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CHAPTER 3 CHEMICAL BONDS. The world around us is composed almost entirely of compounds and mixture of compounds. Most of the pure elements also contain.

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Presentation on theme: "CHAPTER 3 CHEMICAL BONDS. The world around us is composed almost entirely of compounds and mixture of compounds. Most of the pure elements also contain."— Presentation transcript:

1 CHAPTER 3 CHEMICAL BONDS

2 The world around us is composed almost entirely of compounds and mixture of compounds. Most of the pure elements also contain many atoms bound together, for example, diamond is a native form of carbon, in which a large number of carbon atoms are bound. In compounds, atoms are held together by forces known as chemical bonds. Electrons play a key role in chemical bonding.

3 There are three ideal types of chemical bonds: - ionic bond (between metals and nonmetals); - covalent bond (between nonmetals); - metallic bond (between metallic atoms).

4 The ionic bond is a type of chemical bond based on the electrostatic attraction forces between ions having opposite charges. It can form between electropositive and electronegative elements, e.g. between metal and non-metal ions. The metal, with a few electrons on the last shell, donates one or more electrons to get a stable electron configuration and forms positively charged ions (cations). These electrons are accepted by the non-metal to form a negatively charged ion (anion) also with a stable electron configuration. The electrostatic attraction between the anions and cations causes them to come together and form a bond.

5 Example: the formation of ionic bond between Na and Cl. For the sodium atom the electron configuration is: 1s 2 2s 2 2p 6 3s 1 The first and second shells of electrons are full, but the third shell contains only one electron. When this atom reacts, it gains the configuration of the nearest rare gas in the periodic table: Ne 1s 2 2s 2 2p 6 Na atom loses one electron from its outer shell: Na → Na+ + e -

6 The chlorine atom has the configuration 1s 2 2s 2 2p 6 3s 2 3p 5 It gains one electron and realizes the stable electron configuration of Ar:1s 2 2s 2 2p 6 3s 2 3p 6 Cl + e -  Cl - When sodium and chlorine react, the outer electron of the sodium atoms are transferred to the chlorine atoms to produce sodium ions Na + and chlorine ions Cl -, which are held together by the electrostatic force of their opposite charges. NaCl is an ionic compound.

7 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 3s 2 3p 5 1s 2 2s 2 2p 6 1s 2 2s 2 2p 6 3s 2 3p 6 NaCl formation

8 NaCl formation may be illustrated showing the outer electrons only (Lewis symbol): In a similar way, a calcium atom may lose two electrons to two chlorine atoms forming a calcium ion Ca 2+ and two chloride ions Cl -, that is calcium chloride CaCl 2 :

9 In sodium chloride, the ionic bonds are not only between a pair of sodium ion Na + an chlorine ion Cl -, but also between all the ions. These electrostatic interactions have as a result the formation of NaCl crystal. We write the formula of sodium chloride as NaCl, but this is the empirical formula. The sodium chloride crystal contains huge and equal numbers of Na + and Cl - ions pocket together in a way that maximizes the electrostatic forces of the oppositely charged ions. Figures 3.2 and 3.3 show the crystal lattice of NaCl and LiBr.

10 Sodium chloride crystal

11 Lithium bromide crystal

12 Covalent bonds The covalent bond is a type of chemical bond formed by sharing pairs of electrons between atoms. When two electronegative atoms react together, ionic bonds are not formed because both atoms have a tendency to gain electrons. In such cases, an stable electronic configuration may be obtained only by sharing electrons. First, consider how chlorine atoms Cl react to form chlorine molecules Cl 2 :

13 Each chlorine atom shares one of its electrons with the other atom. The electron is shared equally between both atoms, and each atom in the molecule has in its outer shell 8 electrons – a stable electronic configuration corresponding to that of Ar. In a similar way a molecule of carbon tetrachloride CCl 4 is made up of carbon and four chloride atoms:

14 The carbon atom shares all its four electrons and the chlorine atoms share one electron each. The carbon atom forms 4 covalent bonds with 4 chlorine atoms. In this way, both the carbon and all four chlorine atoms attain a stable electronic structure. The sharing of a single pair of electrons results in a single covalent bond, often represented by a dash sign, so chlorine molecule may be written as follow:Cl — Cl carbon tetrachloride

15 For oxygen molecule O 2, there are two pairs of electrons shared between the O atoms (double covalent bond):O ═ O In nitrogen molecule (N 2 ) each nitrogen atom shares three electrons. The sharing of three pairs of electrons between two atoms leads to a triple covalent bond N ≡ N Coordinate bond A molecule of ammonia NH 3 is made up of one nitrogen and three hydrogen atoms:

16 The nitrogen atom forms three bonds and the hydrogen atoms one bond each. In this case, one pair of electrons is not involved in bond formation and this is called a lone pair of electrons. It is possible to have a shared electron pair in which the pair of electrons comes just from one electron and not from both. Such bond is called coordinate covalent bond. Even though the ammonia molecule has a stable configuration, it can react with hydrogen H + by donating the lone pair of electrons, forming the ammonium ion NH 4 + :

17 Partial ionic character of covalent bonds In the chlorine molecule Cl – Cl the pair of electrons of the covalent bond is shared equally between both chlorine atom. Because there is not a charge separation toward one of the Cl atoms, Cl 2 molecule is nonpolar.

