Physical Chemistry RP CH 172 Lecture Notes 2023 PDF
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University of Mines and Technology (UMaT), Tarkwa
2023
Johannes Ami
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These lecture notes cover Physical Chemistry, specifically targeting students of Chemical and Petrochemical Engineering at the University of Mines and Technology (UMaT). The course content includes fundamental topics like equation of state, thermodynamics, and solutions, along with electrochemistry and solvent extraction processes. The notes were compiled in June 2023 by Dr. Johannes Ami.
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University of Mines and Technology (UMaT) Department of Chemical and Petrochemical Engineering PHYSICAL CHEMISTRY RP/CH 172 Lecture Notes Complied by: Johannes Ami (PhD) JUNE, 2023...
University of Mines and Technology (UMaT) Department of Chemical and Petrochemical Engineering PHYSICAL CHEMISTRY RP/CH 172 Lecture Notes Complied by: Johannes Ami (PhD) JUNE, 2023 COURSE INSTRUCTOR: Johannes Ami, PhD (KNUST), MSc (KNUST), BSc (KNUST) Department of Chemical and Petrochemical Engineering, University of Mines and Technology, Ghana Emails: [email protected] or [email protected] Phone: +233 (0) 24 352 99 91 ASSESSMENT OF STUDENTS: Students’ assessment will be in two (2) forms: 1. Continuous Assessment - [40 marks] o Class Attendance o Quizzes/Assignments/Presentations 2. End of Semester Examinations – [60 marks] The results of continuous assessment shall be made known to students at least one week before the start of end of semester examinations. ASSESSMENT OF COURSE LECTURER: At the end of the course, each student will be required to evaluate the course and the lecturer’s performance by answering a questionnaire specifically prepared to obtain the views and opinions of the students about the course and lecturer. Please be sincere. COURSE CONTENT (SYLLABUS): CHAPTER ONE: EQUATION OF STATE CHAPTER TWO: INTRODUCTION TO THERMODYNAMICS CHAPTER THREE: SOLUTIONS CHAPTER FOUR: ELECTROCHEMISTRY AND ELCETROLYSIS CHAPTER FIVE: SOLVENT EXTRACTION PROCESSES LEARNING OUTCOMES FOR PHYSICAL CHEMISTRY: Upon completion of this course, students should be able to: Explain and apply concepts of physical chemistry. Understand equation of state. Define the commonly used terms in thermodynamics. Explain the first law of thermodynamics. Solve numerical problems based on the enthalpy. Describe solutions and ways of expressing concentration. Know Henry´s Law and its application. Describe the kinds of solutions. Understand the principles of electrochemistry and electrolysis and their applications. Define relevant concepts and parameters in solvent extraction processes. Know the principles and methods for liquid-liquid extraction. Know which liquid properties are essential when using these methods. ii TABLE OF CONTENTS TABLE OF CONTENTS.......................................................................................................iii CHAPTER ONE...................................................................................................................... 1 EQUATION OF STATE......................................................................................................... 1 Learning Outcomes.............................................................................................................. 1 1.1 Introduction to Physical Chemistry............................................................................. 1 1.2 Equation of State (EoS)................................................................................................. 2 1.3 Gases................................................................................................................................ 2 1.3.1 General Characteristics of Gases.............................................................................. 3 1.3.2 Parameter of a Gas.................................................................................................... 3 1.4 Gas Laws......................................................................................................................... 5 1.4.1 Boyle’s Law.............................................................................................................. 5 1.4.2 Charles’ Law............................................................................................................. 6 1.4.3 The Combine Gas Law............................................................................................. 8 1.4.4 Avogadro’s Law...................................................................................................... 10 1.5 The Ideal Gas Equation............................................................................................... 12 1.6 Dalton’s Law of Partial Pressures.............................................................................. 13 1.7 Graham’s Law of Diffusion......................................................................................... 15 1.8 Kinetic Molecular Theory Of Gases........................................................................... 17 1.9 Deviations from ideal behaviour................................................................................. 22 1.9.1 Compressibility Factor............................................................................................ 23 1.9.2 Effect of Pressure Variation on Deviations............................................................ 23 1.9.3 Effect of Temperature on Deviations...................................................................... 24 1.9.4 Explanation of Deviations – Van der Waals Equation........................................... 24 1.10 Liquefication Of Gases – Critical Phenomenon...................................................... 30 CHAPTER TWO................................................................................................................... 32 INTRODUCTION TO THERMODYNAMICS................................................................. 32 Learning Outcomes............................................................................................................ 32 2.1 Some Commonly Used Terms..................................................................................... 32 2.1.1 System and Surrounding......................................................................................... 32 2.2 State of a System........................................................................................................ 33 2.2.1 Properties of a System............................................................................................ 34 2.3 Types of Processes..................................................................................................... 34 2.4 Standard States........................................................................................................... 36 2.5 Exothermic and Endothermic Reactions................................................................... 36 2.6 Thermochemical Equations........................................................................................ 37 2.7 The First Law of Thermodynamics............................................................................ 38 iii 2.7.1 Internal Energy (U)................................................................................................. 38 2.7.2 Heat (q) and Work (w)............................................................................................ 39 2.7.3 Work of Expansion................................................................................................. 39 2.7.4 Enthalpy (H)............................................................................................................ 40 2.7.5 Relation between ∆ H and ∆ U............................................................................... 40 2.8 Standard Enthalpy of Reactions................................................................................. 42 2.8.1 Enthalpy of formation (∆f H°)................................................................................. 42 2.8.2 Enthalpy of Combustion (∆comb H°).................................................................... 42 2.8.3 Enthalpy of Neutralization (∆neut Hº)...................................................................... 43 CHAPTER THREE............................................................................................................... 44 SOLUTIONS.......................................................................................................................... 44 Learning Outcomes............................................................................................................ 44 Having completed this chapter, students will be able to:............................................... 44 3.2 Types of Solutions........................................................................................................ 45 3.3 Ways of expressing concentration.............................................................................. 45 3.3.1 Calculation of Molarity........................................................................................... 48 3.4 Solutions of gases in gases........................................................................................... 52 3.5 Henry’s law................................................................................................................... 53 3.6 Solutions of Liquids in Liquids................................................................................... 56 3.6.1 Solubility of Completely Miscible Liquids............................................................. 56 3.6.2 Solubility of Partially Miscible Liquids.................................................................. 56 3.7 Vapour Pressures of Liquid–Liquid Solutions.......................................................... 57 3.7.1 First Type of Mixtures of Miscible Liquids............................................................ 57 3.7.2 Second Type of Mixtures of Miscible Liquids (Minimum boiling point azeotropic solutions).......................................................................................................................... 58 3.7.3 Third Type of Mixtures of Miscible Liquids.......................................................... 59 3.7.4 Azeotropes are Mixtures and not Pure Compounds............................................... 59 3.8 Vapour Pressure of Mixtures of Non–miscible Liquids........................................... 59 3.9 Solutions of Solids in Liquids...................................................................................... 60 3.10 Solubility–Its Equilibrium Concept......................................................................... 