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

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 and pressure, contain equal numbers of molecules.

i.e.Avogadro's Law: V [varies] n, where n = the amount of gas (measured in moles).

The mole is defined as the amount of a substance which contains as many elementary species as there are atoms in 12 g of the carbon-12 isotope. A mole contains 6.02205 x 1023 particles, where NA is defined as Avogadro's constanti.e. NA = 6.02265 x 1023 mol-1.

Ideal Gases

An ideal gas is a theoretical concept, a gas which obeys the gas laws perfectly. If the three gas laws are combined, the resulting equation is the equation of state of an ideal gas:

(a) Boyle's Law: V [varies] 1/p

(b) Charles's Law: V [varies] T

(c) Avogadro's Law: V [varies] n (a), (b) and (c) [??] V [varies] (Tn)/p

[??] = pV [varies] nT

[??] = pV = knT,

where k is another constant of proportionality called the Universal Gas Constant, R.

[Ideal Gas Equation:] pV = nRT

where p = pressure of the gas (measured in bar); V = volume of the gas (measured in dm3); n = amount of gas (measured in mol); T = temperature of the gas (measured in K); R = Universal Gas Constant = 0.08314 dm3 bar K-1 mol-1 (or 8.314 J K-1 mol-1). The above equation can also be modified to take into account changes in temperature, ΔT, changes in volume, ΔV, and changes in the coefficients of gaseous reagents, Δvg, respectively, i.e. pV = nRΔT, pΔV = nRT and pV = Δvg,RT. vg represents the coefficients in a chemical equation. For example, in the reaction N2(g) + 3H2(g) [right arrow] 2NH3(g), Δvg = [summation][v(Gaseous products)] - [summation][v(Gaseous reactants)] = [(2)]-[(1) + (3)] = -2.

Molar Volume

1 mole of an ideal gas measured at 25 °C and 1 bar pressure occupies 24.8 dm3 (where 1 dm3 = 1000 cm3).

i.e. 1 mole of an ideal gas at 25 °C and 1 bar pressure occupies 24.8 dm3.

Example: Calculate the amount of gas in moles in 2000 cm3 at 25 °C and 1 bar pressure.

Solution:

At 25 °C and 1 bar pressure, 1 mole of an ideal gas occupies 24.8 dm3

i.e. 24.8 dm3 = 1 mol

1 dm3 = (1/24.8) mol

2000 cm3 = 2 dm3 = (2/24.8) mol = 0.081 mol

Answer: 0.081 mol.

KINETIC THEORY OF GASES

Kinetic energy is the energy a body possesses by virtue of its motion. The molecules of gases travel at high velocity and hence have high kinetic energy. The kinetic theory of gases is used to explain the observed properties of gases, of which Brownian motion provides good evidence. Brownian motion is the irregular zigzag movement of very small particles suspended in a liquid or gas. If tobacco smoke is placed in a small cell, well-illuminated on a microscope stage, the tiny particles appear to be moving at random, as shown in Figure 1.2(c). This is due to the smaller invisible molecules of the air, colliding with the particles of the smoke. This is Brownian motion.

The kinetic theory of gases is a model proposed to account for the observed properties of gases. In order that the model is applicable, certain assumptions are made. For this reason, gases are categorised into two types: (a) ideal gases (as defined previously) and (b) real gases (defined as non-ideal gases).

Assumptions of the Kinetic Theory for an Ideal Gas

1. Gases consist of tiny molecules, which are so small and so far apart that the actual size of the molecules is negligible in comparison to the large distance between them.

2. The molecules are totally independent of each other, i.e. there are no attractive or repulsive forces between the molecules.

3. The molecules are in constant random motion. They collide with each other and with the walls or sides of the container, which changes the direction of linear motion.

4. For each elastic collision, there is no net loss of kinetic energy. However, there may be transfer of energy between the particles in such a collision.

5. The average kinetic energy, of all the molecules is assumed to be proportional to the absolute temperature T (measured in K) i.e. kinetic energy [varies] T.

Validity of the Assumptions of the Kinetic Theory for an Ideal Gas

Assumption 1 This is largely true, since the compressibility of gases is very high. However, at high pressures, when a gas is highly compressed, it is not valid to state that the physical size of gas molecules is practically negligible compared with the distances between the molecules.

