General Chemistry (CHEM 101) PDF

General Chemistry CHEM (101) First Level Students i INDEX Contents CHAPTER I.................................................................................................................................... - 3 - The Gaseous State....................................................................................................................... - 3 - The Ideal Gas Laws:................................................................................................................. - 4 - Boyle's Law (1962)................................................................................................................... - 4 - Charles (1787) and Gay-Lussac Law (1802)............................................................................. - 5 - The PT behavior of Ideal Gases............................................................................................... - 6 - Equation of State for an Ideal Gas.......................................................................................... - 7 - The Numerical Value of the Gas Constant "R"....................................................................... - 8 - Energy Units............................................................................................................................. - 8 - Some Properties of Gas Mixtures- Dalton's Law of Partial Pressure (1801)....................... - 10 - Graham’s Law of effusion...................................................................................................... - 11 - The Kinetic Theory of Ideal or Perfect Gases........................................................................ - 12 - Postulates of the Kinetic Theory:.......................................................................................... - 13 - Fundamental Equation of Kinetic Theory............................................................................. - 16 - Deduction of Gas Laws:......................................................................................................... - 20 - Heat Capacity of Gases.......................................................................................................... - 24 - Real or Imperfect Gases........................................................................................................ - 29 - Equation of Van der Waal's................................................................................................... - 29 - Explanation of the deviation or real gases on the basis of.................................................. - 31 - Van der Waal's equation....................................................................................................... - 31 - Liquefaction of Gases............................................................................................................ - 33 - Critical Constants of Gases.................................................................................................... - 33 - The Joule Thomson Effect..................................................................................................... - 35 - CHAPTER II................................................................................................................................. - 36 - Liquids, Solids and Change of States......................................................................................... - 36 - General Properties of liquids................................................................................................ - 37 - Heat of Vaporization:............................................................................................................ - 41 - Vapor Pressure....................................................................................................................... - 45 - ii Le Chatelier's principle.......................................................................................................... - 51 - Vapor Pressure Curves for Liquids........................................................................................ - 53 - Boiling Point........................................................................................................................... - 57 - Freezing Point........................................................................................................................ - 61 - Surface tension...................................................................................................................... - 63 - Factors affecting surface tension.......................................................................................... - 67 - Measurement of Surface Tension......................................................................................... - 67 - Capillary rise method............................................................................................................ - 67 - Example:................................................................................................................................. - 70 - Viscosity................................................................................................................................. - 71 - Measurement of absolute viscosities................................................................................... - 73 - Factors affecting viscosity..................................................................................................... - 74 - Example...................................................................................................................................... - 75 - Pertinent phenomena............................................................................................................... - 75 - Solid State.................................................................................................................................. - 77 - General Properties..................................................................................................................... - 77 - Crystal Lattice............................................................................................................................ - 78 - Binding forces in Crystals...................................................................................................... - 80 - Hydrogen Bonds........................................................................................................................ - 83 - Vapor pressure of solid............................................................................................................. - 84 - Heat capacities of Solids........................................................................................................ - 84 - Isomorphism.......................................................................................................................... - 85 - Solved Problems........................................................................................................................ - 86 - CHAPTER III................................................................................................................................ - 88 - Solutions.................................................................................................................................... - 88 - Types of solutions:................................................................................................................. - 88 - Concentration units:.............................................................................................................. - 89 - The solution processes.......................................................................................................... - 92 - Heats of Solution:.................................................................................................................. - 98 - Solution of Liquids in Liquids.............................................................................................. - 100 - Solution of Solids in Liquids:............................................................................................... - 103 - Solubility and Temperature:............................................................................................... - 107 - iii The Effect of Pressure on Solubility:................................................................................... - 109 - Vapor Pressures of Solutions:............................................................................................. - 112 - COLLIGATIVE PROPERTIES OF SOLUTIONS......................................................................... - 120 - Osmosis and Osmotic Pressure:.......................................................................................... - 122 - Semi permeable Membranes:............................................................................................. - 122 - Measurement of Osmotic Pressure:................................................................................... - 124 - Laws of Osmotic Pressure:.................................................................................................. - 125 - CHAPTER I................................................................................................................................ - 128 - GENERAL ATOMIC STRUCTURE............................................................................................... - 128 - Electrons.............................................................................................................................. - 128 - The Nucleus.......................................................................................................................... - 129 - Atom Mass........................................................................................................................... - 131 - Atomic weight...................................................................................................................... - 132 - Mass Number....................................................................................................................... - 133 - Atomic Number................................................................................................................... - 134 - CHAPTER