61 CHAPTER FOUR.............................................................................................................. 65 ELECTROCHEMISTRY AND ELECTROYSIS.......................................................... 65 Learning Outcomes............................................................................................................ 65 Having completed this chapter, students will be able to:............................................... 65 4.1 Introduction to Electrochemistry............................................................................... 65 4.2 Redox Reactions........................................................................................................... 65 4.3 Terminology............................................................................................................... 66 iv 4.3.1 Redox Illustration.................................................................................................... 66 4.3.2 Oxidation Number.................................................................................................. 67 4.3.2.1 Rules for Assigning an Oxidation Number...................................................... 67 4.4 ELECTROLYSIS........................................................................................................ 75 4.4.1 Electrolytes............................................................................................................. 75 4.5 Mechanism of Electrolysis......................................................................................... 76 4.6 Electrochemical Cells................................................................................................ 77 4.6.1 Voltaic Cells............................................................................................................ 78 4.6.2 Daniel Cell.............................................................................................................. 79 4.7 Application of Electrolysis........................................................................................ 80 4.8 Faraday´s laws of electrolysis.................................................................................... 80 CHAPTER FIVE................................................................................................................... 84 SOLVENT EXTRACTION PROCESSES.......................................................................... 84 5.1 Introduction.................................................................................................................. 84 5.2 Uses of solvent extraction process.............................................................................. 85 5.3 The properties solvents used for solvent extraction.................................................. 85 5.4 Distribution coefficient................................................................................................ 86 5.5 Amount remain extracted........................................................................................... 87 5.6 Equations in Solvent Extraction................................................................................. 87 5.7 Factors affecting solvent extraction........................................................................... 90 5.8 Solvent extraction methods of Metal ions.................................................................. 90 5.8.1 Chelate formation.................................................................................................... 90 5.8.2 Ion association method........................................................................................... 91 5.8.3 Solvation................................................................................................................. 91 5.8.4 Synergic extraction................................................................................................. 91 5.9 Solvent extraction techniques..................................................................................... 91 5.9.1 Batch extraction...................................................................................................... 91 5.9.2 Countercurrent extraction....................................................................................... 91 5.10 Features essential for an extractant......................................................................... 92 5.11 Applications of solvent extraction............................................................................ 92 Reference Materials............................................................................................................... 93 v CHAPTER ONE EQUATION OF STATE Learning Outcomes Having completed this chapter, students will be able to: Explain and apply concepts of physical chemistry. Understand equation of state. Review the gas laws. Know, derive the ideal gas equation, and use it to solve mathematical problems. 1.1 Introduction to Physical Chemistry Physical Chemistry is the quantitative and theoretical study of the properties and structure of matter, and their relationship to the interaction of matter and energy. Thus, it is concerned with the measurement, description, and prediction of the characteristics of chemical systems and the interactions with each other with respect to the transfer of mass and energy. Physical Chemistry is needed to interpret the kinetics and understand the mechanism of reactions and to work out the most effective conditions under which it should take place in an industrial setting. The problems of concern in physical chemistry are as follows: Chemical equilibrium – Thermodynamics Structure and behaviour of atoms and molecules - Quantum Mechanics & Spectroscopy Rates and mechanisms of Chemical changes - Chemical Kinetics & Dynamics Behaviour, nature and properties of ions in solutions – Electrochemistry There are basically two (2) approaches to the study of physical chemistry. These are: 1. Synthetic Approach This begins into the structure and behaviour of matter in its finest known state of subdivision and gradually progresses from electrons to atoms to molecules and to the state of aggregates and how chemical reactions occur. 2. Analytical Approach This starts with matter or chemicals as we find them in the laboratory and gradually works its way back to the finer state of sub-division as we require them to explain experimental results. 1 1.2 Equation of State (EoS) An EoS is the relation between functions of state, such as temperature (T), pressure (P), volume (V), internal energy or specific heat. It characterizes the state of matter of a material under a given set of physical conditions. EoSs are used to describe gases, fluids, fluid mixtures and solids. In mineral physics an EoS is used to determine mineral composition at depth (i.e., P and T) and depicts how the volume or equivalently density of a material vary as a function of depth, i.e., as a function of P and T. It reflects atomic structure, chemical bonding and stability of a material. The simplest equation of state is the one for ideal gases: PV = nRT where n is the number of moles and R is a constant, the gas constant R = 8.31451 J K-1 mol-1. For a volume of ideal gas (V) experiencing an external pressure (P), there will be a related increase in temperature of the gas given by PV / R. However, at high pressure and at low temperature, real gases deviate from the ideal gas law and the equation needs to be modified. The EoS for solids and liquids also needs to be modified, more complex EoSs are necessary and various equations have been proposed. For most solids and liquids, the effect of pressure is much greater than the effect of temperature, over the range of temperatures and pressures relevant to the Earth. Therefore, it is easier to consider initially an isothermal EoS and then introduce a thermal expansion coefficient as temperature correction. All EoS are based on assumptions. They are tested by their ability to reproduce the experimental data. Here we will focus on the most commonly used equations. We will start by examining the effect of pressure, i.e. the isothermal EoS. 1.3 Gases All matter exists in three states: gas, liquid and solid. A gas consists of molecules separated wide apart in empty space. The molecules are free to move about throughout the container. A liquid has molecules touching each other. However, the intermolecular space, permit the movement of molecules throughout the liquid. A solid has molecules, atoms or ions arranged in a certain order in fixed positions in the crystal lattice. The particles in a solid are not free to move about but vibrate in their fixed positions. Of the three states of matter, the gaseous state is the one most studied and best understood. We shall consider it first. 2 1.3.1 General Characteristics of Gases 1. Expansibility Gases have limitless expansibility. They expand to fill the entire vessel they are placed in. 2. Compressibility Gases are easily compressed by application of pressure to a movable piston fitted in the container. 3. Diffusibility Gases can diffuse rapidly through each other to form a homogeneous mixture. 4. Pressure Gases exert pressure on the walls of the container in all directions. 5. Effect of Heat When a gas, confined in a vessel is heated, its pressure increases. Upon heating in a vessel fitted with a piston, volume of the gas increases. The above properties of gases can be easily explained by the Kinetic Molecular Theory which will be considered later in the chapter. 1.3.2 Parameter of a Gas A gas sample can be described in terms of four parameters (measurable properties): (1) the volume, V of the gas (2) its pressure, P (3) its temperature, T (4) the number of moles, n, of gas in the container The Volume, V The volume of the container is the volume of the gas sample. It is usually given in litre (l or L) or millilitres (ml or mL). The SI unit for volume is cubic metre (m3) and the smaller unit is decimeter3 (dm3). The Pressure, P 3 The pressure of a gas is defined as the force exerted by the impacts of its molecules per unit surface area in contact. The pressure of a gas sample can be measured with the help of a mercury manometer (Fig. 1.1) Similarly, the atmospheric pressure can be determined with a mercury barometer (Fig. 1.2). Fig. 1.1 A mercury manometer Fig. 1.2 A mercury barometer. In Fig. 1.1, the gas container is connected with a U-tube containing Hg having vacuum in the closed end. The difference in Hg height in two limbs gives the gas pressure in mm Hg. Fig. 1.2 is a long tube (80 × 1 cm) filled with Hg inverted into dish containing Hg. The atmospheric pressure is equal to 760 mm Hg column supported by it at sea level. The pressure of air that can support 760 mm Hg column at sea level, is called one atmosphere (1 atm). The unit of pressure, millimetre of mercury, is also called torr. The SI unit of pressure is the Pascal (Pa). Temperature, T The temperature of a gas may be measured in Centigrade degrees (°C) or Celsius degrees. The SI unit of temperature is Kelvin (K) or Absolute degree. The centigrade degrees can be converted to kelvins by using the equation. K = °C + 273 The Kelvin temperature (or absolute temperature) is always used in calculations of other parameters of gases. Remember that the degree sign (°) is not used with K. The Moles of a Gas Sample, n The number of moles, n, of a sample of a gas in a container can be found by dividing the mass, m, of the sample by the molar mass, M (molecular mass). 