Assumption 2 This assumption is approximate, since gases diffuse or spread out to occupy all space available to them, i.e. there must be no appreciable binding force between the molecules. However, van der Waals forces exist (intermolecular forces of attraction and repulsion), and with polar molecules, other attractive forces exist.

Assumption 3 This assumption is valid, as shown by Brownian motion.

Assumptions 4 and 5 These assumptions are true, since in any elastic collision, kinetic energy is not lost.

Conclusion of the Kinetic Theory of Gases

The kinetic theory of gases is a good approximation, used to explain the behaviour of real gases. The theory has to be modified at very high pressures and in the presence of van der Waals forces, and for polar molecules where stronger intermolecular forces of attraction are involved.

THE GENERAL GAS EQUATION

Consider a gas at temperature T1, pressure p1 and volume V1, and another gas at temperature T2and pressure p2. The volume of the latter gas can be determined easily using the equation:

p1 V1/T1 = p2 V2/T2

A useful way of remembering this is 'peas and Vegetables go on the Table'! In fact, given any five of the above variables, the sixth can be evaluated using the equation. This will form the basis of the worked example at the end of this chapter.

Standard State

The standard state of a body is the most stable state of that body at 25 °C and 1 bar pressure (its symbol is °, e.g. E°, ΔH°, ΔS°, defined later in Chapters 2, 3 and 6 respectively).

Standard State — Most Stable State — S.S. — 25 °C and 1 bar pressure.

IMPORTANCE OF UNITS IN PHYSICAL CHEMISTRY

In physical chemistry questions, the International system of units (SI) should be used. In any problem, one of the first steps is to convert all units to SI; note especially that temperature must always be given in kelvins, never degrees centigrade:

Remember: T (K) = T (°C) + 273

e.g. 25 °C is equal to 298 K since (25 + 273) K = 298 K. Table 1.2 lists (a) the basic SI units, (b) the derived SI units, and (c) some examples of non-SI units, which are commonly used in physical chemistry.

A GENERAL WORKING METHOD TO SOLVE NUMERICAL PROBLEMS IN PHYSICAL CHEMISTRY

1. Read the question carefully — do not be put off by the sheer length or intricacy of a question. If you do not see the wood for the trees immediately — don't worry — all will be revealed if you use a stepwise systematic approach! Just break down the question, step by step!

2. If a chemical equation is involved, identify all the species present, along with their states, i.e. (s), (1), (aq) or (g). This is particularly relevant in thermodynamics, equilibrium and electrochemistry questions.

3. Write down the balanced chemical equation if necessary, including the stoichiometry factors, vA, vB, VC and vD respectively: VA A + VB B [right arrow] VC C + vD D.

4. Identify all the data given in the question, including any constants which are involved, e.g. R = 8.314 J K-1 mol-1, 1 F = 96 500 C mol-1 (defined later), 1 mole of a gas behaving ideally at 25 °C and 1 bar pressure occupies 24.8 dm3.

5. Convert all units to the one system, i.e. change °C to K, hours to seconds, etc. Watch out especially for (a) standard state conditions, e.g. E°, ΔH° ΔS° and ΔG°; parameters: 25 °C (298 K) and 1 bar pressure, and; (b) R= 0.08314 dm3 bar K-1 mol-1 (used when the pressure is expressed in bar) or more generally R = 8.314 J K-1 mol-1. For example, suppose you were asked to calculate the volume occupied by 0.5 mol of N2(g) at 280 K and 0.93 bar, using the ideal gas equation, i.e. p V = nRT, and rearranging, V= nRT/p. Since p, the pressure of the gas, is given in bar, R= 0.08314 dm3 bar K-1 mol-1 must be used, not R = 8.314 J K-1 mol-1. Therefore, V = (0.5 mol) x (0.08314 dm3 bar K-1 mol-1 x (280 K)/0.93 bar = 12.52 dm3.

6. Identify the unknown in the question, i.e. what quantity is being looked for specifically? Do not be afraid to sketch a simple diagram if this identifies the problem for you!

7. Write down all relevant formulae. The question will most likely involve just one or, less likely, more than one, of these equations.