II............................................................................................................................... - 136 - Electronic Configuration of the Atom..................................................................................... - 136 - Bohr's Postulates and Bohr's Theory.................................................................................. - 140 - Calculation of the Energy of the Electron........................................................................... - 144 - Structure of the Hydrogen Atom According to Bohr.......................................................... - 147 - Sommerfeld Theory............................................................................................................. - 149 - Modern Theory of Atomic Structure.................................................................................. - 153 - Spin Quantum Number “mS”............................................................................................... - 159 - Electronic Distribution and Possible Number of Electrons in Each Level in the atom...... - 162 - The exact Electronic Configuration and Electronic Symbols for the Different Elements. - 165 - CHAPTER III.............................................................................................................................. - 175 - Periodic TABLES and PERIODIC LAW....................................................................................... - 175 - Modern Periodic Table........................................................................................................ - 179 - Types of Elements Based on Electronic Structures............................................................ - 184 - CHAPTER IV.............................................................................................................................. - 195 - TRENDS OF PROPERTIES IN PERIODIC TABLE......................................................................... - 195 - I) Atomic Size................................................................................................................... - 195 - iv II) Ionization Potential (I.P).............................................................................................. - 201 - III) Electron Affinity. (E.A)............................................................................................. - 204 - IV) Electronegativity...................................................................................................... - 205 - v Units and Dimensions We are deal with two kinds of unite international system units (Sl) units and Gram centimeters units. Dimensions SI units Gram. Cm. units 1. Mass (m) Kg. grm 2. Time (t) Sec. Sec. 3. Length (l) Meter (m) Cm 4. Electric Current Ampere (A) (A) 5. Temperature (T) Kelvin (K) K 6. Area (A2) m2 Cm2 7. Volume (V) m3 Cm3 Velocity (v) 8. m / Sec. cm I Sec. (distance/time) Acceleration 9. m / Sec2 cm / Sec2 (Velocity/time) Density (d) 1 0. Kg. / m3 grm / cm3 (Weight /volume) Force (F) Kg. m / Sec2 grm. cm. / Sec2 11. (mass*acceleration) Newton (N) Dyne Energy (E) Kg.m2/sec2 grm. cm2 / Sec2 1 2. (Force*length) Joule (J) Erg Kinetic Energy mv2 ½Kg. m2 /Sec2 ½ grm cm2 / Sec2 1 3. Joule Erg 14. Momentum Kg. m / Sec grm cm. / Sec. Mass * velocity 2 (grm. cm. / sec2 ) / crn2 2 Pressure (P) (Kg. m / Sec ) / m 1 5. 2 grm. / cm. Sec2 (Force /Area) Kg. /m.Sec (Pascal Pa) (atmosphere) -1- Different temperature scale Degree Degree Degree Property Celsius Kelvin Fahrenheit 0 C K 0F 1. Boiling Point of 1000C. 373 K 2120 F Water 2. Body Temperature 370C 310 K 98.60 F 3. Room Temperature 250C 298 K 77 0 F 4. Freezing Point 00 C 273 K 320 F 1. Kelvin is called the absolute temperature 2. The size of a Celsius degree on the Fahrenheit scale is only (100/180) or (5/9), so: 0 F = 32 + 1.8 0 c Problems: 1. Melting point (M. P.) of an alloy is 2240C, what is the M. P. in degree Fahrenheit? 2. B. P, of He is - 452 0 F, converts to degree Celsius. 3. Hg found as liquid at room temperature, it melts at 238.9 0C, Convert to Kelvin's. 4. Convert the following: a. 327.5 0 C to 0 F b. 172.9 0 F to 0 c c. 77 K to 0 c -2- CHAPTER I The Gaseous State Matter exists in three physical states gaseous, liquid, and solid. A gas has no internal boundary. It expands to fill any container completely regardless of the size or shape of the container. A liquid has one internal boundary surface. It fills its container below its surface regardless of the shape of the container. A solid is rigid, that is, it bounds itself internally in all dimensions. It needs no external container. The properties of these three states of matter are related to the concept of the structure of matter, and a study of them will help to become more familiar with chemical behavior of matter. The physical behavior of a gas is, to a first approximation, independent on its chemical composition and is determined, instead, by variables volume, pressure, temperature, and the number of moles of the substance. The equation of state of the system is mathematical relationship between the values of these four properties. Only three of these must be specified to describe the state; the fourth can be calculated from the equation of state, which is obtained from knowledge of the experimental behavior of the system. -3- The Ideal Gas Laws: An ideal gas may be defined as a gas to which the laws of Boyle, Gay- Lussac and Avogadro are applicable temperature. Boyle's Law (1962) It states that "At constant temperature the volume of a definite mass of a gas is inversely proportional to the pressure", that is: V α 1/ P or PV = constant = Kl (1) Where, V is the volume, P the pressure and kl is a proportionality constant depends on experimental conditions (t0C), the gas nature, and mass of gas. According to equation (1) it follows that for the same amount of gas: PlVl = P2V2 = Kl or PI / P 2 = V 2 / V l The plot of pressure P versus volume V give a hyperbolic curve (isotherm) as shown in Figure (1). -4- Charles (1787) and Gay-Lussac Law (1802) The law states that "at constant pressure, the volume of any gas expands by the same fraction of its volume at O O C for every I O C rise of temperature", thus, if V0 is the volume of a given mass of gas at O OC and Vt is that at tOC, thus we can say: Vt = Vo (1 + αI t).......... (3) where, α is the coefficient of cubical expansion, should be the same for all gases. The most recent observations indicate that: α = 1 / 273.16 , then equation (3) becomes Vt = Vo [+ (t / 273.16) ] or Vt = Vo / 273.16 (273.16 + t)........ (4) If this relationship remains applicable at very low temperature, it is evident that the volume of an ideal gas must become zero when temperature is -273.16 0 C this is known as "absolute zero" of the ideal gas scale of temperature. This addition of -273.160 to the centigrade temperature gives the so-called absolute scale, or "Kelvin Scale". Absolute represented by T0K, (4) may, therefore, be written as: Vt = (VO/TO) or (VI/ Tt) = (VO/TO) (5) For a given mass of an ideal gas at constant pressure, therefore, the quantity (V/T) is constant or one can say, -5- V / T = k2.................. (6) where, K2 is a proportionality constant depends on experimental conditions, the gas nature, (P atm) and mass of gas. Figure (2) represents the relation (isobars). All slopes are equal to (V o/ 273.16). The PT relation gives the isochors or isometrics. Fig. (2) The PT behavior of Ideal Gases It follows from equations (1) and (6) that for a given mass of an ideal gas PV/T = constant = k3 (7) where, k3 is a proportionality constant depends on the gas nature and mass of the gas. The dependence of volume on both variable -6- (pressure and temperature) can be illustrated by a point on a three- dimensional surface, described by the variables P, V and T. Figure (3). Fig. (3) Equation of State for an Ideal Gas The assumption of Avogadro's law, that equal volumes of all gases, under the same conditions of temperature and pressure, contain equal numbers of molecules, leads to the conclusion that in equation (7) should be independent on the nature of the gas, provided one mole is considered. The volume occupied by this quantity of an ideal gas at a given temperature and pressure will always be the same, so that for 1 mole (PV / T) must be a universal constant, irrespective by the symbol "R", provided it behaves ideally, PV = nRT............ (8) R is known as the molar gas constant. For "n" moles of an ideal gas, equation (8) becomes: PV = nRT............ (9) -7- Where, 𝑤𝑡.𝑜𝑓 𝑡ℎ𝑒 𝑔𝑎𝑠 𝑝𝑒𝑟.𝑔. 𝑚/𝑔 =......... (10) 𝐼𝑡𝑠 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 𝑀.𝑤𝑡. The density "d" can be expressed as (m / V). Rearrangement of eq. (9) gives: (m / M) RT or M = (m /V)(RT / P) or M = (d / P) RT..... (11) The Numerical Value of the Gas Constant "R" The value of "R" varies according to the units used for volume and pressure in the generalization equation. The numerical value for the gas constant can be obtained from the result that at 1 atm, and OOC (S.T.P.) one mole of a gas that behaves ideally occupies 22.414 liters. Substituting these data in the generalization equation we get: 𝑃𝑉 1. 𝑹= 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒 𝑇 1 𝑎𝑡𝑚 𝑥 22.44 𝐿𝑖𝑡𝑟𝑒 𝑹= = 0.082 𝐿𝑖𝑡𝑟 𝑎𝑡𝑚 K-1 Mole-1..... (12) 273.16 This value of R is used in calculation of gas laws. Energy Units If pressure is written as force per unit area and the volume as area time's length, one can write as -8- 𝑓𝑜𝑟𝑐𝑒 𝑃𝑉 = 𝑥 𝑎𝑟𝑒𝑎 𝑥 𝐿𝑒𝑛𝑔𝑡ℎ = 𝑓𝑜𝑟𝑐𝑒 𝑥 𝐿𝑒𝑛𝑔𝑡ℎ 𝑎𝑟𝑒𝑎 The dimensions of force time's length are those of energy. It follows that R has the dimensions of energy per Kelvin per mole (eq. 12). A numerical value of R involving an energy units will now obtain. For the value of R in egg: it is necessary to express the pressure in cm.grm.sec. Units, so that 1 atm., i. e. 76 cm Hg becomes: - 76 x 980.6 x 13.595 = 1.0132 x 106 dynes/cm2, and the volume of 1 mole of ideal gas is 22414 ml or c.c. (cm3), Hence, R equal: 10132 𝑋 106 𝑥 22414 a. 