Lecture Notes 2 nd Series: Phase Chemistry Condensation of gases Let s begin by reminding ourselves of what happens when we cool a gas down the kinetic energy of the molecules decreases. Eventually the kinetic energy of the molecules is insufficient to overcome the intermolecular attractive force between the molecules and the gas stops behaving ideally. Instead it behaves like a real gas. If we continue to lower the temperature then, as the kinetic energy decreases further, the intermolecular forces between the molecules increase and the molecules stop moving in a chaotic manner. Instead they arrange themselves to maximise the intermolecular attractions. If the molecules attract each other they no longer fill the entire volume of the container i.e. they cease to be in the gaseous phase. Instead they are said to be in a CONDENSED PHASE, or liquid phase. The temperature this happens at is called the boiling point temperature, T b, of the substance. If, at a given temperature, we increase the pressure of our system then we will force the molecules to get closer together. When this happens we help the condensation process take place. Therefore we could also liquify a gas simply by increasing the pressure at a given temperature. It should therefore be obvious that the boiling point temperature must be pressure dependent. The higher the pressure is as we cool a gas down, the more readily condensation takes place, and therefore the higher is the boiling point temperature i.e. the condensation takes place at a higher average kinetic energy of the molecules. Therefore if we plot the boiling point temperature of a substance as a function of pressure it looks like this: Boiling point temperature Pressure More commonly we plot the pressure on the y-axis and the temperature on the x-axis. When we do this it is called a PHASE DIAGRAM. Pressure LIQUID VAPOUR T b = boiling point temperature Temperature We normally call the gas phase the vapour phase. To the right-hand side of the line in the diagram the substance is in the vapour phase and, to the left-hand side, it is in the liquid phase. 17
Question 2.1 Draw the graph of: T (y-axis) against P (x-axis) for the boiling point temperature of a substance and label where the liquid exists and where the vapour exists. In the phase diagram it is important to realise what is present when we are on the line. On the line we are at the boiling point temperature and both the liquid and the vapour are present. They are in equilibrium with each other. That is, the rate of condensation of the gas exactly equals the rate of vapourisation of the liquid. vapour vapour liquid evaporation liquid condensation We can think of this in terms of the kinetic energy of the gas molecules exactly balancing the potential energy of the intermolecular forces in the liquid phase. Again we describe this as a dynamic equilibrium as both processes condensation and vapourisation are continuously happening but at the same rate. Question 2.2 At a constant pressure, which of the following pairs of substances has the higher boiling point temperature? Explain your answer in each case. (a) H 2 or N 2 (b) CO 2 or Ar (c) H 2 O or HCl The boiling point temperature of a substance that is normally listed in tables is the boiling point temperature at standard pressure. The boiling point temperature at this pressure is called the normal boiling point temperature. Question 2.3 Read off the normal boiling point temperature ( o C) for the substance represented in the diagram below: Pressure/kN m -2 50 100 150 50 100 150 200 250 300 Temperature /K Question 2.4 Comment on the following statement: The boiling point temperature of water is 100 o C. 18
Liquids Liquids behave quite differently from gases. Instead of having absolute free movement as in a gas, the molecules in a liquid have a profound effect on each others movement. Although they are free to flow past each other the molecules cannot move independently due to intermolecular forces. The molecules are said to have lost a degree of freedom of movement when going from the gaseous state to the liquid state. The liquid phase of a substance is limited to temperatures between the freezing point temperature and the boiling point temperature. When a liquid freezes the molecules lose another degree of freedom of movement and are effectively locked in a fixed position. Degrees of Freedom A molecule in the gaseous state has three degrees of freedom of movement. It can rotate and vibrate and translate (i.e. move in a linear fashion). When a gas liquifies, the molecules lose their rotational degree of freedom. The attractive forces operating between molecules effectively hinder the molecules ability to spin. When a liquid solidifies the molecules lose their translational motion, i.e. they are in fixed positions within the solid structure. The molecules in a solid can however still vibrate. Vibrations in molecules consist of the atoms moving further apart and then closer together periodically. Therefore when we talk of bond lengths we are referring to the average distance the atoms are apart. When we cool a liquid down we eventually reach a temperature where the molecules lose their ability to flow past each other due to a further decrease in kinetic energy. They instead become part of a rigid structure, a solid. This temperature is called the freezing point temperature of the liquid. Is the freezing point temperature dependent on the pressure? Our argument for the pressure