Titel: Working Method Approach for Introductory Physical Chemistry Calculations (RSC Paperbacks)
Verlag: Royal Society of Chemistry
Erscheinungsjahr: 1997
Sprache: Englisch
ISBN-10: 0854045538
ISBN-13: 9780854045532
Einband: Softcover
Zustand: New

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Inhaltsangabe

A Working Method Approach for Introductory Physical Chemistry Calculations is a concise inexpensive introduction to first year chemistry that is aimed at students who are weak in chemistry or have no chemistry on entry to university. Such students usually find physical chemistry the most difficult part of the chemistry course, and within this section numerical problem solving is an additional difficulty. The text should also be invaluable to first year intending chemists. This text provides an introduction to physical chemistry and the gas laws, followed by chapters on thermodynamics, chemical equilibrium, electrochemistry and chemical kinetics. Each section involves a brief introduction followed by a representative examination question, which is broken down into a proposed working method. Both short multiple-choice questions and related full examination-type questions are included. This book will prove invaluable to students who need encouragement in a logical approach to problem solving in physical chemistry, teaching them to think for themselves when faced with a problem.

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A Working Method Approach for Introductory Physical Chemistry Calculations

Numerical and Graphical Problem Solving

By Brian Murphy, Clair Murphy, Brian Hathaway

The Royal Society of Chemistry

Chapter 1 Introduction to Physical Chemistry: Acids and Bases The Gas Laws, Numerical and Graphical Problem Solving, 1, 
Chapter 2 Thermodynamics I: Internal Energy, Enthalpy, First Law of Thermodynamics, State Functions, and Hess's Law, 17, 
Chapter 3 Thermodynamics 11: Enthalpy, Heat Capacity, Entropy, the Second and Third Laws of Thermodynamics and Gibbs Free Energy, 24, 
Chapter 4 Equilibrium I: Introduction to Equilibrium and Le Ch�telier's Principle, 36, 
Chapter 5 Equilibrium 11: Aqueous Solution Equilibria, 47, 
Chapter 6 Electrochemistry I: Galvanic Cells., 63, 
Chapter 7 Electrochemistry 11: Electrolytic Cells, 89, 
Chapter 8 Chemical Kinetics I: Basic Kinetic Laws, 113, 
Chapter 9 Chemical Kinetics II: The Arrhenius Equation and Graphical Problems, 128, 
Answers to Problem, 145, 
Further Reading, 147, 
Periodic Table of the Elements, 148, 
Subject Index, 149,

CHAPTER 1

Introduction to Physical Chemistry: Acids and Bases, The Gas Laws and Numerical and Graphical Problem Solving

INTRODUCTION

This chapter is a brief introduction to many of the assumptions made in the remainder of this text and the basis of physical chemistry type problems. The spider diagram of Figure 1.1 illustrates the various sections.

STATES OF MATTER

Mutter is the chemical term for materials. There are three states of matter: the solid phase (s), the liquid phase (1) and the gaseous phase (g). In the solid phase, all the atoms or molecules are arranged in a highly ordered manner [Figure 1,2(a)], whereas in the liquid phase [Figure 1.2(b)], this ordered structure is not as evident. In the gaseous phase [Figure 1.2(c)), all the particles are moving at high velocity, in random motion. The disorder or entropy, S, is at its maximum in the gaseous phase [Figure 1.2(c)].

If a species is dissolved in water, it is said to be in the aqueous phase (aq), and the symbol can be represented as a subscript, e.g. HCl(aq).

ACIDS AND BASES

An acid is a proton (H+) donor and a base is a proton acceptor, e.g. (OH-). Examples of acids include HCl, H2SO4, HNO3, HCN and CH3CO2H. A monoprotic acid is an acid with one replaceable proton, e.g. HCl (eA = 1); a diprotic acid is an acid with two replaceable protons e.g. H2SO4 (eA = 2) etc., where eA is the number of reactive species. A dilute acid is an acid which contains a small amount of acid dissolved in a large quantity of water, whereas a concentrated acid is an acid which contains a large amount of acid dissolved in a small quantity of water.