18 On the contrary, in HCl molecule, there is a shift of electrons toward the chlorine atom which is more electronegative than hydrogen one. Such molecules, in which a charge separation exists is called polar molecule or dipole molecule The polar molecule of hydrochloric acid

19 The magnitude of the effect described above is denoted through the dipole moment μ. The dipole moment is the product of the magnitude of the charges (δ) and the distance separating them (d): μ = δ · d The symbol δ suggests small magnitude of charge, less than the charge of an electron ( · C ). The magnitude of 3.34 · Cm means Debye ( D ): 1D = 3.34 · C m

20 The hydrochloric acid molecule has a dipole moment μ=1.03 D and the distance between H and Cl atoms is 136 pm ( 136 · m ). A charge δ will be: The charge δ is about 16% of the electron charge (1.602 · C ). We can say therefore that the covalent H – Cl bond has about 16% ionic character.

21 Metallic bond Metals tend to have high melting and boiling points suggesting strong bonds between the atoms. Sodium has the electronic structure 1s 2 2s 2 2p 6 3s 1. When sodium atoms come together, the electron in the 3s atomic orbital of one sodium atom shares space with the corresponding electron of a neighbouring atom to form a molecular orbital in the same way that a covalent bond is formed. The difference, however, is that each sodium atom is touched by eight other sodium atoms, and the sharing occurs between the each atom and 3s orbitals of all the eight other atoms. And each oh these eight is in turn touched by eight sodium atoms and so on.

22 All of the 3s orbitals of all the atoms overlaps to give a vast number of molecular orbitals which extend over the whole piece of metal. There have to be huge numbers of molecular orbitals because any orbital can only hold two electrons. The electron can move freely within these molecular orbitals and so each electron becomes detached from its parent atom. The electrons are said to be delocalised. The metal is held together by the strong forces of attraction between the positive nuclei and the delocalised electron. This may be described as “ an array of positive ions in a sea of electrons “.

23 Metallic bond

24 The “ free “ electrons of the metal are responsible for the characteristic metallic properties: ability to conduct electricity and heat, malleability (ability to be flattened into sheets), ductility (ability to be drawn into wires) and lustrous appearance. Intermolecular bonds Van der Waals forces Intermolecular forces are attractions between one molecule and neighboring molecules. All molecules are under the influence of intermolecular attractions, although in some cases those attractions are very weak. These intermolecular interactions are known as van der Waals forces. Even in a gas like hydrogen (H 2 ), if you slow the molecules down by cooling the gas, the attractions are large enough for the molecules to stick together in order to form a liquid and then a solid.

25 In hydrogen’ s case the attractions are so weak that the molecules have to be cooled to 21 K (-252  C) before the attractions, are enough to consider the hydrogen as a liquid. Helium’ s intermolecular attractions are even weaker – the molecules won’ t stick together to form a liquid until the temperature drops to 4 K ( -269  C). Attractions are electrical in nature. In a symmetrical molecule like hydrogen, however, these doesn’t seem to be any electrical distortion to produce positive or negative parts. But that’ s only true in average. In the next figure the symmetrical molecule of is represented.

26 H 2 symmetrical molecule

27 The even shading shows that on average there is no electrical distraction. But the electrons are mobile and at any one instant they might find them selves towards one out if the molecule. This end of the molecule becomes slightly negative (charge -  ). The other end will be temporarily short of electrons and so becomes slightly positive (+  ) as we can see in the next figure. An instant later the electrons may well have moved up to the other end, reversing the polarity of the temporary dipole of molecule.

28 Temporary dipole of H 2 This phenomena even happens in monoatomic molecules of rare gases, like helium, which consists of a simple atom. If both the helium electrons happen to be on one side of the atom at the same time, the nucleus is no longer properly covered by electrons for that instant.

29 Temporary dipole of He

30 The question is how temporary dipoles give intermolecular bonds? Imagine a molecule which has a temporary polarity being approached by one which happens to be non- polar just at that moment. Induced dipole

31 As the right hand molecule approaches, its electrons will tend to be attracted by the slightly positive end of the left hand one. This sets up on induced dipole in the approaching molecule, which is orientated in such a way that the  + end of one is attached to the  - end of the other. Dipole-dipole attraction

32 An instant later the electrons in the left hand molecule may well have up the other end. In doing so, they will repel the electrons in the right hand one. The polarity of both molecules reverses, but there is still attraction between  - end and  + end. As long as the molecules stay close to each other, the polarities will continue to fluctuate in synchronization so that the attraction is always maintained. This phenomena can occur over huge numbers of molecules. The following diagram shows how a whole lattice of molecules could be held together in a solid.

33 Molecular distribution in a solid The interactions between temporary dipoles and induced dipoles are known as van der Waals dispersion forces.

34 Now, let us consider a molecule like HCl. Such a molecule has a permanent dipole because chlorine is more electronegative than hydrogen. These permanent dipoles will cause the HCl molecules to attract each other rather than if they had to rely only on dispersion forces. It’s important to realize that all molecules experience dispersion forces. Dipole-dipole interactions are not an alternative to dispersion forces. They occur in addition to them. Molecules which have permanent dipoles will have boiling points higher than molecules which have only temporary fluctuating dipoles. Surprisingly, dipole-dipole attractions are fairly minor compared with dispersion forces, and their effect can be seen if we compare two molecules with the same number of electrons and the same size.

35 For example, the boiling points of ethane (CH 3 -CH 3 ) and fluoromethane (CH 3 F) are: K (-88.7  C), respectively K (-78.5  C). The molecule of ethane is symmetric while that of fluoromethane has permanent dipole.

36 Hydrogen bond If we plot the boiling points the hydride of the elements of groups 15, 16 end 17 we find that the boiling point of the first elements in each group is abnormally high.

37 In the cases of NH 3, H 2 O and HF there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break. These relatively powerful intermolecular forces are described as hydrogen bonds.