𝑅 = = 8.314 𝑥 107 𝑒𝑟𝑔 𝐾 −1 𝑚𝑜𝑙 −1 373 b. For electrochemical processes, and Joule = 107 erg ∴ 𝑅 = 8.314 𝐽 𝐾 −1 𝑚𝑜𝑙−1 c. For thermochemistry scale ⸪ 1 cal. = 4.184 J 1 𝑐𝑎𝑙 ∴ 𝑅 = 8.314 𝐽 𝐾 −1 𝑚𝑜𝑙 −1 𝑥 = 1.987 𝑐𝑎𝑙 𝐾 −1 𝑚𝑜𝑙 −1 4.184 𝐽 ≅ 2 𝑐𝑎𝑙 𝐾 −1 𝑚𝑜𝑙−1 -9- Some Properties of Gas Mixtures- Dalton's Law of Partial Pressure (1801) Equation (9) can be applied also to gases that are mixtures of different components, on the basis that the total pressure exerted by a mixture of gases is equal to the sum of the pressures which each component would exert if placed separately into the container. Dalton's law introduces the term partial pressure to denote the pressure exerted by one component of gaseous mixtures. The total pressure "P" is then the sum of the partial pressures of the components, i. e., P = P1 + P2 +P3 + ……........... (16) The ideal-gas law can then be applied to each component to give: 𝑅𝑇 𝑅𝑇 𝑅𝑇 𝑅𝑇 𝑅𝑇 𝑃𝑇 = 𝑛1 +𝑛2 + 𝑛3 +..... = 𝛴𝑛1 = 𝑛......... (17) 𝑉 𝑉 𝑉 𝑉 𝑉 where, n = ∑ni is the total number or moles of the gas mixture in the volume V. In dealing with gas mixtures, it is frequently necessary to be able to express the fraction which one component contributes to the total mixture Two of the most convenient ways of doing this are the use of' the pressure fraction (Pi/P) and the mole fraction (n/n), so - 10 - 𝑅𝑇 𝑃𝑖 𝑛𝑖 ( 𝑉 ) 𝑛𝑖 = 𝑅𝑇 = = 𝑋𝑖 ……………. (18) 𝑃 𝑛( ) 𝑛 𝑉 Furthermore, 𝑛1 𝑛2 𝑛3 + + + ⋯ … = 𝛴𝑋𝑖 = 1 ……….. (19) 𝑛 𝑛 𝑛 𝑃 and ∑ 𝑃1 = 1 Graham’s Law of effusion The process of which a gas moves from a higher to a lower pressure through a porous wall or very small diameter tube is known as diffusion, If the process consists of a molecular rather than bulk flow through an orifice the ward effusion is used. The rate with which the gas effuses under given conditions, is a property characteristic of the gas. Graham in 1829 found that, at constant temperature and constant pressure, the rate of effusion of various gases were inversely proportional to the square roots of densities of the gas. If "𝛾" is the rate of effusion and "d" is the density, then for gas 1 and 2, is written as: 𝛾1 𝑑 = √𝑑2 ……………. (20) 𝛾2 1 - 11 - From equation (Il), the density ratio of two gases9 at the same pressure and temperature is seemed to be equal to the ratio of the molar masses of the two gases, 𝛾1 𝑀 = √𝑀2 ……………. (21) 𝛾2 1 The Kinetic Theory of Ideal or Perfect Gases The suggestion that a gas consists of minute particles is continuous movement was first made by Bernoulli in 1738, In the following century, the idea was developed by Clausuis, Maxwell, Boltzmann and others into the kinetic theory of gases, which was intended to explain the gas laws. One aspect of observed gas behavior which gives the strongest clue to the nature of gases is phenomenon known as Brownian motion. This motion was observed by the botanist Robert Brown for pollen grains in aqueous suspension. In its broadest sense, the Brownian motion is irregular zigzag movement of extremely minute particles when suspended in liquid or a gas. The particle does not settle to the bottom of the container but moves continually and shows no sign of coming to rest. The smaller the suspended particle the more violent is the permanent condition of irregular motion. The - 12 - higher the temperature the more vigorous is the movement of the suspended particle. The Brownian motion gives a powerful support for the suggestion that matter consists of extremely small particles which are ever is motion. Like any theory, the kinetic theory represents a model which is proposed to account for an observed set of facts. In order that the model is practical, certain simplifying assumption (or postulates) must be made. The validity of each assumption and the reliability of the whole model can be checked by how well the experimental facts are explained. Postulates of the Kinetic Theory: The fundamental postulates on which the theory is based may be summarized as follows: 1. Gases consist of minute discrete particles usually called molecules. All molecules of the same gas are considered to have the same mass and size. Such properties depend on the chemical nature of the gas. 2. The molecules are so small that the actual volume of the molecules is negligible compared to total volume of the container. In other words, the molecules are considered as point masses of no volume. 3. The molecules are on the average so far apart that they exert no attractive force on each other. In other wards that molecules are completely independent of each other. - 13 - 4. The molecules are in rapid random straight-line continuous motion (Brownian motion) in all directions thus colliding with each other and with each other and with the walls of container. 5. The gas pressure is caused by the bombardment of the walls of the container by molecules. 6. The molecules are assumed to be perfectly elastic. In each collision, it is assumed that there is no net loss of energy, although there may be a transfer of energy between the particles in the collision. 7. At a particular instant in any collection of gas molecules, different molecules have different speeds and therefore different energy. However, the average kinetic energy of all molecules is assumed to be directly proportional to the absolute temperature. The last assumption has two parts: i. That there is a distribution of kinetic energies, and ii. That the average kinetic energy is proportional to the absolute temperature. The distribution of energies comes about as the result of molecular collisions, which continually change the speed of a particular molecule. A given molecule may move along with a certain speed until it hits another, to which it loses some of its kinetic energy. This exchange of kinetic energy between neighbors is constantly going on, so that it is only the total kinetic energy of a gas sample that stay the same, provided that no energy is added to the sample from outside, - 14 - i.e. no heating. The total kinetic energy of a gas is made up of the contributions of all molecules, each or which may be moving at different speed. At a particular instant a few molecules may be stand still with no kinetic energy; a few may have high kinetic energy; most will have kinetic energy near the average. The situation is represented by Figure (4), which indicates the usual distribution of kinetic energy in a gas sample at a temperature T l. Each point on the curve gives the fraction of the molecules having the corresponding value of the kinetic energy. Fig, (4) The temperature of the gas may be raised by addition of heat. This added heat increases the speed of the molecules and therefore the average kinetic energy. The dotted curve in Figure (4), describes the situation at the Temperature T2. As the temperature is raised, the molecules have a higher average kinetic energy than at lower - 15 - temperature. Thus, temperature serves to measure the average kinetic energy. Fundamental Equation of Kinetic Theory Consider a container in the form of a cube whose side is "L" cm. let us introduce "n" molecules of a certain gas in this container; each molecule has a mass "m". Suppose the average velocity of the molecule = C cm/s. The molecules move in all directions (along x, y, and z axes), and the velocity of this molecule may be resolved into three components along the three axes, as in Figure (5). Fig. (5) If the components of the velocity are designated C x, Cy and Cz respectively, the relation between the velocity "C" and the three components is given by: 𝐶 2 = 𝐶𝑥2 + 𝐶𝑌2 + 𝐶𝑧2 For simplicity, one may study the molecular motion on one axis, e.g., x-axis only. In other words, we are going to consider a molecule - 16 - moving two parallel sides of the cube. This molecule will travel a distance "2L cm" in order to make a single collision (or impact) with one of the two parallel sides or surface of the cube. Therefore, the number of impacts on one single side along x- axis per second 𝐶𝑥 (𝑐𝑚/𝑠 ) 𝑛. 𝑜𝑓 𝑖𝑚𝑝𝑎𝑐𝑡 = = = 𝑖𝑚𝑝𝑎𝑐𝑡/ 𝑆 2𝐿 (𝑐𝑚 / 𝑖𝑚𝑝𝑎𝑐𝑡) The number of impacts on the two parallel sides along x-axis per second is 𝐶𝑥 𝐶𝑥 𝑛. 𝑜𝑓 𝑖𝑚𝑝𝑎𝑐𝑡 = 2 ( ) = ( ) 𝑖𝑚𝑝𝑎𝑐𝑡 / 𝑠 2𝐿 𝐿 Since the molecule is perfectly elastic, it follows that when it comes to collide with a velocity “Cx”, it will rebound with the same velocity “Cx”, it will rebound with the same velocity but with opposite sign, i.e. with "Cx cm/s". The momentum of the molecule is defined as the product of mass by velocity. Therefore, one gets for the momentum before and after collision: Momentum before impact = mCx g. cm/s Momentum after impact = -mCx g. cm/s Change in momentum per molecule per single impact along x-axis: Momentum / molecule / impact = m Cx - (-m Cx) = 2 m Cx g. cm/s Let us then calculate the change in momentum for a single molecule, still on the x-axis per second. This will be equal to the change in - 17 - momentum per impact multiplied by the number of impacts per second, thus: 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝐶𝑥 𝐶𝑥2 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛. 𝑠 = 2𝑚𝐶𝑥 ( ) = 2𝑚 ( ) 𝑔. 𝑐𝑚./𝑠 2 𝑚𝑜𝑙 𝐿 𝐿 If the same procedure is performed along the y and z axis, one gets 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝐶𝑦 𝐶𝑦2 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛. 𝑠 = 2𝑚𝐶𝑦 ( ) = 2𝑚 ( ) 𝑔. 𝑐𝑚./𝑠 2 𝑚𝑜𝑙 𝐿 𝐿 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝐶𝑧 𝐶𝑧2 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛. 𝑠 = 2𝑚𝐶𝑧 ( ) = 2𝑚 ( ) 𝑔. 𝑐𝑚./𝑠 2 𝑚𝑜𝑙 𝐿 𝐿 Then, the change of momentum per molecule per second along the three axes given by: 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 2𝑚 2 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛.𝑠 = (𝐶𝑥 + 𝐶𝑌2 + 𝐶𝑧2) 𝑚𝑜𝑙 𝐿 2𝑚 2 = 𝐶 𝑔. 𝑐𝑚/𝑠 2 𝐿 This quantity represents the force exerted by a single molecule on all surfaces. The total force will be that exerted by all molecules (number). However, these molecules do not have the same velocity; the first molecule has a velocity "CI". The second "C2" and so on until "Cn". Therefore, the total force exerted by all molecules becomes: 2𝑚 2𝑚 2𝑚 2𝑚 2𝑚 𝑇𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 = 𝐶12 + 𝐶22 + 𝐶32 + ⋯ 𝐶𝑛2 = (𝐶12 + 𝐶22 + 𝐶32 ) 𝐿 𝐿 𝐿 𝐿 𝐿 - 18 - The quantity between brackets represents the sum of the squares of the individual molecular velocities. If this sum is divided by the number "n" of molecules we get the mean-squares of velocity, thus, 𝐶12 + 𝐶22+ 𝐶32 + …. … ….. +𝐶𝑛2 = 𝐶2 𝑛 Therefore, the total force becomes: 2𝑚𝑛𝐶 2 𝑇𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 = 𝐿 The pressure "P" is the force exerted per unit area, and since the area of the area of the cube is 6L2, one gets 𝑇𝑜𝑡𝑎𝑙 𝐹𝑜𝑟𝑐𝑒 2𝑚𝑛𝐶 2 1 1 𝑚𝑛𝐶 2 1 𝑚𝑛𝐶 2 𝑝= = 𝑋 2= 𝑋 = 𝑋 6𝐿2 𝐿 6𝐿 3 𝐿3 3 𝑉 Where L3 is the volume of container. Hence, 1 𝑃𝑉 = 𝑚𝑛𝐶 2............... (1) 3 This is the fundamental equation of the kinetic theory. The kinetic energy of a molecule of a mass "m" moving with a velocity "C 2” IS given by: 1 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 = 𝑚𝑛𝐶 2 2 1 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑓𝑜𝑟 𝑛 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 𝑚𝑛𝐶 2 = 𝐾. 