dependence of the boiling point temperature was based upon the fact that an increase in pressure results in an increase in the density of the system. An increase in the density of the system favours liquid formation, therefore the boiling point temperature increases with increasing pressure. When going from a liquid to a solid do we change the density of the system? Well, of course, if the molecules are packed closer together in a solid than in a liquid then the density increases upon freezing. However, this difference in density between the solid and liquid phases is not nearly as big as the difference in density between a gas and a liquid. Therefore we would expect the freezing point temperature to be dependent upon pressure but to a smaller degree than the boiling point temperature. If we increase the pressure of the system we encourage formation of the denser phase. Thus as pressure increases the liquid freezes sooner as we cool it down and the freezing point temperature increases. If we plot this on a phase diagram: Pressure SOLID LIQUID T f = freezing point temperature Temperature 19
Question 2.5 Is ice more or less dense than liquid water? Question 2.6 Keeping your answer to question 5 in mind, plot the P vs T graph for T f for H 2 O. At T f we are on the solid/liquid line in the phase diagram. The two phases are in equilibrium with each other. To the left of the line is the solid phase and to the right of the line is the liquid phase. If we now look at a phase diagram for a pure substance then we would expect to see two equilibrium lines the solid/liquid line and the liquid/vapour line. As they have different gradients, at some point the two lines intersect as shown below. P solid liquid vapour T The point of intersection is the lowest pressure at which the liquid can exist. So below this pressure what is the physical state of the substance? Sublimation Below the minimum pressure required for a liquid, the substance can only exist as either a solid or a vapour. There is a phase change which describes the transition of a solid directly into a gas without having to go through the liquid phase. This process is called sublimation. The opposite change, from a gas to a solid, is called deposition. The temperature this change occurs at is the sublimation temperature, T sub. Again, because the densities of the two phases are different, T sub is pressure dependent. If we add the sublimation equilibrium line to our phase diagram we have the following: Pressure solid liquid vapour X Temperature The point marked X on the diagram is where all three equilibrium lines meet. It is the only point on the diagram where we have all three phases present. This point is called the TRIPLE POINT. 20
Vapour Pressures The solid/vapour and liquid/vapour lines are very important when understanding the concept of the vapour pressure of a substance. Let s take an imaginary substance with the following phase diagram: Pressure/atm 0 0.05 0.10 50 100 150 200 250 300 Temperature/ o C If we place some of our substance in the liquid phase into an evacuated container (i.e. P ~ 0) at 150 o C, is our substance happy to stay as a liquid? The answer of course is no. At P ~ 0 and T = 150 o C on the phase diagram we are in the region where our substance wants to be a vapour. Therefore our substance will start to vapourise and molecules will go from the liquid phase to the vapour phase. As this happens, of course, the pressure inside our container will increase as more gas molecules will collide with the sides of the container. Will this process go on indefinitely until all the liquid turns into vapour? There are, in fact, two possible outcomes here. From the phase diagram we can see that if we have both the liquid and vapour phases present at equilibrium then we must be on the equilibrium liquid-vapour line. At T= 150 o C this requires the pressure of the vapour to be approximately 0.05 atm. Therefore if the pressure of the vapour reaches 0.05 atm then the liquid remaining, and the vapour that has formed, will be at equilibrium i.e. the rate of condensation equals the rate of vapourisation. If, however, the liquid all evaporates and the pressure of the vapour has still not reached 0.05 atm then only the vapour will be present in the container. When the liquid and the vapour are at equilibrium then the pressure within the container is known as the VAPOUR PRESSURE of the liquid. The vapour pressure of the substance is temperature dependent but, for a given temperature (and as long as the liquid is still present at equilibrium), the pressure exerted by the vapour is constant. Question 2.7 1 L of a liquid of a pure substance is placed in an evacuated container. At equilibrium both the liquid and vapour are present and the vapour pressure is 164 Torr. What would be the vapour pressure at equilibrium when 2 L of the same liquid are placed in the same container at the same T? Thus the liquid/vapour equilibrium line is the vapour pressure of the substance as a function of temperature. We have already shown, however, that this line is the boiling point temperature at various pressures. Are these two interpretations of the same line consistent? Let s return to the pressure dependence of the boiling point temperature but this time from the perspective of a liquid in an open test tube. For a liquid to boil molecules must escape from the liquid phase and enter the vapour phase. 21