4 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑔𝑎𝑠 𝑠𝑎𝑚𝑝𝑙𝑒 (𝑚) Mole of gas (n) = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 (𝑀) 1.4 Gas Laws The volume of a given sample of gas depends on the temperature and pressure applied to it. Any change in temperature or pressure will affect the volume of the gas. As results of experimental studies from 17th to 19th century, scientists derived the relationships among the pressure, temperature and volume of a given mass of gas. These relationships, which describe the general behaviour of gases, are called the gas laws. 1.4.1 Boyle’s Law In 1660 Robert Boyle found out experimentally the change in volume of a given sample of gas with pressure at room temperature. From his observations he formulated a generalization known as Boyle’s Law. It states that: at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved. Fig. 1.3 Boyle’s law states that at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure. If the pressure is doubled, the volume is halved. If P1, V1 are the initial pressure and volume of a given sample of gas and P2, V2 the changed pressure and volume, we can write 5 P1V1= k= P2V2 Or P1V1=P2V2 This relationship is useful for the determination of the volume of a gas at any pressure, if its volume at any other pressure is known. (a) (b) Fig. 1.4 Graphical representation of Boyle's law. (a) a plot of versus for a gas sample is hyperbola; (b) a plot of versus 1/P is a straight line. The Boyle’s law can be demonstrated by adding liquid mercury to the open end of a J-tube. As the pressure is increased by addition of mercury, the volume of the sample of trapped gas decreases. Gas pressure and volume are inversely related; one increases when the other decreases. Fig. 1.5 Demonstration of Boyle’s law 1.4.2 Charles’ Law In 1787 Jacques Charles investigated the effect of change of temperature on the volume of a fixed amount of gas at constant pressure. He established a generalization which is called the 6 Charles’ Law. It states that: at constant pressure, the volume of a fixed mass of gas is directly proportional to the Kelvin temperature of absolute temperature. If the absolute temperature is doubled, the volume is doubled. Charles’ Law may be expressed mathematically as V α T (P, n are constant) Or V = kT where is a constant. Or V/T = T If V1, T1 are the initial volume and temperature of a given mass of gas at constant pressure and V2, T2 be the new values, we can write Using this expression, the new volume V2, can be found from the experimental values of V1, T1 and T2. Fig. 1.6 Charles law state that at constant pressure, the volume of a fixed mass of gas is directly proportional to the absolute temperature. 7 Fig. 1.7 Graph showing that at constant pressure, volume of a given mass of gas is directly proportional to the Kelvin temperature. 1.4.3 The Combine Gas Law Boyle’s Law and Charles’ Law can be combined into a single relationship called the Combined Gas Law. The combined law can be stated as: for a fixed mass of gas, the volume is directly proportional to kelvin temperature and inversely proportional to the pressure. If k be the proportionality constant, If the pressure, volume and temperature of a gas be changed from P1, V1 and T1 to P2, T2 and V2, then 8 This is the form of combined law for two sets of conditions. It can be used to solve problems involving a change in the three variables P, V and T for a fixed mass of gas. Solved Problem 25.8 litre of a gas has a pressure of 690 torr and a temperature of 17°C. What will be the volume if the pressure is changed to 1.85 atm and the temperature to 345 K. Solution 9 1.4.4 Avogadro’s Law Let us take a balloon containing a certain mass of gas. If we add to it more mass of gas, holding the temperature (T) and pressure (P) constant, the volume of gas (V) will increase. It was found experimentally that the amount of gas in moles is proportional to the volume. That is, Thus, for equal volumes of the two gases at fixed T and P, number of moles is also equal. This is the basis of Avogadro’s Law which may be stated as: equal volumes of gases at the same temperature and pressure contain equal number of moles or molecules. If the molar amount is doubled, the volume is doubled. Fig. 1.8 Avogadro’s law states that under equal conditions of temperature and pressure, equal volumes of gases contain an equal number of molecules. The Molar Gas Volume It follows as a corollary of Avogadro’s Law that one mole of any gas at a given temperature (T) and pressure (P) has the same fixed volume. It is called the molar gas volume or molar volume. In order to compare the molar volumes of gases, chemists use a fixed reference 10 temperature and pressure. This is called standard temperature and pressure (abbreviated, STP). The standard temperature used is 273 K (0°C) and the standard pressure is 1 atm (760 mm Hg). At STP we find experimentally that one mole of any gas occupies a volume of 22.4 litres. To put it in the form of an equation, we have 1 mole of a gas at STP = 22.4 litres 11 1.5 The Ideal Gas Equation We have studied three simple gas laws: These three laws can be combined into a single more general gas law: This is called the Universal Gas Law. It is also called Ideal Gas Law as it applies to all gases which exhibit ideal behaviour i.e., obey the gas laws perfectly. The ideal gas law may be stated as: the volume of a given amount of gas is directly proportional to the number of moles of gas, directly proportional to the temperature, and inversely proportional to the pressure. Introducing the proportionality constant R in the expression (1) we can write The equation (2) is called the Ideal-gas Equation or simply the general Gas Equation. The constant R is called the Gas constant. The ideal gas equation holds fairly accurately for all gases at low pressures. For one mole (n = 1) of a gas, the ideal-gas equation is reduced to The ideal-gas equation is called an Equation of State for a gas because it contains all the variables (T, P, V and n) which describe completely the condition or state of any gas sample. If we know the three of these variables, it is enough to specify the system completely because the fourth variable can be calculated from the ideal-gas equation. The Numerical Value of R. From the ideal-gas equation, we can write We know that one mole of any gas at STP occupies a volume of 22.4 litres. Substituting the values in the expression (1), we have 12 It may be noted that the unit for R is complex; it is a composite of all the units used in calculating the constant. If the pressure is written as force per unit area and volume as area times length, from (1) Hence R can be expressed in units of work or energy per degree per mole. The actual value of R depends on the units of P and V used in calculating it. The more important values of R are listed in Table 1.0. Table 1.0 Value of R in different units 1.6 Dalton’s Law of Partial Pressures John Dalton visualised that in a mixture of gases, each component gas exerted a pressure as if it were alone in the container. The individual pressure of each gas in the mixture is defined as its Partial Pressure. Based on experimental evidence, in 1807, Dalton enunciated what is commonly known as the Dalton’s Law of Partial Pressures. It states that: the total pressure of a mixture of gases is equal to the sum of the partial pressures of all the gases present (Fig. 1.8). 13 Fig. 1.8 Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by each gas. The pressure of the mixture of O and N (Tanks) is the sum of the pressures in O and N tanks. Mathematically the law can be expressed as Ptotal = P1 + P2 + P3... (V and T are constant) where P1, P2 and P3 are partial pressures of the three gases 1, 2 and 3; and so on. Dalton’s Law of Partial Pressures follows by application of the ideal-gas equation PV = n RT separately to each gas of the mixture. Thus, we can write the partial pressures P1, P2 and P3 of the three gases In the words, the total pressure of the mixture is determined by the total number of moles present whether of just one gas or a mixture of gases. Solved Problem 1 What pressure is exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273K in a 10 litre vessel? Solution 14 Sample Problem 2 A sample of oxygen is collected by the downward displacement of water from an inverted bottle. The water level inside the bottle is equalised with that in the trough. Barometeric pressure is found to be 757 mm Hg, and the temperature of water is 23.0°C. What is the partial pressure of O2? Vapour pressure of H2O at 23°C = 19.8 mm Hg. Solution The total pressure inside the bottle is 1.7 Graham’s Law of Diffusion When two gases are placed in contact, they mix spontaneously. This is due to the movement of molecules of one gas into the other gas. This process of mixing of gases by random motion of the molecules is called Diffusion. Thomas Graham observed that molecules with smaller masses diffused faster than heavy molecules. 15 Fig. 1.9 A light molecule diffuses quicker than a heavy molecule. In 1829 Graham formulated what is now known as Graham’s Law of Diffusion. It states that: under the same conditions of temperature and pressure, the rates of diffusion of different gases are inversely proportional to the square roots of their molecular masses. Mathematically the law can be expressed as where r1 and r2 are the rates of diffusion of gases 1 and 2, while M1 and M2 are their molecular masses. When a gas escapes through a pin-hole into a region of low pressure of vacuum, the process is called Effusion. The rate of effusion of a gas also depends, on the molecular mass of the gas. (a) Diffusion (b) Effusion Fig 1.10 (a) Diffusion is mixing of gas molecules by random motion under conditions where molecular collisions occur. (b) Effusion is escape of a gas through a pinhole without molecular collisions. Dalton’s law when applied to effusion of a gas is called the Dalton’s Law of Effusion. It may be expressed mathematically as 16 The determination of rate of effusion is much easier compared to the rate of diffusion. Therefore, Dalton’s law of effusion is often used to find the molecular mass of a given gas. Solved Problem 1 If a gas diffuses at a rate of one-half as fast as O2, find the molecular mass of the gas. Solved Problem 2 50 ml of gas A effuse through a pin-hole in 146 seconds. The same volume of CO2 under identical conditions effuses in 115 seconds. Calculate the molecular mass of A. Solution 1.8 Kinetic Molecular Theory Of Gases Maxwell and Boltzmann (1859) developed a mathematical theory to explain the behaviour of gases and the gas laws. It is based on the fundamental concept that a gas is made of a large number of molecules in perpetual motion. Hence the theory is called the kinetic molecular theory or simply the kinetic theory of gases (The word kinetic implies motion). The kinetic theory makes the following assumptions. 