38 We can observe that in each of these molecules:  the hydrogen is attached directly to one of the most electronegative elements, causing the hydrogen to acquire a significant amount of positive charge;  each of the elements to which the hydrogen atom is attached is not only significant, but also has one “active“ lone pair of elements. Lone pairs of the 2nd level have the elements contained in a relatively small volume of space which therefore has a high density of negative charge. Lone pair at higher levels are more diffuse and not so attractive to positive particles.

39 Let’s consider two water molecules coming close together:

40 The slightly  + charge of hydrogen is strongly attracted to the lone pair end as a result a coordinate bond is formed. This is a hydrogen bond. Hydrogen bond is significantly stronger than a dipole- dipole interaction, but has about a tenth of the strength of an average covalent bond. In liquid water, hydrogen bonds are constantly broken and reformed. In solid water each water molecule can form hydrogen bond surrounding water molecules as we can see in the next figure.

41

42 This is why the boiling pint of water is higher than of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen atom has only one lone pair. As well, in hydrogen fluoride, the number of hydrogen atoms is not enough to form a three- dimensional structure.

43 CHAPTER 4 GAS LAWS

44 In a gas the molecules are in a permanent and chaotic motion. Each particle travels in random directions at high speed until it reaches another one, when it is deflected, or until it collides with the wall of the vessel. This movement is called Brownian motion and the gas phase is a completely disordered state. The thermodynamic state of a gas is characterized by its pressure, its volume, and its temperature. The relationship between the pressure, volume, temperature and amount of gas are called gas laws.

45 Pressure is measured as force per unit area. The SI unit for pressure is Pa (Pascal). However, several other units are commonly used. The table below shows the conversion between these units: Units of Pressure 1 Pa1 N·m -2 =1 kg·m -1 ·s -2 1 atm ·10 5 Pa 1 atm760 torr (mmHg) 1 bar10 5 Pa

46 Volume is related between all gases by Avogadro’s hypothesis, which states: Equal volumes of gases, at the same temperature and pressure contain equal numbers of molecules. From this, one can derive the molar volume of a gas, that is the volume occupied by one mole of gas under certain conditions. This values, at 1atm and 0°C is: V M = L·mole -1 Temperature is a measure of how much energy the particles have in a gas.

47 1. Boyle’s law This law was discovered by Robert Boyle (1662) and describes the relationship between the gas pressure and volume. The volume occupied by a given amount of gas is inversely proportional to the pressure at constant temperature: where: p – is the pressure (Pa); V – is the volume (m 3 ); k – is a constant.

48 where p 1 and V 1 are the pressure and the volume in another state, at the same temperature. If we represent this relationship we obtain a set of curves with a shape called equilateral hyperbola, corresponding to a particular temperature. Boyle’s law may be written as the relationship:

49

50 The explanation of Boyle’s law is based on the fact that the pressure exerted by a gas arises from the impact of its molecules to the walls of the vessel. If the volume is halved, the density of molecules is doubled. In a given interval of time twice as many molecules strike the walls and so, the pressure is doubled in accord with Boyle’s law. This law is universal in the sense that it applies to all gases without reference to their chemical composition.

51 2. Charles’s law The volume of a given amount of gas, at constant pressure, increases proportionally to the temperature: where: V – is the gas volume (m 3 ); T – is the temperature (K); k – is a constant. For two different states, at the same gas pressure, the relationship becomes: k

52 If we represent the relationship, we obtain a set of straight lines for each pressure considered. The point of intersection between the straight lines and the temperature axis is the same: °C. It is the temperature at which the volume of a gas would become zero, called absolute zero temperature (0 K).

53 3. Gay Lussac’s law The pressure of a given amount of gas at constant volume increases proportionally to the temperature: where: p – is the pressure at temperature T; T – is the temperature (K); k – is a constant. For two different states, at the same gas volume, the relationship becomes: k

54 With the addition of Avogadro’s law we obtain the ideal gas law These three laws were combined to give the combined gas law: where: n is the amount of substance expressed in mole; R – universal gas constant (8.314 J·mole -1 ·K -1 ). For a mole of gas, the relationship becomes: V M is the molar volume.

55 4. Dalton’s law of partial pressures Studies of gaseous mixtures showed that each component behaves independently of the others. The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each component: The partial pressure of a gas is the pressure that the gas would exert if it were alone in the container.

56 CHAPTER 5 SOLUTIONS

57 Solutions are homogeneous mixtures formed by two or more components. For one solution, we distinguish the component that dissolves, called solvent, and the compound that is dissolved, called solute. The notion of solution is not limited to a certain state of aggregation of the substances. There can be liquid, solid or gaseous solutions.

58  a gas dissolved in a liquid – carbon dioxide in water;  liquid dissolved in liquid – ethanol in water;  solid dissolved in liquid – sodium chloride in water, naphthalene in benzene. Solid solutions: the most important are metal alloys, but in this category are included only the alloys which are homogenous mixtures. Liquid solutions are: Gaseous solutions: are gas mixtures, like air. Gases, regardless of their chemical nature, are miscible in any proportion.

59 The most common dissolving agent is water; it can dissolve many solid, liquid or gaseous substances. Other usual dissolving agents are: ethanol, ethyl ether, toluene, chloride derivatives and others. Substances are dissolved in solvents differently. For example, fats are not dissolved in water, but are well dissolved in petrol; iodine is barely dissolved in water, but is well dissolved in alcohol. Solvents

60 Dissolving is a consequence of molecular movement. When a solid substance is introduced in water, its particles (molecules or ions) interact with the water molecules, are separated from the solid and diffuse inside the solution. The higher the number of particles separated in the time unit, the faster the dissolving process. The finely divided substances, having a higher surface area in contact with the solvent, are dissolved faster than massive substances. Also, agitation and temperature intensifies the dissolving process.