The prevailing atmospheric pressure however bears down on the liquid preventing the molecules from escaping into the vapour phase. For molecules to escape, the liquid must exert an equal but opposite pressure to the external pressure. Only when the liquid exerts this pressure does the liquid boil. When the liquid does exert this pressure then it will boil. What is the maximum pressure a liquid can exert at any given temperature? Its vapour pressure! The vapour pressure of a liquid is then the pressure it exerts at its boiling point temperature. Therefore both interpretations of the line are correct. If one wishes to know the boiling point temperature at a given pressure we simply read across from the y- axis of the phase diagram at the appropriate pressure until we reach the liquid/vapour equilibrium line and then read off the boiling point temperature from the x-axis. If we wish to know the vapour pressure of a liquid at a given temperature, we simply read up from the appropriate temperature on the x-axis until we reach the liquid/vapour equilibrium line and read the vapour pressure off the y-axis. Solids also have vapour pressures but these are generally much lower than for liquids as we can see from the solid/vapour line on the phase diagram. The vapour pressure of a solid is called the sublimation vapour pressure. These however tend to be largely theoretical values as the rate at which a solid evaporates (i.e. losing molecules to the gaseous phase) is very slow and equilibrium is seldom reached in practise. It must also be emphasised that if a pure liquid is placed in an open container and starts to vapourise, equilibrium is never reached between the vapour and the liquid. The vapour diffuses out of the open container and eventually all of the liquid vapourises. Therefore if you leave water on a watchglass one evening and come back in the morning, there will be no liquid left on the watchglass. (In South Africa it would all have evaporated; in Scotland it would all have frozen). A dynamic equilibrium can only be established in a closed system. Phase diagrams are a representation of the equilibrium situation. Therefore they are describing what happens in a closed system. If we put a pure liquid into an evacuated container at constant temperature, what will happen? The system approaches equilibrium by molecules evaporating from the surface of the liquid until the vapour pressure of the liquid at that temperature is reached. What now happens if we raise the temperature of the system to a higher value? More molecules have enough kinetic energy to escape from the surface of the liquid and therefore more molecules are present in the vapour phase. Once again a dynamic equilibrium is established at this new temperature. The rate of condensation still equals the rate of evaporation but the vapour pressure above the liquid has increased to the value appropriate to the higher temperature. A result of this is that the vapour phase has become denser i.e. more molecules are present in a given volume. If we continue heating the system then the vapour pressure continues rising and the vapour phase becomes denser and denser. Therefore as heating continues the density of the vapour increases until it eventually equals the density of the remaining liquid. Above this temperature the distinction between the liquid and vapour phases disappears. Technically, as the substance is filling the container, we have a gas present, however it is commonly called a supercritical fluid. The temperature this transition occurs at is known as the critical temperature. At temperatures above the critical temperature we no longer have a liquid. Therefore the liquid/vapour line must terminate at the critical temperature, T c. Associated with T c is the critical pressure, P c. 22
Critical point P c T c Temperature There is no such theoretical end to the solid/liquid equilibrium line. Strictly speaking a vapour and a gas are not the same thing although the two words are commonly used interchangeably. A vapour can be liquified by increasing the pressure at constant temperature. A gas cannot be liquified by increasing the pressure at a constant temperature. At temperatures above T c one cannot liquify a gas by simply increasing the pressure as the liquid cannot exist at temperatures above T c. At temperatures below or equal to T c a vapour can be liquified by increasing the pressure. Pressure Pressure vapour gas Temperature T c 23
Solutions The phase chemistry described so far has been for the behaviour of a pure substance. What if we now mix substances? What effect does that have on the physical properties of the substances involved? If we add one substance to another and a homogeneous mixture results then we have a SOLUTION. A homogeneous mixture is one in which the composition is the same throughout the solution. There is also only one phase throughout the mixture. Up to now you have probably considered a solution to be something dissolved in water. This is in fact only one type of solution an AQUEOUS SOLUTION. Water is called the solvent and the dissolved substance the solute. Many more types of solution exist. The solvent doesn t have to be water. Most organic materials won t dissolve in water and one must use an organic solvent such as hexane or benzene. Indeed, the solvent doesn t even have to be a liquid! Given that a solution is simply a homogeneous mixture then the air we breathe is a solution! We can imagine N 2 (g) being the solvent and the other gases the solutes (as N 2 (g) is the