17 Assumptions of the Kinetic Molecular Theory (1) A gas consists of extremely small discrete particles called molecules dispersed throughout the container. The actual volume of the molecules is negligible compared to the total volume of the gas. The molecules of a given gas are identical and have the same mass (m). Fig. 1.11A gas is made of molecules dispersed Fig. 1.12 Actual volume of the gas in space in the container. molecules is negligible. Fig. 1.13 Gas molecules are in constant Fig. 1.14 Molecules move in straight line motion in all possible directions and direction on collision with another molecule o or wall container (2) Gas molecules are in constant random motion with high velocities. They move in straight lines with uniform velocity and change direction on collision with other molecules or the walls of the container. Pool table analogy is shown in Fig.1.15. 18 Fig. 1.15 Gas molecules can be compared to billiard balls in random motion, bouncing off each other and off the sides of the pool table. (3) The distance between the molecules are very large and it is assumed that van der Waals attractive forces between them do not exist. Thus, the gas molecules can move freely, independent of each other. (4) All collisions are perfectly elastic. Hence, there is no loss of the kinetic energy of a molecule during a collision. (5) The pressure of a gas is caused by the hits recorded by molecules on the walls of the container. 1 (6) The average kinetic energy (2 𝑚𝑣 2 ) of the gas molecules is directly proportional to absolute temperature (Kelvin temperature). This implies that the average kinetic energy of molecules is the same at a given temperature. How Does an Ideal Gas Differ from Real Gases? A gas that confirms to the assumptions of the kinetic theory of gases is called an ideal gas. It obeys the basic laws strictly under all conditions of temperature and pressure. The real gases as hydrogen, oxygen, nitrogen etc., are opposed to the assumptions (1), (2) and (3) stated above. Thus: (a) The actual volume of molecules in an ideal gas is negligible, while in a real gas it is appreciable. (b) There are no attractive forces between molecules in an ideal gas while these exist in a real gas. 19 (c) Molecular collisions in an ideal gas are perfectly elastic while it is not so in a real gas. For the reasons listed above, real gases obey the gas laws under moderate conditions of temperature and pressure. At very low temperature and very high pressure, the clauses (1), (2) and (3) of kinetic theory do not hold. Therefore, under these conditions the real gases show considerable deviations from the ideal gas behaviour. Derivation of Kinetic Gas Equation Starting from the postulates of the kinetic molecular theory of gases we can develop an important equation. This equation expresses PV of a gas in terms of the number of molecules, molecular mass and molecular velocity. This equation which we shall name as the Kinetic Gas Equation may be derived by the following clauses. Let us consider a certain mass of gas enclosed in a cubic box (Fig. 1.16) at a fixed temperature. Suppose that: the length of each side of the box = l cm the total number of gas molecules = n the mass of one molecule =m the velocity of a molecule =v The kinetic gas equation may be derived by the following steps: (1) Resolution of Velocity v of a Single Molecule Along X, Y and Z Axes According to the kinetic theory, a molecule of a gas can move with velocity v in any direction. Velocity is a vector quantity and can be resolved into the components vx, vy, vz along the X, Y and Z axes. These components are related to the velocity v by the following expression. v2 = v2x + v2y + v2z Now we can consider the motion of a single molecule moving with the component velocities independently in each direction. (2) The Number of Collisions Per Second on Face A Due to One Molecule Consider a molecule moving in OX direction between opposite faces A and B. It will strike the face A with velocity vx and rebound with velocity – vx. To hit the same face again, the 20 molecule must travel l cm to collide with the opposite face B and then again l cm to return to face A. Therefore, Fig. 1. 16 Resolution of velocity into Fig. 1.17 Cubic box showing molecular components vx, vy, and vz collision along axis (3) The Total Change of Momentum on All Faces of the Box Due to One Molecule Only 21 (4) Total Change of Momentum Due to Impacts of All the Molecules on All Faces of the Box (5) Calculation of Pressure from Change of Momentum; Derivation of Kinetic Gas Equation 1.9 Deviations from ideal behaviour An ideal gas is one which obeys the gas laws or the gas equation PV = RT at all pressures and temperatures. However, no gas is ideal. Almost all gases show significant deviations from the ideal behaviour. Thus, the gases H2, N2 and CO2 which fail to obey the ideal-gas equation are termed nonideal or real gases. 22 1.9.1 Compressibility Factor The extent to which a real gas departs from the ideal behaviour may be depicted in terms of a new function called the Compressibility factor, denoted by Z. It is defined as 𝑃𝑉 Z = 𝑅𝑇 The deviations from ideality may be shown by a plot of the compressibility factor, Z, against P. For an ideal gas, Z = 1 and it is independent of temperature and pressure. The deviations from ideal behaviour of a real gas will be determined by the value of Z being greater or less than 1. The difference between unity and the value of the compressibility factor of a gas is a measure of the degree of nonideality of the gas. For a real gas, the deviations from ideal behaviour depend on pressure and temperature. This will be illustrated by examining the compressibility curves of some gases discussed below with the variation of pressure and temperature. 1.9.2 Effect of Pressure Variation on Deviations Fig. 1.18 shows the compressibility factor, Z, plotted against pressure for H2, N2 and CO2 at a constant temperature. Fig. 1.18 Z versus P plots for H, N and CO2 at 300 K. At very low pressure, for all these gases Z is approximately equal to one. This indicates that at low pressures (up to 10 atm), real gases exhibit nearly ideal behaviour. As the pressure is increased, H2 shows a continuous increase in Z (from Z = 1). Thus, the H2 curve lies above the ideal gas curve at all pressures. For N2 and CO2, Z first decreases (Z < 1). It passes through a minimum and then increases continuously with pressure (Z > 1). For a gas like CO 2 the dip in the curve is greatest as it is most easily liquefied. 23 1.9.3 Effect of Temperature on Deviations Fig 1.19 shows plots of Z or PV/RT against P for N2 at different temperatures. It is clear from the shape of the curves that the deviations from the ideal gas behaviour become less and less with increase of temperature.At lower temperature, the dip in the curve is large and the slope of the curve is negative. That is, Z < 1. As the temperature is raised, the dip in the curve decreases. At a certain temperature, the minimum in the curve vanishes and the curve remains horizontal for an appreciable range of pressures. At this temperature, PV/RT is almost unity and the Boyle’s law is obeyed. Hence this temperature for the gas is called Boyle’s temperature. The Boyle temperature of each gas is characteristic e.g., for N2 it is 332 K. Conclusions From the above discussions we conclude that: (1) At low pressures and fairly high temperatures, real gases show nearly ideal behaviour and the ideal-gas equation is obeyed. (2) At low temperatures and sufficiently high pressures, a real gas deviates significantly from ideality and the ideal-gas equation is no longer valid. (3) The closer the gas is to the liquefaction point, the larger will be the deviation from the ideal behaviour. Fig. 1.19 The closer the gas is to the liquefaction point; the larger will be the deviation from the ideal behaviour 1.9.4 Explanation of Deviations – Van der Waals Equation van der Waals (1873) attributed the deviations of real gases from ideal behaviour to two erroneous postulates of the kinetic theory. These are: 24 (1) the molecules in a gas are point masses and possesses no volume. (2) there are no intermolecular attractions in a gas. Therefore, the ideal gas equation PV = nRT derived from kinetic theory could not hold for real gases. van der Waals pointed out that both the pressure (P) and volume (V) factors in the ideal gas equation needed correction in order to make it applicable to real gases. Volume Correction The volume of a gas is the free space in the container in which molecules move about. Volume V of an ideal gas is the same as the volume of the container. The dot molecules of ideal gas have zero-volume and the entire space in the container is available for their movement. However, van der Waals assumed that molecules of a real gas are rigid spherical particles which possess a definite volume. Fig. 1. 20 Volume of a Real gas. The volume of a real gas is, therefore, ideal volume minus the volume occupied by gas molecules (Fig. 1.20). If b is the effective volume of molecules per mole of the gas, the volume in the ideal gas equation is corrected as: (𝑉 − 𝑏) For n moles of the gas, the corrected volume is: (𝑉 − 𝑛𝑏) where b is termed the excluded volume, which is constant and characteristic for each gas. 25 Fig. 1.21 Excluded volume for a pair of gas molecules. Excluded volume is four times the actual volume of molecules. The excluded volume is not equal to the actual volume of the gas molecules. In fact, it is four times the actual volume of molecules and can be calculated as follows. Let us consider two molecules of radius r colliding with each other (Fig. 1.21). Obviously, they cannot approach each other closer than a distance (2r) apart. Therefore, the space indicated by the dotted sphere having radius (2r) will not be available to all other molecules of the gas. In other words the dotted spherical space is excluded volume per pair of molecules. Thus, Therefore, in general, excluded volume of the gas molecules is four times the actual volume of molecules. Pressure Correction A molecule in the interior of a gas is attracted by other molecules on all sides. These attractive forces cancel out. But a molecule about to strike the wall of the vessel is attracted by molecules on one side only. Hence it experiences an inward pull (Fig. 1.22). Therefore, it strikes the wall with reduced velocity and the actual pressure of the gas, P, will be less than the ideal pressure. If the actual pressure P, is less than Pideal by a quantity p, we have 26 P = Pideal - p Or Pideal = P + p Fig. 1.22 (a) A molecule about to strike the wall has a net inward pull; (b) A molecule in the interior of gas has balanced attractions. p is determined by the force of attraction between molecules (A) striking the wall of container and the molecules (B) pulling them inward. The net force of attraction is, therefore, proportional to the concentration of (A) type molecules and also of (B) type of molecules. That is, Van der Waals Equation 27 Determination of a and b Fig. 1.23 The striking molecule A is pulled inward by molecules B which reduces the velocity of A and causes the decrease of pressure. 