𝐸. 2 - 19 - The right-hand side of equation (1) may be multiplied by 2/2 to get: 2 1 2 𝑃𝑉 = (2 𝑚𝑛𝐶 2 ) 3 𝐾. 𝐸................... (2) 3 This gives the relation between PV and the kinetic energy of "n" molecules. Deduction of Gas Laws: In this section, the different gas laws are deduced on the basis of the fundamental equation (1), 1. Boyle's Law: The energy is directly proportional to the absolute temperature, thus 1 𝑚𝑛𝐶 2 = 𝐾. 𝐸. = (𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡)𝑇.............. (3) 2 Form equations (2) and (3), one gets 2 2 𝑃𝑉 = 𝐾. 𝐸. = (𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝑇................ (4) 3 3 Therefore, at constant temperature T, PV = constant The same result follows directly from (1) by putting C2 as constant at constant temperature T 2. Charles Law: Rearranging (4) as: 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉=( )𝑇 𝑃 Therefore, if P is constant, V is directly proportional to the temperature T. - 20 - 3. Pressure -Temperature Law Rearranging equation (4) as 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃=( )𝑇 𝑣 It is clear that at constant V, the pressure is directly proportional to the temperature T. 4. General Gas Law This follows immediately from (4) where (2/3) constant is R for one mole of the gas. One may also start with equations (1) and (2), and in order not to confuse the number of molecules with the number of moles, let us use the symbol n' for the first and n for the second. Equation (1) and (2) become 1 𝑃𝑉 = 𝑚𝑛′𝐶 2 3 and, 2 1 2 𝑃𝑉 = (2 𝑚𝑛′ 𝐶 2 ) = 𝐾. 𝐸...... (5) 3 3 For n’ = 1, one can write 2 1 2 2 𝑃𝑉 = ( 𝑚𝐶 2 ) = 𝐾. 𝐸. (𝑓𝑜𝑟 𝑠𝑖𝑔𝑙𝑒 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒) 𝑐𝑜𝑛𝑠𝑡. 𝑇 3 2 3 3 = 𝐾. 𝑇; Where k is the Boltzmann’s constant. For one mole (n' = Avogadro's number N): - 21 - 2 1 𝑃𝑉 = ( 𝑚𝑁𝐶 2 = 𝑁𝐾𝑇 = 𝑅𝑇 3 2 For "n" moles (n' = nN) 2 1 ∴ 𝑃𝑉 = ( 𝑚𝑛𝑁𝐶 2 ) = 𝑛𝑁𝐾𝑇 = 𝑛𝑅𝑇.............. (6) 3 2 5. Avogadro's Principle: It states that equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules, Since the pressure and volume are the same, one may write for two gases. P1 V1 = P2 V2 From this and equation (5) 𝟐 𝟏 𝟐 𝟏 ( 𝒎𝟏 𝒏′𝟏 𝑪𝟐𝟏 ) = ( 𝒎𝟐 𝒏′𝟐 𝑪𝟐𝟐 ) 𝟑 𝟐 𝟑 𝟐 𝟐 𝟐 Or, n’1 (the kinetic energy of 1 molecule of the first gas) and n’2 𝟑 𝟑 (the kinetic energy of 1 molecule of the second gas. Since the temperature is constant, the average kinetic energy per molecule is the same and it follows that: n’1 = n’2 ; as advanced by Avogadro. 6. Gerham’s Law: 3𝑃𝑉 From equations (1) and (5) 𝐶2 = 𝑚𝑛′ - 22 - 𝑚𝑛′ Since the density is equal to (𝑑 = ), it follows that 𝑉 3𝑃 𝐶2 = 𝑑 3𝑃 Therefore, √𝐶 2 = √ 𝑑 The quantity on the left-hand side is the square root of the mean square velocity. The rate of diffusion depends on this quantity, and if the rates of diffusion of the two gases is compared at the same temperature and pressure (as in Gorham’s law) one gets: 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑔𝑎𝑠 3𝑃 3𝑃 𝑑2 = √ ÷ √ = √ 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑 𝑔𝑎𝑠 𝑑1 𝑑2 𝑑1 The square root of the mean square velocity (Known as root mean square velocity) may e obtained from equation (6) for one mole gas, thus: 3 𝑅𝑇 3 𝑅𝑇 √𝐶 2 = √ 𝑚𝑁 = √ … … … … … … (7) 𝑀 Where, mN is the mass of N molecules (N = Avogadro’s number), i.e. the molecular weight M. again using equation (7) for comparing the diffusion rate of two gases, we can have: 𝑅𝑎𝑡𝑒 (1) 3 𝑅𝑇 3 𝑅𝑇 𝑀2 = √ ÷ √ = √ 𝑅𝑎𝑡𝑒 (2) 𝑀1 𝑀2 𝑀1 - 23 - Equation (7) is useful for the calculation of the root mean square velocity. For example, the velocity of O2 at 25°C is given by: 3 𝑋 8.314 𝑋 107 𝑋 298 √𝐶 2 = √ = 4.82 𝑋 104 𝑐𝑚/𝑠 32 “R” is expressed here in “erg / degree / mole” in order to get the velocity in cm/s. The root mean-square velocity is not the only way of expressing the velocity of a molecule. Another expression is the average or mean velocity, which is the arithmetic mean of all velocities, thus C = (C1 + C2 + C3 + … … … … … … Cn) / n The root means square velocity and the average velocity are not exactly identical. Heat Capacity of Gases The specific heat is defined as "the amount of heat in calories required raising the temperature of one gram of the substance one degree centigrade". The molar heat capacity is "the amount of heat in calories required raising the temperature of one mole of the substance one degree centigrade". Therefore, the molar heat capacity is equal to the specific heat multiplied by the molecular weight. It is a fact that a substance has two specific heats, one measured at constant pressure - 24 - and one at constant volume. For solid or a liquid, the difference between the two is usually negligible. Since the effect of heat on the velocity is small. But in the case of gas, the expansion which takes place on heating is large, and hence the two values of specific heats are widely different. To heat a gas, energy must be supplied or added. The energy added must appear as kinetic energy of the molecules, or as work done by the gas in expanding against external pressure, or as both. In order to a gas to perform work, it must expand, for work is always the product of a force and a displacement If the volume of a gas is holding constant, there is no displacement, and no work is done. Therefore, any energy we add to a gas at constant volume must appear as kinetic energy of the molecules. Consider one mole of an ideal mono atomic gas (contain one atom per molecule, and hence the only type of motion is translational motion). If the kinetic energy of this molecule is represented by E, the kinetic theory gives: 2 3 𝑃𝑉 = 𝐸 = 𝑅𝑇 ∴𝐸= 𝑅𝑇 3 2 Therefore, if we increase the energy by an amount ΔE at constant volume, there must be a corresponding increase of temperature by ΔT, 3 ∆𝐸 = 𝑅(∆𝑇) 2 - 25 - But (ΔE / ΔT) is the increase in energy per degree per mole, or the molar heat capacity at constant volume "C v” Therefore, we can say that, ∆𝐸 3 𝐶𝑣 = = 𝑅 ∆𝑇 2 3 Thus, Cv for an ideal mono atomic gas is R or about 3 cal/ degree/ 2 mole. When the temperature is raised at constant pressure, the kinetic energy of the gas increases and the gas does work of expansion against the atmosphere by virtue of its increase in volume. The work of expansion is given by: Work = w = PΔV writing, PV = RT At constant pressure P, the increase in volume AV corresponds to an increase in T by ΔT since R is constant, thus PΔV = RΔT The extra heat capacity due to the expansion of the gas is (PΔV)/ΔT = R Therefore, if the heat capacity at constant P is represented by C p , one may write So, Cp = heat capacity due to increase of kinetic energy + extra heat capacity due to expansion Cp = CV + R = (3/2)R + R = (5/2) R - 26 - γ is the ratio of (Cp/Cv) γ = (Cp/Cv) = 5/3 = 1.67 This result applies only to mono atomic gases like the inert gases (He, Ne, Ar, Kr, Xe). In this case the only type of motion is translational motion (the molecule moves as a whole from place to place). This motion can be resolved into three independent velocity components (along x, y and z axes), Since (Cv = 3/2 R), we can inter that each of the three independent translational motions contributes ( 1/2) R to the molar heat capacity. This result arises from the principle of equipartition of energy deduced by Maxwell and Boltzmann. According to this principle, the energy imparted to a gas should be equally distributed between every degree of freedom. A degree of freedom is a possible mode of motion of a molecule, and we have three degree of freedom for translational motion. Gases of all kinds have translational and therefore they all have (3/2) RT corresponding to this motion as well as any extra energy due to other types. In addition to the three translational motions, a diatomic molecule can rotate about its center of the mass in two mutually perpendicular and independent ways or modes. In other words, there are two degrees of freedom for rotation. Assigning (R/2) as the heat capacity contribution of each of these motions, the total contribution of rotation is R. - 27 - For diatomic molecules, there is also the possibility of vibration which occurs along the axis of the molecule. Vibrational energy is both kinetic and potential. Therefore, a diatomic molecule has two vibration degree of freedom: one for kinetic energy and one for potential energy. Therefore, the number of degrees of freedom for diatomic molecule is: Degree of freedom = 3 (translation) + 2 (rotation) + 2 (vibration) Hence, CV = (7/2) R However, if the diatomic molecule is rigid, the contribution of vibration can be neglected. This is true for all practical purposes at ordinary temperatures, thus Cv = (5/2) R , Therefore, 𝟓 𝟕 𝟕 𝑪𝑷 − 𝑪𝑽 + 𝑹 = 𝑹+𝑹= 𝑹 ∴ 𝜸= = 𝟏. 𝟒𝟎 𝟐 𝟐 𝟓 This result agrees with the observed values. At high temperatures, the diatomic molecule cannot remain rigid, and the contribution of vibration becomes significant. For tri atomic gases, the number of degrees of freedom increases, and the ratio of (Cp/Cv) becomes: γ = 1.33 - 28 - Real or Imperfect Gases During the nineteenth century many attempts were made to test Boyle's and the other gas laws. Values of PV were plotted against pressure as shown in Figure 6. If a gas obeys Boyle's law, it will give a horizontal straight line and its behavior is said to be ideal behavior and the extent of deviation depends on the nature of gas and its temperature. By comparing the curve for O2 at Tl and T2 > Tl, it will be clear that the deviation increases with the lowering of temperature. Other gases behave similarly. It may therefore be concluded that real gases approach the ideal gas behavior only under the limiting condition of high temperature and low pressure, PV Equation of Van der Waal's Van der Waal in 1879 made early attempt to introduce additional terms into the simple gas law thus giving an expression - 29 - which describes more exactly the behavior of real gases. The corrections were introduced by Van der Waal, these were as follows: 1. Since the gas molecules possess material volume, it follows then that the volume within which the molecules are free to move will be less than that containing them (volume of container). This volume, which may be called compressible volume, is equal to (V- b), where V is the volume of the container, and b is a term connected with the actual incompressible volume of molecules. 