61 The thermal effect of dissolving The dissolving of the substances is accompanied by a thermal effect: either heat absorption or heat release. For example, dissolving one mole of potassium nitrate in a large quantity of water requires 36 kJ absorbed from the environment. The dissolving process of an ionic substance, like potassium nitrate, consists of two successive processes:  the separation of K + and NO 3 - ions from the crystal lattice, process that requires energy from the exterior,  the solvation of the ions (hydrating when the solvent is water), that takes place with heat release.

62 The dissolving process of potassium nitrate in water: Solvation (hydrating) represents the process of attaching solvent molecules to the separated ions from the crystal lattice.

63 Because the energy absorbed for the extraction of the ions from the crystal lattice is higher than the energy released during the solvation of the ions, the dissolution of potassium nitrate is an endothermic process, meaning that dissolving potassium nitrate in water the solutions will cool down. In case of other ionic substances, like copper sulfate, the dissolution is an exothermic process.

64 Concentration of the solutions Concentration expresses the quantitative relation between the components of the solution. There are several ways of expressing the concentration of solutions: 1. Mass percentage: represents the mass of substance (g) dissolved in 100 g of solution. The relation of mass percent is: [%] Where: m d is the mass of the dissolved substance; m s – the mass of the solution.

65 2. Volume percentage: represents the volume of substance (m 3 ) dissolved in 100m 3 of solution: [%] Where:V d is the volume of the dissolved substance; V s – the volume of the solution. This way of expressing the concentration of solutions is used especially in the case of liquids dissolved in other liquids. 80% (v) ethanol contains 80 volumes of pure ethanol and 20 volumes of water. 80° alcohol means 80% (v).

66 3. Molarity: represents the number of moles of substance dissolved in 1L of solution: [mol  L -1 ] Where: Md is the molecular mass of the dissolved substance; Vs – the volume of the solution (L). 4. Molality: represents the number of moles of substance dissolved in 1kg of solvent: [mol  kg -1 ] Unlike molarity, which depends on temperature, the molality is independent on temperature.

67 5. Molar fraction (mole fraction): is the number of moles of solute divided by the total number of moles of a solution. For a solution that contains n A moles of compound A and n B moles of compound B the mole fraction of compound A in the solution is: Similarly, the mole fraction of compound B is:

68 From these relations results that the sum of the mole fractions of compounds A and B is 1. Similar relations result in the case of solutions that have several compounds. 6. Titer: represents the mass of dissolved substance (expressed in g) that is found in 1mL of solution: This way of expressing concentration is commonly used in analytical chemistry. [g mL -1 ] or [g cm -3 ]

69 Transformation relations between different ways of expressing concentration: Way of expressing concentration Percentage c% Molarity c M Molality c m Percentage c% Molarity c M Molality c m

70 Introducing sodium chloride in a certain amount of water, in small portions and under stirring, it can be seen that at a certain moment, the quantities of NaCl that are added don’t dissolve anymore, they remain in solid state. The solution that at a certain temperature contains the maximum proportion of dissolved substance is called saturated solution. For example, at 20°C, 35.8g NaCl is the maximum quantity of NaCl that can be dissolved in 100g of water. The maximum concentration of the substance in the saturated solution represents the solubility. Solubility

71 The solubility of sodium chloride is 35.8 g/100 g of water at the considered temperature. Solubility depends on the nature of the substances. Substances that at 20°C have the solubility of more than 1 g solute per 100 g solvent are considered soluble. Substances with solubility under this value are considered slightly soluble. Soluble substances in water are: NaCl, KNO 3, AgNO 3, KBr, NaOH, sodium acetate, sulfuric acid, sugar, etc. Slightly soluble substances in water are: AgBr, PbSO 4, Fe(OH) 3, CaCO 3, BaSO 4.

72 The solubility of substances depends on temperature. The variation of solubility with temperature is represented by the solubility curves. The solubility of salts generally increases with increasing temperature. For some solid substances like Ce(SO 4 ) 3 or Ca(OH) 2 the solubility decreases with increasing temperature. The solubility of liquids increases with increasing temperature. The solubility of gases decreases with increasing temperature. The solubility of gases is also influenced by the pressure of the gas above the solution. The higher the pressure of the gas, the higher the solubility.

73 Solubility curves for some solid substances

74 The vapor pressure of solutions Vapor pressure is the pressure of a vapor in equilibrium with its non-vapor phases. The transition of a liquid substance in gaseous state (evaporation) takes place even before reaching boiling point. At the liquid – air interface the molecules of the substance are stopped from leaving the liquid due to the intermolecular forces which are orientated towards the mass of the liquid.

75 But, if the kinetic energy of the molecule becomes very large, this molecule can “escape” from the solution and passes in the gaseous state. This phenomenon is reversible, and at the interface there is a dynamic equilibrium, when the number of molecules that passes from liquid to air is equal to the number of molecules that passes from air to liquid. This means that at equilibrium, the gaseous state is saturated with the molecules of the liquid substance. The vapor pressure is an indication of a liquid's evaporation rate. A substance with a high vapor pressure at normal temperatures is often referred to as volatile.

76 The temperature at which the vapor pressure of a liquid becomes equal to atmospheric pressure (or in case of closed spaces – the pressure above the liquid) is the boiling temperature. Vapor pressure of mixtures If a non-volatile substance is dissolved in a solvent, the vapor pressure of the solution is smaller than the one of the pure solvent, at the same temperature. The relative drop of vapor pressure is given by: p 0 = the vapor pressure of the pure solvent p = the vapor pressure of the solvent above the solution.