28 Solved Problem Calculate the pressure exerted by 1.00 mole of methane (CH4) in a 250 mL container at 300 K using van der Waals equation. What pressure will be predicted by ideal gas equation? 29 Solution 1.10 Liquefication Of Gases – Critical Phenomenon A gas can be liquefied by lowering the temperature and increasing the pressure. At lower temperature, the gas molecules lose kinetic energy. The slow moving molecules then aggregate due to attractions between them and are converted into liquid. The same effect is produced by the increase of pressure. The gas molecules come closer by compression and coalesce to form the liquid. Andres (1869) studied the P – T conditions of liquefaction of several gases. He established that for every gas there is a temperature below which the gas can be liquefied but above it the gas defies liquefaction. This temperature is called the critical temperature of the gas. The critical temperature, Tc, of a gas may be defined as that temperature above which it cannot be liquefied no matter how great the pressure applied. The critical pressure, Pc, is the minimum pressure required to liquefy the gas at its critical temperature. The critical volume, Vc, is the volume occupied by a mole of the gas at the critical temperature and critical pressure. Tc, Pc and Vc are collectively called the critical constants of the gas. All real gases have characteristic critical constants. At critical temperature and critical pressure, the gas becomes identical with its liquid and is said to be in critical state. The smooth merging of the gas with its liquid is referred to as the critical phenomenon. Andrews demonstrated the critical phenomenon in gases by taking example of carbon dioxide. At critical point 30 𝑅𝑇𝑐 𝑎 Pc = - 𝑉𝑐−𝑏 𝑉𝑐 2 Limitations of Van der Waal’s Equation 1. It is valid over a wide range of pressures and temperatures. 2. It fails to give exact agreement with experimental data at high pressures and low temperatures. 31 CHAPTER TWO INTRODUCTION TO THERMODYNAMICS Learning Outcomes Having completed this chapter, students will be able to: Define the commonly used terms in thermodynamics. Explain the first law of thermodynamics. Solve numerical problems based on the enthalpy. 2.1 Some Commonly Used Terms In this lesson you would come across some frequently used terms. Let us understand the meaning of these terms first. 2.1.1 System and Surrounding If we are studying the reaction of two substances A and B kept in a beaker, the reaction mixture of A and B is a system and the beaker and the room where it is kept are surroundings as shown in figure 11.1 System Surroundings A+B Fig 2.1: System and surroundings System is the part of the physical universe which is under study, while the rest of the universe is surroundings. There are basically, three types of systems which follows discussions below. You know that hot tea/milk (let us call it a system) kept in a stoppered thermos flask remains hot for a couple of hours. If this flask is made of perfect insulating material, then there would be no exchange of matter or energy between the system and the surroundings. We call such a system an isolated system. 32 Isolated system is a system which can exchange neither matter nor energy with the surroundings. If we keep hot tea/milk in a stoppered stainless steel flask, it will not remain hot after some time. Here energy is lost to the surroundings through the steel walls, but due to stopper, the matter will not be lost. We call this system a closed system. Closed system is a system which can exchange energy but not matter with the surroundings. If we keep stainless steel flask or thermos flask open, some matter will also be lost due to evaporation along with energy. We call such a system an open system. Plants, animals, human beings are all examples of open systems, because they continuously exchange matter (food, etc) and energy with the surroundings. Open system is a system which can exchange both energy and matter with surroundings. 2.2 State of a System We describe the state of a system by its measurable properties. For example, we can describe the state of a gas by specifying its pressure, volume, temperature etc. These variable properties are called state variables or state functions. Their values depend only on the initial and final state of the system and not on the path taken by the system during the change. When the state of a system changes, the change depends only on the initial and the final state of the system (Fig 11.2). Final State I II III A (p1, T1) Initial State T Fig. 2.2: Change of state from initial state to final state through three paths I, II and III. The difference p2– p1and T2– T1 are independent of the path since pressure and temperature are state functions. 33 State functions are those functions which depend only on the state of the system. Change in state of a system is defined by giving the initial and the final state of the system. We can understand it by considering another example. We travel from one point to another. The distance travelled depends on the path or the route we take. But the separation between these two points on the earth is fixed. Thus, separation is a state function, but not the distance travelled. 2.2.1 Properties of a System As stated earlier, the measurable properties of a system are called state variables. They may be further divided into two main types. (i) Extensive property or variable is one whose value depends upon the size of the system. For example, volume, weight, heat, etc. (ii) Intensive property or variable is one whose value is independent of the size of the system. For example, temperature, pressure, refractive index, viscosity, density, surface, tension, etc. You may note that an extensive property can become an intensive property by specifying a unit amount of the substance concerned. For example, mass and volume are extensive properties, but density (mass per unit volume) and specific volume (volume per unit mass) are intensive properties. 2.3 Types of Processes Let us first understand what we mean by a process. Suppose we want to raise the temperature of the system. We may do it by heating it. Here, heating is the process. The method of bringing about a change in state is called process. Processes could be of different types. The different types of processes are explained below. i) Isothermal process: Ice melts at 273 K and 1 atm pressure. The temperature does not change as long as the process of melting goes on. Such processes are examples of isothermal process. We can define isothermal process as follows. 34 When the temperature of the system remains constant during various operations, then the process is said to be isothermal. This is attained either by removing heat from the system or by supplying heat to the system. For an isothermal process dT = 0 So, Ice melts at 273 K and 1 atm pressure. The temperature does not change as long as the process of melting goes on. ii) Adiabatic Process If an acid is mixed with a base in a closed thermos flask, the heat evolved is retained by the system. Such processes are known as adiabatic processes because the thermos flask does not allow exchange of heat between the system and the surroundings. Adiabatic process can be defined as follows: In an adiabatic process there is no exchange of heat between the system and the surroundings. Thus, in adiabatic processes there is always a change in temperature. For an adiabatic process dq = 0 iii) Isobaric Processes Those processes which take place at constant pressure are called isobaric processes. For example, heating of water to its boiling point and its vaporisation take place at the same atmospheric pressure. These changes are, therefore, designated as isobaric processes and are said to take place isobarically. For an isobaric process dp = 0 iv) Isochoric Processes Those processes in which the volume remains constant are known as isochoric processes. The heating of a substance in a non-expanding chamber is an example of isochoric process. For isochoric processes dV = 0 v) Cyclic Process When a system in a given state goes through a number of different processes and finally returns to its initial state, the overall process is called a cycle or cyclic process. For a cyclic process dE = 0, dH = 0 35 2.4 Standard States You have seen that a system is described by the state variables. In order to compare the energies for different compounds, a standard set of conditions is chosen. This refers to the condition of 1 bar pressure at any specified temperature, with a substance in its most stable form. 2.5 Exothermic and Endothermic Reactions (i) Add a few cm3 of dilute hydrochloric acid in a test tube containing a few pieces of granulated zinc and observe the evolution of a gas. Feel the test tube. It would be hot. (ii) You must have observed that when some water is added to quick lime to prepare white wash, a lot of heat is evolved. (iii) When a fuel like cooking gas or coal is burnt in air, heat is evolved besides light. Many chemical reactions lead to release of energy (heat) to the surroundings. We call these type of reactions as exothermic reactions. Exothermic reactions are those reactions which proceed with the evolution of heat. Let us now consider the following reactions: (i) Add a small amount of solid ammonium chloride in a test tube half-filled with water. Shake and feel the test tube. It will feel cold. (ii) Similarly repeat this experiment with potassium nitrate and feel the test tube, it will feel cold. (iii) Mix barium hydroxide with ammonium chloride in small quantities in water taken in a test tube. Feel the test tube. It will be cold. In all the above processes we see that heat is absorbed by the system from the surroundings. Such reactions are called endothermic reactions. Endothermic reactions are those reactions which proceed with the absorption of heat from the surroundings. 