2. Since forces of attraction exist between the molecules of real gases the effect of these forces must be taken into consideration. A molecule in the bulk of the gas will be subjected to equal and opposing attraction forces in all directions from surrounding molecules. As the molecule is about to bombard the wall, thus contributing to the pressure of the gas, it will be subjected to an inward pull which causes a lowering of the pressure of the gas. The pressure P measured for a real gas (where attraction force is present) will therefore be less than that measured for an ideal gas (where attraction is absent). A correction of the observed pressure of' a real gas that takes account of intermolecular attraction should therefore be necessary. Obviously, the correction to be introduced will depend on the number of molecules, in the surface layer and also on the number in the bulk of the gas, such that: - 30 - 𝟏 Forces of attraction α number of molecules at the surface α 𝑽 1 Forces of attraction number of molecules in the bulk α 𝑉 1 For a given mass of a gas, therefore; Forces of attraction α 𝑉2 𝑎 Thus, force of attraction = 𝑉2 𝑎 Hence the actual pressure of a real gas be given by = 𝑃 + 𝑉2 Introducing the corrected values of volume and pressure for a real gas in the general gas law, it follows that 𝑛2 𝑎 (𝑃 + 2 ) (𝑉 − 𝑛𝑏) = 𝑅𝑇 𝑉 This is the Van der Waal equation as applied to 1 mole of gas. For "n" moles. 𝑛2 𝑎 (𝑃 + 2 ) (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 𝑉 The forces of attraction between molecules which give rise to the term (a/V2) in Van der Waals equation are often called "Van der Waal forces". The values of the Van der Waals constants "a" and "b" depend on the nature of the gas. Representative results are shown below. Explanation of the deviation or real gases on the basis of Van der Waal's equation We can explain the curve (PV vs. P) for real gases like CO2, N2 and H2 as follows. The incompressible volume and the force of attraction in Van - 31 - der Waals equation are important but the importance of each factor differs according to the difference in the pressure and volume. 𝑎 1. At low pressure (i. e. big volumes the force of attraction ( 2 ) is 𝑉 important but (b) the incompressible volume can be neglected the 𝑛2 𝑎 equation will be (𝑃 + )𝑉 = 𝑉2 𝑅𝑇 𝑎 ∴ 𝑃𝑉 + = 𝑅𝑇 𝑉 𝑎 Or, 𝑃𝑉 = 𝑅𝑇 − 𝑉 𝑎 The value of PV decreases by a value equal to 𝑉 In the figure (PV vs. P) for N2 or CO2 the value of PV decreases firstly with increase in pressure. 2. At high pressure (i. e, small volumes) the value of b (incompressible volume) cannot be neglected but the value of force of 𝑎 attraction(𝑉2 ) can be neglected, the equation will be P (V – b) = RT or, PV – Pb = RT or, PV = RT + Pb The value of (PV) increases with increase - 32 - in pressure as in case of hydrogen (H2) Liquefaction of Gases Critical Constants of Gases At sufficiently low temperature as a gas may be made to liquefy by applying pressure, thus reducing the volume, and bringing the molecules close enough together so that inter molecular attractive forces will be effective. All gases have been liquefied in this way. However, there is a temperature above which it is impossible to liquefy a gas no matter how great pressure is applied. This temperature is called the "critical temperature". The volume occupied by one mole (of gas or liquid) at the critical pressure is the critical volume. The significance of the critical temperature may be realized by considering Figure (8) in which the pressure "P" is plotted against volume "V" for carbon dioxide over a range of temperature from 0 0 to 500C. as reported by Andrews in 1861. These results indicate, the isotherm at "48.1OC" is almost a hyperbola, which is the form it should take if the gas followed Boyle's law, i.e. perfect behavior. The isotherm at "31.1OC" shows a short horizontal period, The isotherms at lower temperatures are divided into 3 portions, with the horizontal one increasing in length as the temperature gets lower. Now, consider the isotherm at "21.50C”, the portion "AB" represents a steady decrease in volume as the pressure of the gas increases. At point "B" liquefaction starts and the horizontal portion "BC" shows rapid reduction in the volume which - 33 - accompanies the change of state from gas to liquid. At point "C" liquefaction is complete and since liquids are less compressible than gases, the portion "CD" shows little volume changes and is almost vertical. Within the parabola indicated by the dotted line, gas and liquid exist in equilibrium. In all parts of the diagram to the right of the parabola is a gas. To the left of the parabola, carbon dioxide exists as a liquid. The temperature of "31.1OC", was called by Andrews the critical temperature. It is represented that "point at which the cohesive (Van der Waal) forces which hold the molecules of the liquid just balance those of kinetic motion which disperse the molecules of the gas". Gas and liquid are thus in equilibrium. Fig. (7) - 34 - The Joule Thomson Effect Joule and Thomson allowed air under high pressure to escape through a porous plug into a region of low pressure and noticed that air was cooled. All gases except hydrogen and inert gases show this cooling effect at ordinary temperature. However, when hydrogen and inert gases are cooled below certain temperature, they also show the effect. This cooling effect of most gases upon escaping from a high pressure into vacuum or a region of low pressure is called "Joule Thomson effect". This effect is due to the fact that work must be done to overcome the intermolecular attraction forces (Van der Waal forces) in real gas. The energy necessary for much work is drawn from the kinetic energy of the molecules with a consequent lowering in temperature. - 35 - CHAPTER II Liquids, Solids and Change of States As heat energy is added to a solid there is an increase in the kinetic energy of the molecules or ions that occupy the lattice sites. The particles move about more and more violently, until finally the attractive forces are no longer able to hold them in the lattice and the solid gets melt, forming a liquid. A gas can also be condensed to a liquid by sufficiently lowering its temperature or under appropriate conditions, sufficiently lowering its temperature or, under appropriates conditions, sufficiently increasing its pressure. Lowering the temperature of a gas brings about a decrease in the kinetic energy of the molecules. These causes the molecules to slow down, and, at the condensation temperature, the intermolecular attractive forces are able to allow groups of molecules to cling together. Increasing the pressure on a gas causes the molecules together, if the attractive forces become strong enough, condensation occurs. In this chapter we will examine the properties of liquids in terms of the nature of the particles that make up the liquid. We will also look at factors that influence the transition between the three states solid, liquid and gas. - 36 - General Properties of liquids A liquid is composed of molecules that are constantly and randomly moving about, each undergoing many billions of collisions per second. However, strong attractive forces of the dipole-dipole, hydrogen bond, or London type prevent them from moving as freely and as apart as in gas, On the other hand, the molecules of a liquid are not as close together or as structured as they are in a solid. For these reasons liquids exhibits characteristics that place them somewhere between the completely chaotic gaseous state and the well-ordered solid state. 1. Volume and Shape: in a liquid the attractive forces are strong enough to restrict the molecules to move about within a definite volume, but they are not strong enough to cause the molecules to maintain a definite position within the liquid. In fact, the molecules, within the limits of the liquid's volume, are free to move over and around one another, thus allowing liquids to flow. Liquids, therefore, maintain a definite volume but, because of their ability to flow, their shape depends on the contour of the container holding them. 2. Compression and Expansion: in a liquid the attractive forces hold the molecules close together, and increasing the pressure has little effect on the volume because there is little free space into which the molecules may be crowded. Liquids are therefore virtually incompressible. - 37 - Similarly, changes in temperature cause only small volume change (compared to a gas). The increasing molecular motion that accompanies rising temperature tends to increase the intermolecular distance, but this is opposed the strong attractive forces. 