77 Raoult’s law (1877): the relative drop of the vapor pressure of a diluted solution is equal to the molar fraction of the solute in solution: Considering x 1 as the molar fraction of the solute and x 2 the molar fraction of the solvent, the relation can be written: from which results: Considering the fact that x 1 + x 2 =1, we obtain: p = x 2  p o

78 The vapor pressure of a solvent in a solution is directly proportional to its molar fraction. The solutions that respect the Raoult’s law are called ideal solutions. Diluted solutions are approaching the state of ideal solution. If a gaseous substance is dissolved in a liquid solvent, the molecules of gas are dispersed in the mass of the solvent. They can reach the liquid – gas interface, and if their kinetic energy is sufficiently high, they pass in the gaseous state.

79 Equilibrium is reached at a certain concentration of the gas in solution, when the number of the gas molecules that pass from the solution in gaseous state is equal to the number of gas molecules that pass the opposite way. At equilibrium, the solution is saturated in gas. The variation of the solubility of a gas with the pressure is expressed by Henry’s law: the molar fraction of a gas dissolved in a solvent is proportional to the pressure of the gas in equilibrium with the solution: x = k  p

80 Increasing the boiling point of the solutions According to Raoult’s law, by dissolving a non-volatile substance in a solvent, the vapor pressure of the solvent above the solution is smaller than the one above the pure solvent. Thus, the boiling temperature of the solution will be higher than the one of the solvent. The increase of the boiling point of the solution compared to the solvent is proportional to the decrease of the vapor pressure of the solution compared to the solvent:

81 The increase of the boiling point is: T f - the boiling temperature of the solution; T f 0 - the boiling temperature of the solvent The decrease of the vapor pressure is: p 0 is the vapor pressure of the solvent; p – is the vapor pressure of the solution. The variation of the boiling point of the solution depends also on the concentration of the dissolved substance.

82 Vapor pressure and temperature for different concentrations of the solute expressed in molality

83 The increase of the boiling point can be expressed by the relation: where K eb is the ebullioscopic constant, c m – the molal concentration of the solute. The ebullioscopic constant, K eb represents the increase of the boiling point when one mole of substance is dissolved in 1 kg of solvent.

84 For diluted solutions, the ebullioscopic constant does not depend on the nature of the dissolved substance, as it is a characteristic of the solvent. This means that solving the same quantity of substance in a solvent, the increase of the boiling point of the solution will be the same. Ebullioscopic constant for different solvents. SolventH2OH2Ochloroformethanolbenzenediethyl ether K eb

85 Replacing the molarity with its expression, it results: where: m d is the mass of solute (kg); m solv – the mass of solvent (kg); M – the molar mass (kg·mol -1 ). This relation is used for determining the molecular mass of the substances. The research method, based on the experimental determination of the increase of the solutions boiling point, is called ebullioscopy.

86 Decreasing the freezing point of solutions Another consequence of Raoult’s law is the drop of the freezing point of solutions. The decrease of the freezing point is proportional with the molal concentration of the dissolved substance: where:T s is the freezing temperature of the solution; T s 0 – the freezing temperature of the solvent; K cr – the cryoscopic constant.

87 SolventH2OH2OCamphorNaphthaleneBenzeneCyclohexane K cr The research method based on the experimental determination of the decrease of the freezing point of solutions is called cryoscopy. The relation used in cryoscopy for determining the molecular masses of substances is: The cryoscopic constant represents the drop of the freezing point produced by dissolving one mole of substance in 1 kg of solvent.

88 Osmosis and osmotic pressure If we carefully pour water on a copper sulfate solution (blue), we will see at the beginning a clear separation between the blue-coloured copper sulfate solution and the colorless water. Because of the Brownian movement the Cu 2+ and SO 4 2- ions are dislocated from the solution in the water layer and the water in the copper sulfate solution, so that, after a while, a homogenization of the copper sulfate concentration is produced.

89 The effective movement of the chemical species, ionic or molecular, under the influence of the difference of concentration is called diffusion. At equal concentrations the diffusion stops.

90 The diffusion of some chemical species can be prevented using membranes. There are semi-permeable membranes that allow certain molecules or ions to pass through, but prevent the passage of other molecules. The osmosis can be evidenced by the following experience:

91 At the beginning, the liquid from the funnel is at the same level with the liquid in the vessel. In time, the liquid ascends in the gradual tube to a certain level. This happens because water diffuses through the membrane in the sugar solution. The membrane is permeable only for the small water molecules but not for the large sugar molecules. membrane sugar solution water

92 The movement of the solvent through a semi- permeable membrane from the diluted solution into the concentrated solution is called osmosis. The increase of the level stops when the hydrostatic pressure h is sufficiently high to prevent the passage of water. The pressure necessary to stop the diffusion of water is the osmotic pressure. It can be measured by the height of the liquid column. The general osmotic pressure expression was formulated by van’t Hoff:  = cRT

93 where:π is the osmotic pressure (N·m -2 ); c – concentration (mole·m -3 ); R – universal constant of gases; T – thermodynamic temperature (K). The van’t Hoff’s equation is similar to the general equation of ideal gases.