36 2.6 Thermochemical Equations You are familiar with equations for chemical reactions. Now we shall write the chemical equations which will specify heat energy changes and states of the reactants and products. These are called the thermochemical equations. For writing these equations, we follow the conventions listed below: (i) The heat evolved or absorbed in a reaction is affected by the physical state of the reacting sub1stances. Therefore, gaseous, liquid and solid states are represented by putting symbols (g), (l), and (s) alongside the chemical formulae respectively. For example, to represent burning of methane in oxygen, we write CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) + heat In writing thermochemical reactions, we denote the amount of heat evolved or absorbed by a symbol ∆H. The amount of heat evolved or absorbed is written after the equation followed by semicolon. ∆H is negative for exothermic reactions and it is positive for endothermic reactions. For example: An exothermic reaction is written as CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O (l); ∆H= – 891 kJ Whereas an endothermic reaction is written as H2 (g) + I2 (g) 2HI (g); ∆H= 52.2 kJ (ii) In case of elements which exhibit allotropy, the name of allotropic modification is mentioned. For example, C (graphite), C (diamond), etc. (iii) The substances in aqueous solutions are specified using the symbol (aq). For example NaCl (aq) stands for an aqueous solution of sodium chloride. (iv) Thermochemical equations may be balanced even by using fractional coefficients, if so required. The coefficients of the substances of the chemical equation indicate the number of 37 moles of each substance involved in the reaction and the ∆H values given correspond to these quantities of substances. (v) In case the coefficients are multiplied or divided by a factor, ∆H value must also be multiplied or divided by the same factor. In such cases, the ∆H value will depend upon the coefficients. For example, in equation. 1 H2 (g) +2O2 (g) →H2O (g); ∆H= – 242 kJ If coefficients are multiplied by 2, we would write the equation 2H2 (g) + O2 (g) →2H2O (g); ∆H =2 (– 242) = – 484 k 2.7 The First Law of Thermodynamics You have learnt that chemical reactions are accompanied by energy changes. How do we determine these energy changes? You know that we cannot create or destroy energy. Energy only changes from one form to another. This is the observation made by many scientists over the years. This observation has taken the form of first law of thermodynamics. It has been found valid for various situations. We state this law as follows: Energy can neither be created nor destroyed. The total energy of the universe or an isolated system is constant. Mathematically the first law of thermodynamics is stated as: ∆U= q+ w (1.1) where ∆U = change in internal energy, q= heat absorbed by the system, and w= work done on the system. These terms are explained as: 2.7.1 Internal Energy (U) Every system has a definite amount of energy. This amount is different for different substances. It includes translational, vibrational and rotational energies of molecules, energy of electrons and nuclei. The internal energy may be defined as the sum of the energies of all the atoms, molecules or ions contained in the system. It is a state variable. It is not possible to measure the absolute values of internal energy. However, we can calculate the change in internal energy. If the internal energy of the system in the initial state is U1 and that in the final state is U2, then change in internal 38 Energy ∆U is independent of the path taken from the initial to the final state. We can write this change as: ∆U= U2– U1 The internal energy of the system can be changed in two ways (i) Either by allowing heat to flow into the system or out of the system; and (ii) By work done on the system or by the system 2.7.2 Heat (q) and Work (w) Heat and work are not state functions. This is because the values of both q and w depend upon the way in which the change is carried out. Since the law deals with the transfer of heat and work, we assign some signs to these quantities. Anything which increases the internal energy of a system is given a positive sign. Heat given to the system (q) and work done on the system (w) are given positive signs. Let us illustrate this with an example. If a certain change is accompanied by absorption of 50 kJ of heat and expenditure of 30 kJ of work, q= + 50 kJ w= – 30 kJ Change in internal energy ∆U= (+ 50 kJ) + (– 30 kJ) = + 20 kJ Thus the system has undergone a net increase in the internal energy of –20 kJ. Change in the internal energy of the surroundings will be 20 kJ. 2.7.3 Work of Expansion Let us assume that pressure p is constant and the volume of the system changes from V1to V2. The work done by a system is given as w= – p (V2– V1) = – p ∆ V (1.2) (Here we have taken minus sign, because the work is done by the system). Let us substitute the expression given for w in equation 1.1. 39 We get ∆U= q– p ∆ V (1.3) If the process is carried out at constant volume, i.e. ∆ V= 0, then ∆U= qv (1.4) The subscript v in qV denotes that volume is constant. The equation 1.4 shows that we can determine internal energy change if we measure the heat gained or lost by the system at constant volume. However, in chemistry, the chemical reactions are generally carried out at constant pressure (atmospheric pressure). What do we do then? Let us define another state function, called, enthalpy. 2.7.4 Enthalpy (H) For measuring heat lost or gained at constant pressure, we define a new state function Called enthalpy. It is denoted by the symbol H and is given by H= U+ p V (1.5) Enthalpy change, ∆H, is given by ∆H= ∆U+ ∆ (pV) (1.6) or ∆H= ∆U+ p ∆ V+ V ∆ p If the change is carried out at constant pressure, then ∆p= 0. The equation 1.6 will Become ∆H= ∆U+ p ∆V (at constant pressure) (1.7) Substituting the value of ∆ U from equation 1.3 in equation 1.7, we get ∆H= q– p ∆V+ p ∆V= q (at constant pressure) We denote q at constant pressure by q P hence ∆H= qp (1.8) Equation 11.8 shows that by measuring heat lost or gained at constant pressure, we can measure enthalpy change for any process. 2.7.5 Relation between ∆ H and ∆ U For liquids and solids, the difference between ∆rH and ∆rU is not significant but for gasses, the difference is significant as we will see here. Let VA be the total volume of the gaseous reactants, VB the total volume of the gaseous products, nA the number of moles of gaseous reactants, 40 and nB the number of moles of gaseous products, at constant pressure and temperature. Then, using ideal gas law, we can write p VA= nART (1.9) p VB= nBRT (1.10) Subtracting equation 1.9 from equation 1.10, we get p VB–pVA= nBRT –nART = (nB–nA) RT p (VB–VA) = p ∆V = ∆ngRT At constant pressure ∆H = ∆U + p ∆V Therefore ∆H= ∆ U+ ∆ngRT Here ∆ ng = (number of moles of gaseous products) - (number of moles of gaseous reactants) Thus we can find the value of ∆H from ∆U or vice versa. For solids and liquids ∆Vis very small. We can neglect the term p ∆V, hence ∆H is nearly the same as ∆U. Sample Questions 1. Which of the following is false? (a) The reaction H2 (g) + Cl2 (g) →2HCl (g) + 185 kJ is endothermic. (b) Enthalpy change is a state function. (c) Standard state condition for a gaseous system is 1 bar pressure at a specified temperature. 2. For the reaction at 298 K, 1 N2 (g) +3/2H2(g) →NH3 (g); ∆ H= - 46 kJ 2 (a) What is the value of ∆ng?................................................................................................................................ (b) Calculate the value of ∆U at 298 K?................................................................................................................................ 3. Which of the following will increase the internal energy of a system? (a) Heat given to the system (b) Work done by the system................................................................................................................................ 41 2.8 Standard Enthalpy of Reactions Let as denote total enthalpy of reactants as H reactants and total enthalpy of products as H products. The difference between these enthalpies, ∆H, is the enthalpy of the reaction ∆rH = Hproducts – Hreactants When Hproducts is greater than H reactants then ∆H is positive and heat is absorbed in the reaction, and the reaction will be endothermic. For example, H2 (g) + I2(g) →2HI (g); ∆rH= 52.5 kJ When Hproducts is less than Hreactants then ∆H is negative and heat is evolved in the reaction, and the reaction will be exothermic. For example, CH4 (g) + 2O2 (g) →CO2 (g) + 2H2O (l); ∆r H= – 890.4 kJ Enthalpy of a reaction changes with pressure and temperature. It is convenient to report enthalpies of the reactions in terms of standard state of substances as we have defined earlier. When substances are in their standard states, we call the enthalpy of reaction as standard enthalpy of reaction. It is defined as the enthalpy change for a reaction, when the reactants and the products are in their standard states. It is donoted by ∆rH0 2.8.1 Enthalpy of formation (∆f H°) The enthalpy change when one mole of a pure compound is formed from its elements in their most stable states is called the enthalpy of formation and is denoted by ∆f H0. When the reacting elements and the products formed are all in their standard states, the enthalpy change accompanying the chemical reaction is called the standard enthalpy of formation and is denoted by ∆f H 0. By convention, we take the standard enthalpy of formation of an element in its most stable state as zero. For example: C (Graphite) + O2(g) → CO2 (g) ; ∆f H0 = – 393.5 kJ mol–1 This means that carbon dioxide is formed from its elements in their most stable states. Carbon in the form of graphite and at room temperature and gaseous O2 and CO2 being at 1 bar. 2.8.2 Enthalpy of Combustion (∆comb H°) Enthalpy of combustion is the enthalpy change (heat evolved) accompanying the complete combustion of 1 mole of a substance in oxygen at a given temperature and 1 bar pressure. For example: C2H5OH (l) + 3O2 (g) →2CO2 (g) + 3H2O (l); ∆comb H0 = – 1365.6 kJ Enthalpy of combustion of C2H5OH (l) is – 1365.6 kJ mol–1 42 2.8.3 Enthalpy of Neutralization (∆neut Hº) Enthalpy of neutralization is the enthalpy change (heat evolved) when one mole of hydrogen ions (H+) is neutralized by one mole of hydroxyl ions (OH-) in dilute aqueous medium to form water. For example: H+ (aq) + OH–(aq) →H2O (l); ∆neut H = –57 kJ/mol Enthalpy of neutralization of a strong acid with a strong base is always constant having a value of –57 kJ. However, enthalpy of neutralization of strong acid with a weak base or weak acid with a strong base will be different, because of varying degree of ionization of weak acids and bases. 