3. Diffusion: when two liquids mix, the molecules of one liquid diffuse throughout the molecules of the other liquid at a rate much slower than is observed when two gases are mixed. We can observe the diffusion of two liquids by dropping a small quantity of ink into some water, As demonstrated in Figure I (a and b), when the ink drop strikes the water, we see it as a concentrated "dot" which slowly spreads throughout the water. Diffusion takes place because the molecules in both liquids are so close together, each molecule undergoes billions of collisions before traveling very far. The average distance between collisions, called the mean free path, is much shorter in liquids than in gases, where the molecules are relatively far apart. Because of constant interruptions in their molecular paths, liquids diffuse much more slowly than gases. - 38 - (a ) (b) Fig. 1: Diffusion in liquids (a) ink drop placed into water (b) ink has spread throughout the liquid 4. Evaporation: in a liquid the molecules are constantly undergoing elastic collisions, giving rise to a distribution in individual molecular velocities and, of course, kinetic energies, following reasoning similar to that described before, it follows that in a liquid, even at room temperature, and a small percentage of the molecules are moving with relatively high kinetic energies. If some of these faster moving molecules possess enough kinetic energy to overcome the attractive forces operating within the liquid and can escape through the surface into the gaseous state, the liquid evaporates. Figure (2) represents a typical distribution of the kinetic energies of the molecules in a liquid; the shaded area corresponds to the fraction of the total number of molecules that possess sufficient kinetic energy to evaporate. Just as removing the smart students from chemistry class will lower the class - 39 - average on exams, the loss of the higher-energy fraction because of evaporation leads to a lowering of the average kinetic energy of the remaining molecules. Since the temperature is directly proportional molecules. Since the temperature is directly proportional to the average kinetic energy, this result in a decrease in the temperature of a liquid as it evaporates. For example, we have all felt cool after a bath, because the evaporation of water from the body has drawn heat from us. In fact, evaporation of perspiration provides the body with a mechanism for controlling body temperature. Fig, (2) Kinetic energy distribution in a liquid If a liquid, such as water, is to continue to evaporate from a container, heat must constantly be absorbed from the surroundings in Order to replenish the energy taken away by the molecules leaving the liquid. If the surrounding are at a high temperature, heat can be supplied faster than when they are cool. Thus, water evaporates faster on dry hot days than on dray cold days. - 40 - Heat of Vaporization: The molar heat of vaporization, which we will refer to as ΔHvaporization or (ΔHvap), represents "the quantity of energy that must be supplied to one mole of liquid to convert it into one mole of vapor at the same temperature". The Greek letter, Δ, is usually used to symbolize a change; in this case, a change in the heat content (the total quantity of heat energy) of a substance as it undergoes a change from liquid to vapor. This change in heat content is equal to the energy contained in the substance in its final state (vapor) minus the energy that the substance possessed in its initial state (liquid). Thus, ΔHvap = ΔHvapor - ΔHliquid In actual practice, neither ΔHvapor nor ΔHIiquid can be measured; however, their difference (ΔHvap) can be measured. The magnitude of ΔHvap provides a good measure of the strength of the attractive force’s operative in a liquid. In Table (1) we find the values of ΔHvap for C10H22, we observe a steady increase in ΔHvap with an increase in molecular weight. These compounds are non-polar, therefore the only attractive forces that exist between their molecular are "London force". It was pointed out that the strengths of London forces are related, at least in part, to the number of atoms in molecules that contain the same elements. If we take a closer look at the hydrocarbons in Table (1), we find that we proceed from CH4 to C10H22, the length of the carbon chain increases, as illustrated in Fig. - 41 - (3). This has the effect of increasing the number of locations along the molecule where London forces may occur with other molecules. A long chainlike molecule is therefore held in more places than short molecule, and more energy must be supplied to remove such long- chain molecules from the liquid. These results are that as the chain length increases, ΔHvap increases. Another factor that influences the strengths of London forces is molecular size. If we examine molecules of the same general formula, such as the halogens (F2, CI2, Br2), we find that large molecules have a greater ΔHvap than small molecules. As we proceed from F2 to Br2, the atoms that make up the molecules become larger; hence the molecules also become larger. As the size increases, the outer electrons are further from the nuclei and are not held tightly. Because of this the electron cloud of a large molecule is more easily distorted and it is easier to create the instantaneous dipoles that are responsible for London forces. The ease of distortion of the electron cloud is referred to as polarizability. The result of this is that the London forces are stronger between molecules composed of large, easily polarized atoms such as "Br" than between molecules composed of small atoms such as "F". Hence, ΔHvap increases from "F2" to "Br2" - 42 - Table (1): Heats of vaporization and boiling points Compound ΔHvap (K. Mol-1) Boiling Point (°C) CH4 9.20 -161 C2H6 14 -89 C3H8 18.1 -30 C4H10 22.3 0 C6H14 28.6 68 C8H18 33.9 125 C10H22 35.8 160 F2 6.52 -188 CI2 20.4 -34.6 Br2 30.7 59 HF 30.2 17 HCl 15.1 -84 HBr 16.3 -70 HI 18.2 -34 H2O 40.6 100 H2S 18.8 -61 NH3 23.6 -33 PH3 14.6 -88 SiH4 12.3 -112 Figure (3): Attractive forces increase with increasing chain length. There are more points along the molecule that can be attracted to other molecules nearby. - 43 - H H H H ǀ ǀ ǀ ǀ ̶ H ̶ C ̶ H ̶ ̶ H ̶ C ̶ ̶ C ̶ C̶ H ̶ ǀ ǀ ǀ ǀ H H H H H H H H H H H H H H ǀ ǀ ǀ ǀ ǀ ǀ ǀ ǀ ǀ ǀ ̶- H ̶ C – C – C – C – C - C – C – C – C – C – H- ǀ ǀ ǀ ǀ ǀ ǀ ǀ ǀ ǀ ǀ H H H H H H H H H H Fig. (3) When we look at the hydrogen halides, "HF" through "HI", however, we find that the expected variation of ΔHvap with molecular size is reversed between “HF" and "HCI". In fact, "HF" has a considerably higher heat of vaporization than any of the other "HX" compounds. This anomalous behavior is attributed to the presence of hydrogen bonding. This bonding is a particularly strong dipole - dipole interaction that can occur when hydrogen is bound covalently to a small, very high electronegative element. We see the same inverted order of Hvap for "H2O" and "H2S" and for "H2O" and "NH3" but not for "H2S” and "PH3”. Oxygen, fluorine, and nitrogen are very small atoms and are the most electronegative elements in the periodic table, while the elements below them are much larger and much less electronegative. Thus, we predict that hydrogen bonding is important only for "H 2O" "HF" and "NH3" Normal behavior is reached in Group "IV A", where ΔHvap for "CH4" is less than ΔHvap for "SiH4". Here neither "CH4" nor - 44 - "SiH4" have any tendency to hydrogen bond because they are non- polar. Vapor Pressure If a liquid evaporates in an open container, eventually all the liquid will disappear because the molecules that have escaped from the liquid into the vapor phase diffuse readily into the atmosphere. If the same quantity of liquid the same temperature is placed in a closed container, what will happen? In this case the volume of a liquid will initially decrease and then eventually become constant. If we monitored the pressure of the gas above the liquid, we would find that it initially increases and then it too levels off at a constant value. These observations can be explained in the following way. The molecules with higher kinetic energies begin to leave the liquid, evaporating into the vapor phase, where they become trapped. The loss of molecules from the liquid must, of course, be accompanied by a volume decrease. In time the space above the liquid become occupied with more and more gaseous molecules, and the pressure of the vapor increases. With the increasing number of chaotically moving gas molecules the number of collisions. One of these walls is the surface of the liquid itself, which will trap any bombarding molecules having low kinetic energies. Thus, condensation as well as evaporation (vaporization) takes place at the surface of the liquid. - 45 - Eventually, the number of molecules in the vapor becomes large enough so that the rate at which the gas condenses exactly equals the rate at which the liquid evaporates, and no further change in either the volume of the liquid or the pressure exerted by its vapor is observed. To emphasize this point again, vaporization and condensation are still taking place, but with no change in the liquid volume or the vapor pressure. At this point the liquid is said to be in dynamic equilibrium with its vapor. The pressure exerted by the quantity of vapor above the liquid, when equilibrium is established, is called the equilibrium vapor pressure of the liquid. The vapor pressure of a liquid, quite naturally, depends on the case with which its molecules can leave the liquid and enter the vapor state. In liquids where the intermolecular attractive forces are strong, the vapor pressure will be low; and, in liquids where the attractive forces are weak, the vapor pressure will be high. Since increasing the temperature of the liquid increases the number of molecules possessing sufficient energy to overcome the attractive forces, the vapor pressure must increase with increasing temperature. Thus, whenever the vapor pressure is given, the temperature at which it was measured must also be specified. One way that we can determine the vapor pressure of a liquid is by the use of a barometer, as shown in Figure (4). The height of the mercury in the barometer, before any liquid is added, is measured - 46 - accurately. A liquid whose vapor pressure is to be determined is carefully added to the barometer by means of an eye dropper and allowed to rise to the top of the mercury in the column, as shown in Figure 4 "a, b, c and d" (most liquids are less dense than mercury and will therefore float on the mercury surface). The space above the mercury column in Figure 4 is, for all practical purposes a vacuum and exerts no downward force. The space above the mercury in Figure 4 "b, c and d" is filled with a small amount of liquid and its vapor. As the liquid begins to evaporate, the pressure of the trapped vapor causes the level of the mercury in the column to decrease, when the liquid and vapor are in equilibrium, the height of the mercury column becomes stationary. The total pressure exerted at the reference level outside each barometer will be the atmospheric pressure, "P atm”. The total pressure exerted within the barometer is "P Hg" the pressure due to the pull of gravity on the mercury in the column, plus "P vapor” the pressure exerted by the vapor in equilibrium with its liquid. The additional pressure exerted by the weight of the small amount of liquid on top of