94 CHAPTER 6 CHEMICAL REACTIONS

95 The chemical reaction represents the phenomenon through which one or more substances are transformed in other substances, without affecting the nature of the constituent atoms of the transformed substances. In the environment several reactions can be observed, although most of them have a slow rate. Some examples in this way are rusting of the steel pieces, alcoholic fermentation, green turning of leaves due to the forming of chlorophyll, the ignition of fuels. Chemical reactions can be emphasized through the next manifestations:

96 a) Evolution of gas bubbles If we introduce a piece of zinc in a hydrochloric acid solution, we may observe the hydrogen evolution reaction. Zn + 2HCl  ZnCl 2 + H 2

97 A more violent reaction occurs between sodium and water. The reaction product is also hydrogen. 2Na + 2H 2 O  2NaOH + H 2

98 b) Forming of precipitates By pouring sodium dichromate solution in a lead nitrate solution we observe the appearance of a yellow-coloured precipitate consisting of slightly soluble lead dichromate. Pb(NO 3 ) 2 + Na 2 Cr 2 O 7  2NaNO 3 + PbCr 2 O 7

99 c) Changing of colour Substances absorb light of different wave lengths, so they appear differently coloured. Changing the nature of a substance through a chemical reaction can sometimes lead to color modifications. So, if in a colourless solution of ammonium thiocyanate we pour an iron (III) and ammonia sulphate solution we observe the colouring of the solution in deep red because of the forming of the iron (III) tiocyanate. Sometimes the modifying of colour can be the sign of a physical process, not necessary chemical. 3NH 4 SCN + FeNH 4 (SO 4 ) 2  Fe(SCN) 3 + 2(NH 4 ) 2 SO 4

100 d) Appearance of flame This is another sign that a chemical reaction takes place. An example is the ignition reactions of hydro- carbons. The flame that appears at the Bunsen bulb is the sign of the oxidation reaction of methane with oxygen from air.

101 e) Modification of physical properties of solutions This is another proof of a chemical reaction. Such kind of property is conductivity. If in a vessel with hydrochloric acid solution we add a sodium hydroxide solution, with the help of a conductivity meter one can measure the decreasing of the solution’s conductivity until the complete neutralization of the acid.

102 Thermal effects The chemical reactions take place through the breaking of chemical bonds and the forming of new ones. Therefore, chemical reactions are accompanied by important thermal effects (heat release or absorption). Exothermic reactions = reactions that take place with heat release. Endothermic reactions = reactions that take place with heat absorption.

103 Chemical reactions are represented using chemical equations. Reactants = substances initially involved in a chemical reaction. They are written in the left term of the equation. Reaction products = substances formed in a chemical reaction. They are written in the right term of the equation Because in a chemical reaction, the nature of atoms of the substances is not changed, the chemical equations are equalized so that the number of atoms of a certain element from the left term is equal to the one from the right term.

104 Let’s consider the chemical reaction between hydrogen and chlorine, when hydrochloric acid is formed: H 2 + Cl 2 = 2HCl For the hydrochloric acid we chose the coefficient 2 so that the number of chlorine atoms, as well as the number of hydrogen atoms is not modified. The primary signification of this chemical reaction is that a hydrogen molecule interacts with a chlorine molecule in order to form two molecules of hydrochloric acid. During this transformation, the covalent bonds: H – H and Cl – Cl are broken, and a new bond is formed: H – Cl.

105 The chemical equations have the same properties as mathematical equations. Thus, the equation can be multiplied with Avogadro’s number, and we obtain: N A  H 2 + N A  Cl 2 = 2 N A  HCl The second signification of the chemical equation is: that 1 mole of hydrogen reacts with 1 mole of chlorine to obtain 2 moles of hydrochloric acid.

106 In some situations, in order not to create confusion, chemical formulas of the reactants and the reaction products are followed by the symbol of the aggregation state written between brackets: 2Na (s) + 2H 2 O (l) = 2NaOH (aq) + H 2 (g) The next symbols are used: s – solid, l – liquid, g – gas, aq – aqueous solution.

107 Classification of chemical reactions It is very difficult to choose unique and well defined criteria for the chemical reactions classification. One criterion can be the way the reactants interact in order to form the reaction products. Based on these criteria, we can distinguish:  combination reactions (synthesis),  decomposition reactions,  single displacement reactions,  double displacement reactions.

108 N 2 + 3H 2 = 2NH 3 Fe + S = FeS Ca + Cl 2 = CaCl 2 SO 3 + H 2 O = H 2 SO 4 a) Combination reactions (synthesis) are reactions in which two substances interact to form a single compound. There are many examples for this:

109 CaCO 3 = CaO + CO 2 4NH 4 NO 3 = 3N 2 + N 2 O 4 + 8H 2 O Fe 2 (SO 4 ) 3 = Fe 2 O 3 + 3SO 3 b) Decomposition reactions are transformations in which from one substance, two or more substances are formed:

110 Fe + CuSO 4 = Cu + FeSO 4 Mg + 2H 2 O = Mg(OH) 2 + H 2 Zn + 2HCl = ZnCl 2 + H 2 Cl 2 + 2KI = 2KCl + I 2 c) Single displacement or substitution reactions are transformations in which one element or one group of elements from a combination is replaced with another element or group of elements:

111 Pb(NO 3 ) 2 + 2KI = PbI 2 + 2KNO 3 AgNO 3 + KCl = AgCl + KNO 3 H 2 SO 4 + BaCl 2 = BaSO 4 + 2HCl CaCl 2 + K 2 CO 3 = CaCO 3 + 2KCl d) Double displacement or coupling substitutions are transformations in which two elements or groups of elements are exchanged between two chemical combinations: A special case of double substitution reactions is the reaction between acids and bases: H 2 SO 4 + 2NaOH = Na 2 SO 4 + 2H 2 O

