Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 80

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)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the dissociation of \(\mathrm{I}_{2}(\mathrm{g})\) at about \(1200^{\circ} \mathrm{C}\) \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K_{\mathrm{c}}=1.1 \times 10^{-2} .\) What volume flask should we use if we want 0.37 mol I to be present for every \(1.00 \mathrm{mol} \mathrm{I}_{2}\) at equilibrium?

The standard enthalpy of reaction for the decomposition of calcium carbonate is \(\Delta H^{\circ}=813.5 \mathrm{kJmol}^{-1}\) As temperature increases, does the concentration of calcium carbonate increase, decrease, or remain the same? Explain.

Write an equilibrium constant, \(K_{\mathrm{p}},\) for the formation from its gaseous elements of (a) 1 mol \(\mathrm{NOCl}(\mathrm{g})\) (b) \(2 \mathrm{mol} \mathrm{ClNO}_{2}(\mathrm{g}) ;\) (c) \(1 \mathrm{mol} \mathrm{N}_{2} \mathrm{H}_{4}(\mathrm{g}) ;\) (d) \(1 \mathrm{mol}\) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\)

A mixture of \(\mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) and \(\mathrm{CH}_{4}(\mathrm{g})\) in the mole ratio 2: 1 was brought to equilibrium at \(700^{\circ} \mathrm{C}\) and a total pressure of 1 atm. On analysis, the equilibrium mixture was found to contain \(9.54 \times 10^{-3} \mathrm{mol} \mathrm{H}_{2} \mathrm{S} .\) The \(\mathrm{CS}_{2}\) pre- sent at equilibrium was converted successively to \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and then to \(\mathrm{BaSO}_{4} ; 1.42 \times 10^{-3} \mathrm{mol} \mathrm{BaSO}_{4}\) was obtained. Use these data to determine \(K_{\mathrm{p}}\) at \(700^{\circ} \mathrm{C}\) for the reaction $$\begin{aligned} 2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+& 4 \mathrm{H}_{2}(\mathrm{g}) \\\ & K_{\mathrm{p}} \text { at } 700^{\circ} \mathrm{C}=? \end{aligned}$$

Formamide, used in the manufacture of pharmaceuticals, dyes, and agricultural chemicals, decomposes at high temperatures. $$\begin{array}{r} \mathrm{HCONH}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \\ K_{\mathrm{c}}=4.84 \text { at } 400 \mathrm{K} \end{array}$$ If \(0.186 \mathrm{mol} \mathrm{HCONH}_{2}(\mathrm{g})\) dissociates in a 2.16 Lflask at 400 K, what will be the total pressure at equilibrium?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free