the column is negligibly small. Therefore, at equilibrium in each barometer Patm = PHg + Pvapor In Figure” 4a", Pvap = 0, therefore, Patm = PHg = 101.3 kPa. In Figure 4 "b, c and d" PHg = 98.1kPa , 92.7 kPa, and 28.6 kPa, respectively. Therefore, at 250C the vapor pressure of water is 3,2 kPa, that ethyl - 47 - alcohol is 8.6 kPa, and 28.6 kPa, respectively. Therefore, at 25 0C the vapor pressure of water is 3.2 kPa, that of ethyl alcohol is 8.6 kPa, and that of diethyl ether is 72.7 kPa. Water has the lowest vapor pressure of the three liquids in our example; therefore, it must have the strongest intermolecular attractive forces. Diethyl ether, on the other hand, has the highest vapor pressure of the three liquids, which means that relatively weak attractive forces exist in it. Fig. (4) We have seen that the vapor pressure of a liquid is dependent on the liquid and its temperature. What happens to the vapor pressure when the volume or pressure of the vapor is changed? In Figure “5”, we see an illustration of an apparatus that can be used to demonstrate the effect of volume and pressure changes on liquid- vapor equilibrium. At a constant temperature equilibrium is established, as shown by Figure "5" if, at this same constant - 48 - temperature we allow the vapor to expand by rapidly raising the piston, then the system is no longer in equilibrium. As the piston is withdrawn, creating the larger volume, the number of molecule-wall collisions decreases, causing a decrease in the pressure exerted by the vapor. The rate of condensation at the surface, which depends on the number of collisions between the vapor molecules and the surface of the liquid, must also decrease. Figure (5): Effect of volume change on vapor pressure (a) Equilibrium between liquid and vapor, (b) No equilibrium, rate of evaporation is greater than rate of condensation, (c) No equilibrium; rate of condensation is greater than rate of evaporation. The rate of evaporation, however, remains essentially the same. This means that, at a constant temperature, an increase in the volume of a vapor in equilibrium with its liquid causes more molecules to leave the liquid state than return to it. As the process continues, more and more molecules enter the vapor phase, causing an increase in the pressure exerted by the vapor and a corresponding increase in - 49 - the rate of condensation. After a while enough molecules will be present in the vapor phase so that the rate of condensation will again exactly equal the rate of evaporation, and equilibrium will be reestablished. At this newly established equilibrium the larger volume of gas is now occupied by more molecules. The vapor pressure will be the same as before the volume change occurred, but the volume of the liquid will be slightly smaller. Decreasing the volume of vapor by lowering the piston Fig. "5c" will also disturb the equilibrium. Increasing the pressure of the vapor will cause an increase in the number of molecule-wall collisions. This in turn, will lead to an increase in the rate of condensation but will have little effect on the rate of evaporation. The rate of which the molecules leave the vapor phase, then, will be greater than the rate at which molecules leave the liquid phase. This imbalance in rates causes the pressure exerted by the vapor to decrease and the volume of the liquid to increase. Eventually, the rate of condensation will decrease to point where it exactly equals the rate of vaporization reestablishing equilibrium. At this new equilibrium the smaller vapor volume, caused by the movement of the piston, will be occupied by fewer gaseous molecules. The vapor pressure will have returned to its initial value, and the volume of the liquid will have increased slightly. - 50 - The net result of this discussion is that the vapor pressure of a liquid is independent of the volume of the container, provided that there is some liquid present so that equilibrium can be established. Le Chatelier's principle The dynamic equilibrium between a liquid and its vapor can be represented by the following equation. Liquid ⇌ vapor Here the double half arrows mean that the rates of evaporation and condensation are equal. If we in any way disturb this system so that it is no longer at equilibrium (we say that a "stress" is applied to the system), a change occurs that will, if possible, bring the system back to equilibrium. In our example above, an increase in the volume of the vapor caused the system to no longer be at equilibrium. We saw that more liquid evaporated until equilibrium was reestablished. In equation this corresponds to the process as read from left to right, that is, " Liquid → vapor" and results in a new position of equilibrium in which there is less liquid and more vapor. In this sense the position of equilibrium has shifted to the right when we applied the stress. The action taken by any system at equilibrium when a stress is applied can be described by the principle of Le Chatelier’s, which states that "when a system in a state of dynamic equilibrium is acted upon by some outside stress, the system will, if possible, Shift to a new position of equilibrium in order to minimize the effect of the stress". - 51 - For example, let us apply Le Chatelier's principle to describe what effect pressure changes have on liquid vapor equilibrium. When the stress applied is a pressure decrease caused by an increase in the volume of the container, the system attempts to undergo a change that will return the pressure to its initial value. In this example, the pressure can be increased if more molecules enter to the vapor phase, that is, if some additional liquid is evaporates. After equilibrium has been reestablished, there will be less liquid and more vapor pressure in the container, and we say that the position of equilibrium represented by the equation has shifted to the right. However, if the volume is increased sufficiently, all of the liquid will evaporate, and equilibrium will not be reestablished. This will occur, for example, if the piston in Figure "5" is removed entirely so that the liquid is open to the atmosphere. In a similar fashion, we predict that raising the pressure leads to a decrease in the quantity of vapor and of course, a corresponding increase in the amount of liquid. Thus, we might conclude in general that an increase in pressure on a system at equilibrium favors the production of the denser phase, while a decrease in pressure favors the formation of the less dense phase. The effect of temperature changes on equilibrium can also be described using Le Chatelier’s. Increasing the temperature of a system at equilibrium favors the absorption of energy (an endothermic - 52 - change). In a liquid-vapor equilibrium system this means that a temperature increase causes more liquid to evaporate, because this process absorbs heat. Decreasing the temperature (removing heat), on the other hand, favors the release of energy, i.ee an exothermic change. As the temperature is decreased in a liquid - vapor equilibrium, more molecules condense into the liquid phase, releasing heat and thereby minimizing the effect of the applied stress. In summary, Le Chatelier’s principle predicts that a temperature increase will shift the position of equilibrium in the direction of the endothermic process. Similarly, a decrease in temperature will favor the exothermic change. Vapor Pressure Curves for Liquids We can determine the vapor pressure of liquid as a function of temperature, data accumulated in such experiments performed on water; ethyl alcohol and diethyl ether are illustrated graphically in Figure (6). We can see from the figure that at lower temperatures vapor pressure changes relatively slowly with increasing temperature, while at higher temperatures the changes more rapid. Points along the curve represent combinations of pressures and temperatures that must be satisfied in order for the liquid to be in equilibrium with its vapor. These curves terminate at a temperature called the critical temperature "Tc", above which molecular motion is so violent that substance can exist only as a gas. In other words, the critical - 53 - temperature is that "temperature above which a substance can no longer exist as a liquid, regardless of the applied pressure". "The pressure that must be applied to a substance at its critical temperature in order to achieve liquid-vapor equilibrium is called its critical pressure” Fig. (6) Vapor pressure (along with its variation with temperature) and heat of vaporization are both controlled by the intermolecular attractive forces within a liquid. It should not be surprising, therefore, to find that there is a quantitative relationship among vapor pressure, temperature and ΔHvap. - 54 - Table (2): Critical temperature and pressures of different compounds Tc(°C) Compound Pc (MPa) Methane CH4 -82,1 4.64 Ethane C2H6 32.2 4.88 Benzene C6H6 288.9 4.92 Ammonia 132,5 11.40 Carbon dioxide 31 7.39 Water 374.1 22.06 Helium - 267.8 0.23 If the logarithm of the vapor pressure "log P" is plotted versus the reciprocal of absolute temperature "1/T", a straight line is obtained at least over relatively short temperature ranges, as shown in Figure "6", In general, any straight line can be described by an equation y = b + mx where, m is the slope of the line and b is the intercept of the line with the vertical axis. In the present case we can write log P = b + m (I/T) It can be shown that the slope of the line is related to the heat of vaporization m = (- ΔHvap/2.303R) - 55 - When the value of R used in calculation is 8.314 J. mol-1. K-1, the units of ΔHvap will be J. mol-1 A convenient form of the relationship between ΔHvap and vapor pressure is given by the "Clausius-Clapeyron" equation (named after a German physicist R. Causius and a French engineer, B. P. EO Clapeyron). 