most abundant gas). Brass is a solid solution (of Zn(s) in Cu(s)) as is any other homogeneous alloy. This course will limit itself to liquid solutions and examine what happens to the physical properties of the solvent when one dissolves another substance in it. Solubility Why do substances dissolve in a solvent in the first place? Why doesn t the solute spontaneously separate from the solvent? We know that this doesn t happen. If we take sea-water for instance, NaCl(s) does not spontaneously precipitate out of the water. The sodium ions and the chloride ions are happy to be dissociated from each other and surrounded by water molecules. The process of a substance dissolving must therefore be energetically favourable. If it is not energetically favourable then the substance does not dissolve. One of the principle tenets of thermodynamics is that all spontaneous changes increase the amount of disorder in the universe. This disorder is so important that scientists give it a special term ENTROPY and have even formed a special law to describe it. The 2 nd Law of Thermodynamics: All spontaneous changes increase the entropy of the universe. If we take a solute and add it to a solvent then the disorder, or entropy, of the system increases. Solvent Solute Solution This is the stability gain that a solvent gets from having a solute dissolve in it. However this stability gain is balanced by the fact that for a solute to dissolve two things must happen The intermolecular forces holding the solute particles together must be broken. The intermolecular forces between solvent molecules must also be broken to allow space for the solute molecules to enter. These two points can be illustrated by the example of sodium chloride dissolving in water. 24
Electrostatic interaction between ions solution Hydrogen bonding between water molecules Ion-dipole interactions between ions and water molecules The intermolecular forces in the solution must be strong enough that sufficient stability is gained to compensate for the stability that has been lost by breaking the intermolecular forces in the solute and solvent. Thus for a species to dissolve Gain in stability > Loss in stability Lowering of energy from intermolecular energy required to break solvent interactions between solvent and solute plus the intermolecular forces plus stability gain in increasing disorder energy required to break solute intermolecular forces As there are so many factors involved it is very difficult to predict whether a species is soluble in a particular solvent. However we can state a very important (if unscientific) general principle Like dissolves Like This statement presumes that the solvent-solute interactions are the deciding factor whether a substance is soluble or not. Therefore, for example, ionic substances dissolve in polar solvents due to the strong ion-dipole interactions present in the solution. Similarly, organic solutes dissolve in organic solvents due to the presence of London Forces between the molecules. Hexane, C 6 H 14, for instance, dissolves in pentane, C 5 H 12, as the resulting solution is more disordered than the two separate species and hexane molecules fit well between the pentane molecules. 25
London forces between solute and solvent Thus when predicting whether a solid will dissolve in a particular solvent one must have some idea as to the bonding in the solid. The different types of bonding in solids will be covered later in the course. Molality As we are about to find out in the next section, when we dissolve a solute in a liquid solvent the physical properties of the solvent change. The degree of change depends on the amount of the solute that one adds and the resulting concentration in the solvent. So far we have commonly used molarity to describe the concentration of a species in solution. However, when dealing with physical properties the use of the molar concentration has a distinct disadvantage. Molarity is the moles per unit volume (dm 3 ). However the volume of a system depends upon the temperature. If we heat a solution the kinetic energy of the molecules increases, the volume expands and the molar concentrations of the solutes change. Question 2.8 Does the molar concentration of a solute increase or decrease as we heat a solution? Explain your answer. Physical chemists thus prefer to measure the concentration of a solute as moles per unit mass (kg) of solvent (NB. not per unit mass of solution). This is the MOLAL concentration and the concentration is expressed as the molality, m. m = number of moles of solute (mol) mass of solvent(kg) Question 2.9 Calculate the molality of the following solutions. (i) 17.42 g of NaF in 634 g of H 2 O (ii) 1.34 10-3 mol of Na 2 CO 3 in 55.6 g of H 2 O It is a simple exercise to convert between molality and molarity if one knows the density of the solution. 26