43 CHAPTER THREE SOLUTIONS Learning Outcomes Having completed this chapter, students will be able to: Describe solutions and ways of expressing concentration. Know Henry´s Law and its application. Describe the kinds of solutions. 3.0 Definition of Solution A solution is a homogeneous mixture of two or more substance on molecular level. The constituents of the mixture present in a smaller amount is called the Solute and the one present in a larger amount is called the Solvent. For example, when a smaller amount of sugar (solute) is mixed with water (solvent), a homogeneous solution in water is obtained. In this solution, sugar molecules are uniformly dispersed in molecules of water. Similarly, a solution of salt (Na+, Cl–) in water consists of ions of salt (Na+, Cl–) dispersed in water. 3.1Concentration of a Solution The concentration of a solution is defined as the amount of solute present in a given amount of solution. Concentration is generally expressed as the quantity of solute in a unit volume of solution. 𝑄𝑢𝑎𝑛𝑡𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Concentration = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 A solution containing a relatively low concentration of solute is called Dilute solution. A solution of high concentration is called Concentrated solution. Fig. 3.1 Molecular model of a solution. 44 3.2 Types of Solutions The common solutions that we come across are those where the solute is a solid and the solvent is a liquid. In fact, substance in any three states of matter (solid, liquid, gas) can act as solute or solvent. Thus there are seven types of solutions whose examples are listed in Table 3.1. Table 3.1 Types and examples of solutions Of the seven types of solutions mentioned in Table 3.1, we will discuss only the important ones in detail. 3.3 Ways of expressing concentration There are several ways of expressing concentration of a solution: (a) Per cent by weight (b) Mole fraction (c) Molarity (d) Molality (e) Normality a. Per cent by Weight It is the weight of the solute as a per cent of the total weight of the solution. That is, 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 % by weight of solute = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 × 100 45 For example, if a solution of HCl contains 36 per cent HCl by weight, it has 36 g of HCl in 100 g of solution. SOLVED PROBLEM What is the per cent by weight of NaCl if 1.75 g of NaCl is dissolved in 5.85 g of water. SOLUTION Wt. of solute (NaCl) = 1.75 g Wt. of solvent (H2O) = 5.85 g ∴ Wt. of solution = 1.75 + 5.85 = 7.60 g 1.75 Hence concentration of NaCl % by weight = 7.60 × 100 = 23.0 b. Mole fraction A simple solution is made of two substances: one is the solute and the other solvent. Mole fraction, X, of solute is defined as the ratio of the number of moles of solute and the total number of moles of solute and solvent. Thus, 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Xsolute = 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒+ 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 If n represents moles of solute and N number of moles of solvent, 𝑛 Xsolute = 𝑛+ 𝑁 Notice that mole fraction of solvent would be 𝑁 Xsolvent = 𝑛+ 𝑁 46 Mole fraction is unitless and Xsolute + Xsolvent = 1 SOLVED PROBLEM Calculate the mole fraction of HCl in a solution of hydrochloric acid in water, containing 36 per cent HCl by weight. SOLUTION The solution contains 36 g of HCl and 64 g of H2O 1𝑚𝑜𝑙 𝐻𝐶𝑙 Number of Moles of HCl = (36 g HCl)(36.5𝑔 𝐻𝐶𝑙 ) = 0.99 1𝑚𝑜𝑙 𝐻2𝑂 Number of Moles of H2O = (64 g H2O)( 18𝑔 ) 𝐻2𝑂 = 3.55 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝐶𝑙 XHCl = 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝐶𝑙+ 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻20 47 c. Molarity In current practice, concentration is most often expressed as molarity. Molarity (symbol M) is defined as the number of moles of solute per litre of solution. If n is the number of moles of solute and V litres the volume of solution, 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Molarity = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒𝑠 𝑛 Or M= 𝑉 (𝑖𝑛 𝑙𝑖𝑡𝑟𝑒𝑠) ………………….(1) For one mole of solute dissolved in one litre of solution M = 1 i.e., molarity is one. Such a solution is called 1M (read “1 molar”). A solution containing two moles of solute in one litre is 2M; and so on. As evident from expression (1), unit of molarity is mol litre-1. Fig. 3.2 Molarity 3.3.1 Calculation of Molarity Molarity of a solution can be calculated with the help of the expression (1) if moles of solute (n) and volume V (in litres) are known. When the amount of solute is given in grams and its molecular weight is MW, it can be converted to moles: 1𝑚𝑜𝑙 n = x g 𝑀𝑊 𝑔 𝑚𝑜𝑙 × Substituting in expression (1) 1𝑚𝑜𝑙 1 M = x g × 𝑀𝑊 𝑔 × ……………...(2) 𝑉 From expression (1) can also be found the amount of solute in grams if molarity is given. SOLVED PROBLEM 1 48 What is the molarity of a solution prepared by dissolving 75.5 g of pure KOH in 540 ml of solution. SOLUTION Molecular mass of KOH = 39.1 + 16.0 + 1.0 = 56.1 Calculation of moles of KOH: 1𝑚𝑜𝑙 75.5g KOH × 56.1 𝑔 = 1.35 mol Calculation of volume in litres: 1𝑙𝑖𝑡𝑟𝑒 540ml × 1000𝑚𝑙 = 0.540litres Calculation of Molarity: 𝑛 M=𝑉 1.35𝑚𝑜𝑙 = 0.54 𝑙𝑖𝑡𝑟𝑒 = 2.50M Thus the solution is 2.50 M. SOLVED PROBLEM 2 What weight of HCl is present in 155 ml of a 0.540 M solution? SOLUTION M = 0.540 155 V in litres = 1000 = 0.155litre n=M×V = 0. 540 mol/litre × 0.155 litre = 0.0837 mol of HCl Thus 3.06 g of HCl is present in 155 ml of the given solution. d. Molality Molality of a solution (symbol m) is defined as the number of moles of solute per kilogram of solvent: 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Molality (m) = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚 49 A solution obtained by dissolving one mole of the solute in 1000 g of solvent is called one molal or 1m solution. Notice the difference between molality and molarity. Molality is defined in terms of mass of solvent while molarity is defined in terms of volume of solution. SOLVED PROBLEM What is the molality of a solution prepared by dissolving 5.0 g of toluene (C7H8) in 225 g of benzene (C6H6)? SOLUTION Calculation of number of moles of solute: Molecular mass of toluene = 12 × 7 + 1 × 8 = 92 5 No. of moles 5 g of toluene = = 0.0543 9 225𝑔 Mass of solvent in kg = 1000 = 0.225kg 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 0.0543 Molality = = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑣𝑒𝑛𝑡 𝑖𝑛 𝑘𝑔 0.225 = 0.24 m e. Normality Normality of a solution (symbol N) is defined as number of equivalents of solute per litre of the solution: 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Normality (N) = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒𝑠 Thus, if 40 g of NaOH (eq. wt. = 40) be dissolved in one litre of solution, normality of the solution is one and the solution is called 1N (one-normal). A solution containing 4.0 g of NaOH is 1/10 N or 0.1 N or decinormal SOLVED PROBLEM 5g of NaCl is dissolved in 1000 g of water. If the density of the resulting solution is 0.997 g per ml, calculate the molality, molarity, normality and mole fraction of the solute, assuming volume of the solution is equal to that of solvent. SOLUTION 50 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 Number of moles of NaCl = 𝑀𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 5 = 58.5 = 0.0854 By definition, 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 ×1000 Molality = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 0.0854 = × 1000 1000 = 0.0854m 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑔𝑟𝑎𝑚𝑠 Volume of the solution = 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑔𝑚 𝑝𝑒𝑟 𝑚𝑙 1000+5 = = 1008ml 0.997 = 1.008 litres 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 Now Molarity = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑟𝑒𝑠 0.0854 = = 0.0847 M 1.008 𝑁𝑜.𝑜𝑓 𝑔𝑟𝑎𝑚𝑠 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 And Normality = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑟𝑒𝑠 0.0854 = 1.008 [𝑀𝑜𝑙. 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 = 𝐸𝑞. 𝑤𝑡. 𝑜𝑓 𝑁𝑎𝐶𝑙 ] = 0.0847 N 51 To Calculate Mole Fraction of the Solute 1000 No. of moles of water in 1000 ml = = 55.5 18 [1 𝑚𝑙 𝑜𝑓 𝑊𝑎𝑡𝑒𝑟 = 1 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 ] Total number of moles = No. of moles of solute + No. of moles of solvent = 0.0854 + 55.5 = 55.5854 𝑁𝑜.𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝐶𝑙 Mole fraction of NaCl = 𝑇𝑜𝑡𝑎𝑙 𝑚𝑜𝑙𝑒𝑠 0.0854 = 55.584 = 1.536 × 10–3 3.4 Solutions of gases in gases When a gas is mixed with another gas a completely homogeneous solution results provided of course they do not react chemically. It is obvious that such gaseous solutions will have the following characteristic properties: (1) Complete miscibility. According to the Kinetic Theory, a gas consists of tiny molecules moving about in vacant space and thus when one gas is dissolved in another gas they form a homogeneous solution quite readily. In such a gaseous mixture, the components can be present to an unlimited extent. (2) Dalton’s law of Partial pressures. Since in a gaseous mixture the constituent molecules exist separately, it is obvious that the properties of the mixture will be the sum of properties of the components. Thus Dalton (1842) was the first to show that the total pressure exerted by a gaseous mixture is the sum of the individual or partial pressures of the component gases. If p1, p2, p3... be the partial pressures of the constituents, the total pressure P of the mixture is given the expression P = p1 + p2 + p3 +... 