𝑃1 −∆𝐻𝑣𝑎𝑝 1 1 𝐿𝑜𝑔 = =( − ) 𝑃2 2.303𝑅 𝑇1 𝑇2 In this equation P1 is the vapor pressure when the absolute temperature of the liquid is T1, P2 is the vapor pressure when the absolute temperature of the liquid is T2. Equation can be used to calculate the ΔHvap if the vapor pressure is known at different temperatures. It can also be used to calculate the vapor pressure at a specific temperature provided that ΔH vap and the vapor pressure at some other temperatures are known. Fig. (7) A Plot of log Vp of H2O vs. I/T In temperature range 250C to 500C. - 56 - Boiling Point The boiling point is defined as "that temperature at which the vapor pressure of a liquid is equal to atmospheric pressure, also known as boiling point of the liquid". At this temperature the vapor pressure is high enough to cause vaporization to occur at various points throughout the interior of the liquid. Thus, boiling is accompanied by the formation of bubbles, which form simultaneously at many spots in the liquid. When a bubble is formed within the liquid, the liquid that originally occupied this space is pushed aside and the level of the liquid in the container is forced to rise against the downward pressure exerted by the atmosphere. In other ward, it is the pressure exerted by the vapor inside the bubble that pushes the surface of the liquid up against the atmospheric pressure. This can occur only when the vapor pressure of the liquid becomes equal to the prevailing atmospheric pressure. If it were less, the atmospheric pressure would cause the bubble to collapse. As long as bubbles are forming within the liquid, that is, as the liquid is boiling, the vapor pressure of the liquid is equal to the atmospheric pressure. Since the vapor pressure remains constant, the temperature of the boiling liquid is also stays the same. As increase in the rate at which heat is supplied to the boiling liquid simply causes - 57 - bubbles to form more rapidly. The liquid boils away more quickly, but the temperature does not increase. It is obvious, from the discussion above, that the boiling point of a liquid depends on the prevailing atmospheric pressure. The boiling point of a liquid at 1 atm. (101.325 k Pa) is referred to as its standard of normal boiling point at 1000C. At higher pressures its boiling point is greater, at lower pressures (For example, on a mountain top) its boiling point is less. Boiling points given in reference tabled are always normal boiling points, unless otherwise stated. The temperature at which a liquid boil is another example of a property that gives a good estimation of the strength of the attractive forces operating within the liquid. Liquids whose attractive forces are relatively high have correspondingly high boiling points, while liquids with weak attractive forces boil at relatively low temperature. In Table (2), for example, we see that the trends in boiling points follow trends of ΔHvap. The dependence of boiling point on intermolecular attractive forces is also demonstrated by Figure (7), in which the boiling points of some hydrogen compounds of the elements in groups IVA, VA, VIA and VIIA are compared. Let us look at the compounds of group IVA first because they form a nearly ideal pattern. We see from the figure that as the atomic weights of the elements in group IVA increase, so do the boiling points of their hydrogen compounds. Using reasoning similar to that developed in the discussion of the molar heat of - 58 - vaporization, we know that as the size of the molecules increase from CH4 to SnH4, the London forces also increase. we expect, therefore, that the boiling points of this series of compounds increase, as they actually do, in the direction of increasing molecular weights. Fig. (8): Boiling Points of Hydrogen compounds of Group IVA, VA, VIA and VHA elements Except for the first members of the hydrogen compounds of group VA, VIA and VIIA, the same trend is also observed, that is increasing boiling point with increasing atomic weight of the element in the group. The first member of each of these groups, however, has a relatively high boiling point, much higher than expected from molecular weights alone. They, therefore, must possess attractive forces in addition to those of the London type. The position of each of the first members in relation to the rest of the group can be attributed to the presence of - 59 - hydrogen bonding. We saw that the strongest hydrogen bonds form when hydrogen is bounded to a very electronegative element. Therefore, the contribution of hydrogen bonding is expected to be the strongest for the first member of groups VA, VIA and VIIA and become s relatively unimportant for the remaining members. Methane, CH4 , which is non polar and cannot form hydrogen bonds (the electro negativity of carbon is too low and there are no lone electron pairs on the carbon to which hydrogen bonds can be formed) follows the normal nearly straight line pattern, within its group. The fact that water has a higher boiling point than HE, even though fluorine is more electronegative than oxygen, seems to be because each water molecule is capable of' forming four hydrogen bonds with other H20 molecules while an HF molecule forms only two hydrogen bonds with two other molecules Figure (8), The strength of the four hydrogen bonds in water exceeds that of the two hydrogen bonds in HF even though an HF hydrogen bond is stronger than single hydrogen bond between H20 molecules, The hydrogen bonding in NH3 is much weaker than in H20 or HF because of the considerably lower electro negativity of nitrogen. Thus, even though NH 3 could conceivably from three or four hydrogen bonds, their total strength is so small that NH3 has a lower boiling point than either HF or H2O. - 60 - Fig. (9): Hydrogen bonding in HF and H2O Freezing Point Liquids become solids by removal of heat. As the temperature is lowered, more molecules in the liquid slow down. If the temperature is sufficiently lowered so that the attractive forces cause the slower- moving molecules to become rigidly held in position, the liquid begins to freeze. If the molecules are "frozen" in a well-ordered lattice, a crystalline solid is formed, while if the molecules are frozen in random fashion, an amorphous solid is produced. As a solid forms, the average kinetic energy of the molecules remaining in the liquid increases. This is because the molecules with low kinetic energies of molecules remaining in the liquid increases. This is because the molecules with low kinetic energies have been lost to the solid, Therefore, heat must continually be removed if' freezing is to continue. In a solid, as well as in a liquid, we have a distribution of kinetic energies. When a solid is in contact with its liquid, high energy molecules at its surface can break away from the solid and enter the liquid state. At some particular temperature called the freezing point (or melting point), the rate at which molecules leave the solid to enter the liquid is the same as the rate which molecules are leaving the liquid state to become part of the solid. Thus, at the melting point of a - 61 - solid or the freezing point of a liquid, equilibrium exists between the liquid and solid. The total quantity of heat that must be removed in order to freeze one mole of a liquid is called its molar heat of crystallization. The molar heat of fusion, ΔHfus on the other hand, is equal in magnitude but opposite in sign to the molar heat of crystallization and is defined as "the quantity of heat that must be supplied to melt one mole of a solid". Since heat must be added to melt a solid, the attractive forces must be slightly higher in a solid than in liquid. This is not surprising because it is expected that greater attractive forces are necessary to hold the particles rigidly in place in a solid than are required to keep them within the liquid where they are free to roam about. The magnitude of the molar heat of fusion provides us with a measure of the difference between the intermolecular attractive forces in the solid and the liquid; that is, ΔHfus = Hliquid - Hsolid As before, we cannot actually measure Hliquid or Hsolid but, instead, only their differences, ΔHfus , ΔHfus is always much smaller than the molar heat of vaporization, as shown in Table (3). The reason for this is as follows. When a solid is melts there are relatively small change in the distance between the molecules. As a result, only small energy changes are involved. When a liquid is converted to a gas, however, the intermolecular distance increase tremendously and large energy - 62 - change occur. This means that the quantity of energy (ΔH fus) required for liquid molecules to move apart, forming a gap. Table (3): Heat of Fusion Hfus and Vaporization Hvap Substance Hfus(K.J mol-l ) HVap(K.J mol-l ) Water 5.98 40.6 Benzene 9.92 30.8 Chloroform 12.4 31.9 Diethyl ether 6.86 26.0 Ethyl alcohol 7.61 38.6 Surface tension Due to unbalanced forces between molecules of the liquid at the surface, the latter behave like a stretched elastic membrane containing the molecules of liquid inside. To understand the phenomenon, consider a vessel containing a liquid as shown in Figure (10). It is obvious that a molecule "A" in the interior of the liquid is entirely surrounded by like molecules and therefore is attracted equally in all directions. But a molecule in the surface for example "B", is subjected to an unequal force of attraction that is tending to pull it inward, as the number of' molecules per unit volume is greater in the bulk of the liquid than in the vapor. As a consequence of this inward - 63 - pull the surface of a liquid always tends to contract to the smallest possible area. It is for this reason that the drops of liquids and bubbles of gas in a liquid become spherical, the surface being then a minimum for the given volume. Fig. (10) The surface behaves as if it were in state of tension. In order to extend the area of the surface it is necessary to do work to bring the molecules from the bulk of the liquid into the surface against the inward attractive force. "The work required to increase the area by lcm is called free surface energy and expressed in erg/ cm 2 ". Thus, "the force in dynes which resists attempts to increase the extent of the surface acting at right angle to any line lcm length in the surface is known as the surface tension y". The usual symbol for surface tension is "gamma y" and it possesses the unit "dyne/cm" the surface tension numerically equal the surface energy. - 64 - Surface tension exists whenever two phases come into contact with each other and is sometimes known a

General Chemistry (CHEM 101) PDF

Document Details

Tags

Related

Summary

Full Transcript

Upgrade to continue