Examples of bases include NaOH (eB = 1), KOH (eB = 1), Ba(OH)2 (eB = 2), Ca(OH)2 (eB. An acid combines with a base to form a salt and water:

i.e. ACID + BASE [right arrow] SALT + WATER

e.g. HNO3 + NaOH [right arrow] NaNO3 + H2O

In general, an acid can be represented as HA, where HA [right arrow] H+ + A- or, more precisely, HA + H2O [right arrow] A- + H3O+, since all aqueous protons are solvated by water. Likewise, a base containing hydroxide anions, OH-, can be represented as MOH, where MOH [right arrow] M+ + OH-.

When an acid donates a proton, H+, it is said to form the conjugate base of the acid, i.e. HA (acid) [??] H+ + A- (conjugate base). The conjugate base is a base since it can accept a proton to reform HA, the acid. Similarly, when a base accepts a proton, H+, the conjugate acid of the base is said to be formed, i.e. B (base) + H+ [??] HB+ (conjugate acid). The conjugate acid is an acid since it can donate a proton, H+, and reform the base, e.g. NH4+ (conjugate acid) [??] NH3 (base) + H+.

Ions, Cations, Anions, Oxyanions and Oxyacids

Ions are charged species, e.g. Na+, Cl-, NH4+etc. Cations are positively charged ions, e.g. Na+, NH4+, Mg2+, H3O+etc. Anions are negatively charged species, e.g. OH-, Cl-, O2-etc. A useful way of remembering this is the two n's, i.e. anion = negatively charged ion! An oxyanion, as its name suggests, is an anion containing oxygen, e.g. NO-3, SO2-4etc. An oxyacid is the corresponding acid of the oxyanion, e.g. HNO3 and H2SO4 are the oxyacids of the nitrate and sulfate oxyanions respectively. The oxidation state or oxidation number of the nitrogen, NV and the sulfur, SVI, is the same in both the oxyacid and the oxyanion. Table 1.1 is a summary of some of the common oxyanions and their corresponding oxyacids.

THE GAS LAWS — IDEA OF PROPORTIONALITY

Boyle's Law

Pressure is defined as the force acting on a unit area, i.e. p = F/A. The unit of pressure is the newton per square metre, N m-2 or the Pascal, Pa. At sea level, the pressure due to the weight of the earth's atmosphere is approximately 105 Pa. The bar, is often used as the unit of pressure in problems in physical chemistry, where 1 bar = 105 N m-2 or 105 Pa.

Consider the effect of a piston pressing down on a fixed mass of gas of initial pressure pinitial and initial volume Vinitial, (Figure 1.3).

As the pressure applied by the piston is increased, the volume of the gas decreases (i.e. the space it occupies), if the temperature is kept constant. This is Boyle's Law – the volume of a definite mass of gas at constant temperature is inversely proportional to its pressure.

i.e.Boyle's Law: V [varies] 1/p or V = k/p, where k is a constant of proportionality.

Example: A sample of gas G used in an air conditioner has a volume of 350 dm3 and a pressure of 85 kPa at 25 �C. Determine the pressure of the gas at the same temperature when the volume is 500 dm3.

Solution:

The gas is at constant temperature, and therefore Boyle's Law can be applied:

Hence initially, V1 = k/p1 or k = p1 V1

Hence k = (85 kPa) x (350 dm3)

= 29750 kPa dm3

However finally, p2 = k/V2

= 29750/500

= 59.5 kPa

Answer: Final Pressure = 59.5 kPa

Charles's Law

In contrast, if the pressure is kept constant, the volume of a definite mass of gas will increase, if the temperature is raised. This is Charles's Law: the volume of a definite mass of gas at constant pressure is directly proportional to its temperature.

i.e. Charles's Law: V [varies] T

or V = kT, where k is a different constant of proportionality.

Example: A sample of gas G occupies 200 cm3 at 288 K and 0.87 bar. Determine the volume the gas will occupy at 303 K and at the same pressure.

Solution:

The gas is at constant pressure, and therefore Charles's Law can be applied:

Hence initially, V1 = kT1 or k = V1/T1

Hence k = (200 cm3)/(288 K)

= 0.694 cm3 K-1

Now V2 = kT2, so, V2 = (0.694 cm3 K-1) x (303 K)

= 210.42 cm3

Answer: Final Pressure = 210.42 cm3

Avogadro's Law

This states that equal volumes of gases, measured at the same temperature...

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