52 (3) Like other gas laws, Dalton’s law holds strictly only when the partial pressures are not too high. This law can be experimentally tested by comparing the total pressure of a gaseous mixture with the sum of the individual pressure of each gas before mixing. SOLVED PROBLEM At constant temperature, 250 ml of nitrogen under 720 mm pressure and 380ml oxygen under 650 mm pressure were put into a one-litre flask. What will be the final pressure of the mixture? SOLUTION Since PV = P'V' at constant temperature (Boyle’s Law) PN2 × 1000 = 720 × 250 or PN2 = 180mm Hg PO2 × 1000 = 380 650 or PO2 = 247 mm Hg Total Pressure P = PN2 + PO2 (Dalton’s Law) = 180 + 247 = 427 mm Hg 3.5 Henry’s law The solubility of a gas in a solvent depends on the pressure and the temperature. When a gas is enclosed over its saturated solution, the following equilibrium exists. gas ↔ gas in solution If pressure is increased on the system, the equilibrium will move in the direction which will reduce the pressure (Le Chatelier Principle). The pressure can be reduced by more gas dissolving in solvent. Thus, solubility or concentration of a gas in a given solvent is increased with increase of pressure. A kinetic molecular explanation of the effect of pressure on gas- solution system is illustrated in Fig. 2.3. The relationship between pressure and solubility of a gas in a particular solvent was investigated by William Henry (1803). He gave a generalisation which is known as Henry’s Law. It may be stated as: for a gas in contact with a solvent at constant temperature, concentration of the gas that dissolves in the solvent is directly proportional to the pressure of the gas. 53 Fig. 3.3 (a) Equilibrium exists between the gas molecules and those in solution (b) When pressure is applied, gas volume decreases and concentration in gas phase increases (c) More gas molecules pass into solution to reestablish equilibrium whereby the pressure above the gas is decreased. Mathematically, Henry’s Law may be expressed as C∝P or C=kP where P = pressure of the gas; C = concentration of the gas in solution; and k = proportionality constant known as Henry’s Law Constant. The value of k depends on the of the gas and solvent, and the units pressure and concentration are used. SOLVED PROBLEM 1 The solubility of pure oxygen in water at 20°C and 1.00 atmosphere pressure is 1.38 × 10–3 mole/litre. Calculate the concentration of oxygen at 20°C and partial pressure of 0.21 atmosphere. 54 SOLUTION Calculation of k: C=kP 𝐶 Or k= 𝑃 Substituting values for pure oxygen, 0.00138𝑚𝑜𝑙𝑒 / 𝑙𝑖𝑡𝑟𝑒 0.00138𝑚𝑜𝑙𝑒 / 𝑙𝑖𝑡𝑟𝑒 k= = 1𝑎𝑡𝑚 𝑎𝑡𝑚 Calculation of Concentration of O2 at given pressure C=kP 0.00138𝑚𝑜𝑙𝑒 / 𝑙𝑖𝑡𝑟𝑒 Concentration of O2 = × 0.21atm 𝑎𝑡𝑚 = 2.9 × 10–4 mole/litre SOLVED PROBLEM 2 At 20°C the solubility of nitrogen gas in water is 0.0150 g/litre when the partial pressure of N2 is 580 torr. Find the solubility of N2 in H2O at 20°C when its partial pressure is 800 torr. SOLUTION We know that Henry’s law for a gas at two different concentration and pressures can be written as 𝐶1 𝐶2 =k =k 𝑃1 𝑃2 Since k has the same value in two cases, 𝐶1 𝐶2 = 𝑃2 ………...(1) 𝑃1 In this example, C1 = 0.0150 g/litre C2 = ? P1 = 580 torr P2 = 800 torr Substituting values in equation (1), we have 0.0150𝑔 / 𝑙𝑖𝑡𝑟𝑒 𝐶2 = 800 𝑡𝑜𝑟𝑟 580 𝑡𝑜𝑟𝑟 0800 𝑡𝑜𝑟𝑟 ×.0150𝑔 / 𝑙𝑖𝑡𝑟𝑒 C2 = 580 𝑡𝑜𝑟𝑟 55 = 0.0207 g/litre Thus, solubility of N2 in H2O at 20°C at pressure of 800 torr is 0.0207g/litre. Note. The above solution eliminates the calculation of k as was done in example 1. 3.6 Solutions of Liquids in Liquids The solutions of liquids in liquids may be divided into three classes as follows: We will now proceed to consider the more important properties of these three classes of solutions. 3.6.1 Solubility of Completely Miscible Liquids Liquids like alcohol and ether mix in all proportions and in this respect, they could be compared to gases. The properties of such solutions, however, are not strictly additive, and therefore their study has not proved of much interest. Generally, the volume decreases on mixing but in some cases it increases. Sometimes heat is evolved when they are mixed while in others it is absorbed. The separation of this type of solutions can be effected by fractional distillation. 3.6.2 Solubility of Partially Miscible Liquids A large number of liquids are known which dissolve in one another only to a limited extent e.g., ether and water. Ether dissolves about 1.2% water; and water also dissolves about 6.5% ether. Since their mutual solubilities are limited, they are only partially miscible. When equal volumes of ether and water are shaken together, two layers are formed, one of a saturated solution of ether in water and the other of a saturated solution of water in ether. These two solutions are referred to as conjugate solutions. The effect of temperature on the mutual solubility of these mixtures of conjugate solutions is of special interest. The underlisted systems fall under this category: 56 (1) Phenol-Water system (2) Triethylamine-Water system (3) Nicotine-Water system 3.7 Vapour Pressures of Liquid–Liquid Solutions The study of the vapour pressures of mixtures of completely miscible liquids has proved of great help in the separation of the liquids by fractional distillation. The vapour pressures of two liquids with varying composition have been determined at constant temperature. By plotting the vapour pressure against composition, it has been revealed that, in general, mixtures of the miscible liquids are of three types. Fig. 2.4 Vapour pressure of liquids. 3.7.1 First Type of Mixtures of Miscible Liquids For this type of solutions, the vapour pressure curve exhibits a minimum. If we take a mixture which has an excess of X (more volatile component), we are somewhere at C on the curve. When this is distilled the vapour will contain excess of X and thus the remaining mixture will get richer in Y. Finally, we reach the point D where vapour pressure is minimum and thus boiling point is maximum. Here the mixture will distil unchanged in composition. Exactly similarly if we take a mixture having a greater proportion of Y (point E), on distillation Y will pass over leaving the residue richer in X till in this way the minimum point D is again reached when the mixture will distil unchanged. It is obvious that complete separation of this type of solutions into components is impossible. 57 Fig. 3.5 Vapour pressure of liquids. At best it can be resolved into one pure component and the constant boiling mixture. Solutions of this type which distil unchanged at a constant temperature and show a maximum boiling point are called maximum boiling point azeotropic solutions. The best known example of this type is presented by hydrochloric acid which forms a constant boiling mixture at 110°C and containing 20.24% of the acid. If a mixture of any composition is distilled, either hydrochloric acid or water will pass over, the composition will move to the point of minimum vapour pressure when it distills without any change in composition. 3.7.2 Second Type of Mixtures of Miscible Liquids (Minimum boiling point azeotropic solutions) Here the vapour pressure curve records a maximum at F. At this point the mixture has the highest vapour pressure and, therefore, the lowest boiling point. Thus in this type of solutions the first fraction will consist of a constant boiling mixture with a fixed composition corresponding to the maximum point until whole of one component has been exhausted. After this the temperature will rise and the other component will pass over In this kind of solutions also it is not possible to effect a complete separation by fractional distillation. At best we can resolve it into a constant-boiling mixture and one component in the pure state. Ethanol and water mixtures offer a good example of this type. Ethanol- water mixture containing 95.6 per cent ethanol boils at the minimum temperature 78.13°. Thus it is very difficult to obtain pure absolute alcohol by distillation. This difficulty has, however, been overcome by adding benzene which form a low boiling mixture with water and on distillation it comes over leaving pure ethanol behind. 58 3.7.3 Third Type of Mixtures of Miscible Liquids In this case the vapour pressures of mixtures always lie between the vapour pressures of pure components and thus the vapour-pressure composition curve is a straight line. Suppose we have a mixture containing excess of Y which is represented by point G on the curve. On distillation X component being more volatile will be obtained in greater proportion in the distillate and we gradually travel along the curve AB. The latter fractions will, of course, be poorer in X and richer in Y till we reach the 100 per cent Y- axis, when all the X will have passed over. By repeating the process of distillation with the fresh distillate, which is now richer in X, we can get almost pure component X. Only in this type of solutions we can completely separate the components by fractional distillation. Thus, methyl alcohol-water mixtures can be resolved into pure components by distillation. Liquid mixtures which distil with a change in composition are called zeotropic mixtures. 3.7.4 Azeotropes are Mixtures and not Pure Compounds Although the azeotropes boil at a constant temperature and distil over without change in composition at any given pressure like a pure chemical compound, these cannot be regarded as chemical compounds. The reason is that when the total pressure is changed, both the boiling point and the composition of the azeotrope change whereas for a chemical compound the composition must remain constant over a range of temperature and pressures. 3.8 Vapour Pressure of Mixtures of Non–miscible Liquids In a mixture of non-miscible liquids each component exerts its own vapour pressure independent of others and the total vapour pressure is thus equal to the sum of individual vapour pressures of all the liquids. Consider two liquids in separate compartments. Each one of them would have a certain vapour pressure, say, p1 and p2. When present in a vessel with a common top, they exert their own vapour pressure independently. According to Dalton’s law of partial pressures, the total vapour pressure will be equal to the sum of their individual pressures. 59 Fig. 3.6 Vapour pressure of mixtures of non-miscible liquids (illustration). This generalization, which is the basic principle of steam distillation, has been tested experimentally in several cases. Some of the results obtained by Regnault are given below: The observed vapour pressure of the mixture is a little less than the sum of the individual vapour pressures of water and carbon disulphide and that is to be expected since each liquid is slightly soluble in the other. 3.9 Solutions of Solids in Liquids Solutions of this type are most commonly met with. The process of solution of a solid substance in a solvent is explained by the electrical forces operating between the molecules or ions of the solute and the molecules of the solvent. It is a common observation that polar solutes dissolve easily in polar solvents while they remain insoluble in non-polar solvents. For example, sodium chloride (an electrolyte) is fairly soluble in water which is highly polar solvent, while it is insoluble in a non-polar solvent like chloroform. On the other hand, a non- polar solute does not dissolve in a polar solvent e.g., benzene which is non-polar is insoluble in water. The electrical attraction between the oppositely charged ends of the solute and the solvent molecules results to form a solution. 60 Water being highly polar is one of the best solvents for ionised solutes. An ionic substance, when placed in water furnishes cations (+) and anions (–). These ions are surrounded by solvent molecules with their oppositely charged ends directed towards the ion. The ion enveloped by a layer of the solvent molecules in this manner, is called a Solvated ion or Hydrated