112 Based on the nature of the reactants or products there are:- combustion reactions - hydrolysis reaction - precipitation and complexation reactions a) Combustion reactions: oxygen reacts with a carbon compound containing hydrogen and/or other element like O, S, N. Example: the combustion of hydrocarbons (toluene, methane, acetylene), alcohols (methanol) or sulfur compounds (thiophene) C 6 H 5 -CH 3 + 9O 2 = 7CO 2 + 4H 2 O CH 4 + 2O 2 = CO 2 +2H 2 O

113 2C 2 H 2 + 5O 2 = 4CO 2 + 2H 2 O 2CH 3 OH + 3O 2 = 2CO 2 + 2H 2 O C 4 H 4 S + 6O 2 = 4CO 2 + 2H 2 O + SO 2 The burning of carbon can also be considered a combustion reaction: C + O 2 = CO 2 b) Hydrolysis reaction: the reactant is water; this reactions are frequent in inorganic chemistry as well as in organic chemistry: Al 2 (SO 4 ) 3 + 6H 2 O = 2Al(OH) 3 + 3H 2 SO 4 R-CN + H 2 O = R-CONH 2

114 c) The precipitation and complexation reactions: the classification criteria is the nature of the reaction products: Pb(NO 3 ) 2 + K 2 SO 4 = PbSO 4 + 2KNO 3 CoCl 3 + 6NH 3 = [Co(NH 3 ) 6 ]Cl 3

115 In organic chemistry, the chemical reactions imply usually the breaking and formation of covalent bonds. There are three fundamental types of reactions: substitution, addition and elimination. Generally, the organic molecule that suffers a transformation is called substrate, and the reactant used in it is called reagent. The substitution is the reaction in which an atom or a group of atoms attached to a carbon atom is replaced with another atom or group of atoms: CH 3 -CH 2 -Cl + NaOH = CH 3 -CH 2 -OH + NaCl

116 The addition reaction is the transformation that leads to the increasing of the number of atoms or groups of atoms attached to the carbon atoms of the substrate: HC  CH + HCN => H 2 C=CH-CN The elimination is the reverse of the addition and it leads to the decreases of the number of atoms or groups of atoms attached to the carbon atoms: CH 3 -CH 2 -OH => H 2 C=CH 2 + H 2 O The breaking of the covalent C – C bonds can be interpreted as an elimination reaction.

117 Stoichiometry It is the part of chemistry that has as aim the establishment of the quantitative relations between the reactants and reaction products. The name stoichiometry derives from Greek stoicheon that means element and metron that means measurement. So, stoichiometry is the science of the elements measuring. As it was seen before, the atomic mass unit (uam) was introduced, that represents the 12th part of the mass of the C: 1 uam = · kg

118 Based on the atomic mass unit the relative atomic masses of all elements have been determined. Knowing the atomic masses one can calculate the (relative) molecular masses, as the sum of the relative masses of all the atoms in the molecule. For example, the molecular mass of water is M H2O = 2·1+16=18 M H2SO4 = 2·1+32+4·16=98 M NaCl = =58.5 M CuSO4 = ·16=159.5.

119 In the chemical equations, the stoichiometric coefficients indicate the ratio between the number of molecules of the reactants and reaction products. The mole was initially defined as the mass of substance, expressed in grams, equal to the molecular mass of the substance. Thus, 1 mole of H 2 SO 4 is the quantity of substance that contains 98 g H 2 SO 4. The definition of the mole, as a fundamental unit in the International System of Units, is the following:

120 The mole is the quantity of substance of a system that contains 6.022·10 23 (the Avogadro’s number N A ) elementary particles. Avogadro’s number refers to different elementary particles that can be: molecule, atoms, ions or electrons.

121 Stoichiometric calculation Stoichiometric calculation is based on the law of conservation of mass: In a chemical reaction, the mass of the reactants is equal to the mass of the reaction products. Let us consider the reaction between metallic sodium and water that occurs according to the chemical equation: 2Na + 2H 2 O = 2NaOH + H 2 Atomic masses: Na – 23, H – 1, O – 16.

122 In a vessel filled with sufficiently enough water we introduce 0.23 g sodium. Calculate the quantity (mass) of water that has reacted, as well as the quantities (masses) of sodium hydroxide and hydrogen that have resulted. The quantity of water that has reacted with sodium: 2·23g Na………………………………2·18g H 2 O 0.23g Na………………………………x g H 2 O ________________________________________

123 Similarly, we calculate the mass of the resulted NaOH: 2·23g Na………………………………2·40g NaOH 0.23g Na………………………………x g NaOH _________________________________________ The resulted hydrogen mass: 2·23g Na………………………………2g H2 0.23g Na………………………………x g H2 _____________________________________

124 We can calculate directly the volume of H 2 that results from the reaction in normal conditions of temperature and pressure: 2·23g Na………………………………22.4L H 2 (cn) 0.23g Na………………………………x L H 2 (cn) ___________________________________________

125 CHAPTER 7 CHEMICAL EQUILIBRIUM

126 CHAPTER 7 CHEMICAL EQUILIBRIUM Reversible reactions Reactions that may proceed in both directions are called reversible reactions. H 2 + I 2 2HI Example: The reversible equation is represented using arrows in both ways instead of the equality sign.

127 The law of mass action We consider the reversible reaction: CH 3 -COOH + C 2 H 5 -OH CH 3 COOC 2 H 5 + H 2 O The ratio between the product of the reaction products concentrations and the product of the reactants concentrations, all taken to the power of their stoichiometric coefficients, is constant. K c is the equilibrium constant.

128 For a general reversible reaction: aA + bB mM + nN the expression of the law of mass action is: Le Chatelier’s principle If a dynamic equilibrium is disturbed by changing the conditions (concentrations, temperature and pressure) the position of equilibrium moves to counteract the change.

