) and \(\text{Cl}_2\).
In a system like this, a key parameter is the degree of dissociation, denoted by \(\alpha\), which represents the fraction of the original compound that dissociates. This parameter is crucial as it impacts:
  • The equilibrium composition of the reaction mixture.
  • The partial pressures of the gases involved.
Thus, understanding dissociation reactions is critical for calculating key metrics like the equilibrium constant and for predicting how changes in conditions might alter a reaction's progress. The concept of percentage dissociation is practical in laboratory settings, where controlling reaction conditions can optimize the yield and purity of the desired products.

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Most popular questions from this chapter

A sample of air with a mole ratio of \(\mathrm{N}_{2}\) to \(\mathrm{O}_{2}\) of 79: 21 is heated to 2500 K. When equilibrium is established in a closed container with air initially at 1.00 atm, the mole percent of \(\mathrm{NO}\) is found to be \(1.8 \% .\) Calculate \(K_{\mathrm{p}}\) for the reaction. $$\mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})$$

A 1.100 L flask at \(25^{\circ} \mathrm{C}\) and 1.00 atm pressure contains \(\mathrm{CO}_{2}(\mathrm{g})\) in contact with \(100.0 \mathrm{mL}\) of a saturated aqueous solution in which \(\left[\mathrm{CO}_{2}(\mathrm{aq})\right]=3.29 \times 10^{-2} \mathrm{M}\) (a) What is the value of \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq}) ?\) (b) If 0.01000 mol of radioactive \(^{14} \mathrm{CO}_{2}\) is added to the flask, how many moles of the \(^{14} \mathrm{CO}_{2}\) will be found in the gas phase and in the aqueous solution when equilibrium is re-established? [Hint: The radioactive \(^{14} \mathrm{CO}_{2}\) distributes itself between the two phases in exactly the same manner as the nonradioactive \(\left.^{12} \mathrm{CO}_{2} .\right]\)

In the Ostwald process for oxidizing ammonia, a variety of products is possible- \(\mathrm{N}_{2}, \mathrm{N}_{2} \mathrm{O}, \mathrm{NO},\) and \(\mathrm{NO}_{2}-\) depending on the conditions. One possibility is $$\begin{aligned} \mathrm{NH}_{3}(\mathrm{g})+\frac{5}{4} \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) &+\frac{3}{2} \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \\ K_{\mathrm{p}} &=2.11 \times 10^{19} \mathrm{at} 700 \mathrm{K} \end{aligned}$$ For the decomposition of \(\mathrm{NO}_{2}\) at \(700 \mathrm{K}\) $$\mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \quad K_{\mathrm{p}}=0.524$$ (a) Write a chemical equation for the oxidation of \(\mathrm{NH}_{3}(\mathrm{g})\) to \(\mathrm{NO}_{2}(\mathrm{g})\) (b) Determine \(K_{\mathrm{p}}\) for the chemical equation you have written.

A classic experiment in equilibrium studies dating from 1862 involved the reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) to produce ethyl acetate and water. $$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$ The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content. $$\begin{array}{r} 2 \mathrm{CH}_{3} \mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq}) \rightleftharpoons \\ \mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array}$$ In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol is brought to equilibrium. A sample containing exactly one-hundredth of the equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\) \(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this experiment.

Can a mixture of \(2.2 \mathrm{mol} \mathrm{O}_{2}, 3.6 \mathrm{mol} \mathrm{SO}_{2},\) and \(1.8 \mathrm{mol}\) \(\mathrm{SO}_{3}\) be maintained indefinitely in a \(7.2 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=100\) in this reaction? Explain. $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) $$

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