129 3. Increasing the pressure will shift the equilibrium so that molecules with smaller volume are being formed. Consequences of Le Chatelier’s principle: 1. Increasing the concentration of one of the components will shift the equilibrium in the direction in which this component reacts; 2. Increasing the temperature of the system will shift the equilibrium in the direction of endothermic reaction, so that the heat will be absorbed;

130 Electrolytic dissociation of water The water molecules dissociates according to the reaction: H 2 O + H 2 O H 3 O + + HO - The equilibrium is shifted far to the left. Experimentally, it was determined that at 25°C, only one molecule of water, out of 556,000,000 is dissociated, which means the dissociation degree of water is α = 18· The equilibrium constant for the dissociation reaction of water is:

131 The ionic product of water depends on the temperature. At 25°C, the value of K W is mol L -1. In pure water the concentration of the H 3 O + ions is equal to that of the HO -, which means that at 25°C: mol L -1 Kw is called the ionic product of water

132 In order to express the concentration of the hydrogen ions in aqueous solutions, the notion of pH was introduced by Sörensen (1909): The relation was modified by Bates by replacing the concentration of the hydronium ions with their activity: In the case of diluted solutions, the activity can be considered equal to the concentration

133 Similar to the pH notion the term of pOH was introduced, that is a measure of the concentration of the hydroxyl ions: It is easily demonstrated that, at the temperature of 25°C: pH + pOH = -lg KW = 14

134 Acid – base equilibrium Acids are substances that, in aqueous solutions, release hydrogen ions H +. For example, the hydrochloric acid dissociates in H + and Cl - ions: HClH + + Cl - Bases are substances that, in aqueous solutions, produce hydroxyl ions, like the case of sodium hydroxide, that dissociates in Na + and HO - ions: NaOH Na + + HO -

135 In aqueous solutions, acids like HCl, H 2 SO 4 or HNO 3 are completely dissociated, the dissociation degree is 1. They are called strong acids. Similarly, bases like KOH or NaOH are completely dissociated in solution, reason for which they are called strong bases. Partially dissociated acids in aqueous solution, like CH 3 COOH, HCN or H 2 S, are called weak acids, and partially dissociated bases in solution, like NH 3 or organic amines, are called weak bases. The dissociation degree for weak acids and bases is less than 1.

136 The reaction between an acid and a base is called neutralization reaction and it leads to the formation of a salt and water. For example, the reaction between nitric acid and potassium hydroxide can be represented by the equation: HNO 3 + KOH = KNO 3 + H 2 O The dissociation degree is defined as the ratio between the number of dissociated molecules and the total number of dissolved molecules:

137 We consider an acid that dissociates according to the equation: HA + H 2 O H 3 O + + A - The equilibrium constant is given by the relation: Acidity constant For diluted solutions, the concentration of water can be considered constant and it is included in K. We obtain the acidity constant:

138 Very weak acids: the first acidity constant lower than AcidConstant K 1 HClO (hypochlorous acid)3.2·10 -8 H 3 BO 3 (boric acid)5.8· Weak acids: the first acidity constant between and AcidConstant K 1 H 3 PO 4 (phosphoric acid)7.5·10 -3 CH 3 COOH (acetic acid)1.8·10 -5 H 2 CO 3 (carbonic acid)0.45·10 -6 Strong acids are completely dissociated in aqueous solution; one can not distinguish between their acidity constants.

139 Base constant For a base that, in aqueous solution, dissociates according to the reaction: BOH B + + HO - the base constant is given by the relation: Weak bases, like ammonia, aniline, have the basicity constant below : BaseConstant K b Ammonia NH 3 1.7·10 -5 Aniline C 6 H 5 – NH 2 3.8· Strong bases, like sodium hydroxide, calcium hydroxide, are completely dissociated in water, like in case of strong acids.

140 Calculation of pH for acid and base solutions Monoprotic acids are completely dissociated in aqueous solutions so the hydronium ions concentration is equal to the concentration of the acid. For example, for a mol L -1 solution of HCl, the concentration of the hydronium ions is [H 3 O + ] = mol L -1. The pH of the solution is: pH = -lg[H 3 O + ] = -lg = 3 For a strong base, for example mol L -1 KOH the concentration of hydroxyl ions is [HO - ] = mol L -1. It results that the pOH of the solution is:pOH = -lg[HO - ] = -lg = 3 Considering the relation between pOH and pH one obtains: pH = 14 – pOH = 14 – 3 = 11

141 For concentrations higher than mol L -1 the pH is calculated using the Bates relation because the activity differs from the concentration At very low acid concentrations for the calculation of pH it is necessary to consider the hydronium ions coming from the dissociation of both acid as well as water molecule.

142 For example, the pH of a mol L -1 solution of HCl is not 7 because the hydronium ions result not only from the dissociation of the acid, but from the dissociation of water as well: HCl + H 2 O H 3 O + + Cl - H 2 O + H 2 O H 3 O + + HO - Considering that the concentration of hydrochloric acid in the solution is c and the concentration of hydronium, respectively hydroxyl ions is x, the total hydronium ions concentration will be c+x. The ionic product of water, at the temperature of 25°C, will be: (x + c)x =

143 One obtains a second degree equation: x 2 + cx = 0 Solving the equation one obtains: =1.12  The solution with “minus” in front of the square root has no meaning, since it is negative. The concentration of the hydronium ions will be: [H 3 O + ] = ,12  = 2.12  Thus, the pH of a mol L -1 solution of HCl, will be: pH = -lg[H 3 O + ] = -lg 2.12  = 6.67


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