Example 2.1 Question: The acid that is used in car batteries is 4.27 mol dm -3 aqueous sulfuric acid, which has a density of 1.25 g per millilitre. What is the molality of the acid? Answer: Question 2.10 Calculate the molality of a 10% weight by volume aqueous solution of methanol, CH 3 OH, given its density is 0.98 g ml -1 Question 2.11 What mass (g) of NaCl must be dissolved in 542 g of water in order to prepare a 0.100 m solution? Question 2.12 What is the mass% of Na 2 SO 4 in a 0.222 m aqueous solution? Question 2.13 What mass of oxalic acid dihydrate is required to produce a 0.0554 m solution of oxalic acid in 505 g of water? Colligative Properties The phase transitions that a substance can undergo all involve a re-arrangement of the molecules during the process. Therefore they are all intimately dependent upon the intermolecular forces present. Consider the case, for example, when a liquid evaporates. Molecules escape from the surface of the liquid when they have sufficient kinetic energy to overcome the intermolecular forces present in the liquid. If we now add a solute which disturbs the intermolecular forces of the solvent and increases the entropy of the system, we would then expect to affect the ability of a solvent molecule to escape from the surface of the solution. Thus the solvent s vapour pressure would be expected to change. In fact, the addition of a solute to a solvent changes the solvent s boiling point temperature, melting point temperature and its vapour pressure. Moreover it is found that it is not the identity of the solute which affects the degree of change in the properties of the 27
solvent, but the amount of the solute which has been added. These three properties (boiling point temperature, melting point temperature and vapour pressure) are called colligative properties. A colligative property of a solvent is one which depends upon the number of solute particles in the solution and is independent of the chemical identity of the particles. Before looking at each of the colligative properties in turn, we must first put an initial limitation on our argument. The solute that we add to the solvent will be non-volatile. A volatile substance is one which exerts a measurable vapour pressure. The greater the vapour pressure, the more volatile a substance is said to be. A non-volatile substance, therefore, does not have a measurable vapour pressure. Question 2.14 Consider the following data and determine which is the more volatile; ethanol or methanol. Explain your answer. C 2 H 5 OH vapour pressures Temp, o C 17.7 34.9 63.5 78.4 Vapour Pressure, Torr 20 100 400 760 CH 3 OH vapour pressures Temp, o C -6.0 34.9 49.9 64.7 Vapour Pressure, Torr 20 200 400 760 The Lowering of the Vapour Pressure When a non-volatile solute is added to a volatile solvent, the vapour pressure exerted by the solution is less than that for the pure solvent. To understand why this happens we must remind ourselves that the vapour pressure of a liquid is a measure of the position of equilibrium between the rate of evaporation and the rate of condensation. At low temperature there are fewer molecules in the vapour phase. The rate of condensation equals the rate of evaporation at equilibrium. At higher temperature the rate of evaporation has increased but it is still equal to the rate of condensation at equilibrium. The difference is that at the higher temperature there are more molecules in the vapour phase and therefore the vapour pressure is greater. Why then, at constant temperature, does the presence of non-volatile solute molecules lower the vapour pressure of the solvent? Remember, evaporation must take place from the surface of the liquid phase (whether it s a solvent or a solution). If there are solute molecules present then there are fewer solvent molecules present at the surface, as some of the surface sites are occupied by solute molecules. The rate of evaporation then necessarily decreases as less molecules escape into the vapour phase (the non-volatile solvent molecules never enter the vapour phase). The more solute molecules that are present, the fewer solvent molecules can escape from the surface and the greater is the lowering of the vapour pressure. Note that the identity of the solute molecules does not matter, as long as they are non-volatile. It is the number of the surface sites occupied by the solute molecules that counts. 28
This effect can be quantified by means of Raoult s Law. Raoult s Law: the vapour pressure of the solvent in a solution containing a non-volatile solute is directly proportional to the mole fraction of the solvent in the solution. The mole fraction of the solvent, of course, is a measure of the number of solvent molecules compared with solute molecules. The mole fraction of a species in solution is given the symbol, x. Raoult s Law is very important in industrial chemistry and is certainly one you should remember and be able to apply. It is best remembered as P solution = x solvent P* solvent where P* is the vapour pressure of the pure solvent. If you re struggling to see where this equation comes from, consider two points on th graph of P solution vs x solvent. If x solvent is 1 then P solution = P* solvent P solution If X solvent is zero then the vapour pressure will be zero as we have only a non-volatile species present 0 x solvent 1 As there is a direct proportionality relating P solution to x solvent the graph will look like this. P * solvent P solution 0 x solvent 1 29
Example 2.2 Question: Calculate the vapour pressure (atm) of an aqueous solution at 100 o C which contains 10.0 g of sucrose, C 12 H 22 O 11, in 1.00 10 2 g of water. Answer: Question 2.15 Below are two plots of the liquid/vapour equilibrium line; one for a pure solvent and the other for the pure solvent with some non-volatile solute present. Which is which? Pressure Temperature Question 2.16 Calculate by how much the vapour pressure of the solvent changes when 40.3 g of naphthalene, C 10 H 8, is added to 135 g of benzene, C 6 H 6, at 20 o C. The vapour pressure of benzene at 20 o C is 74.6 Torr. Consider naphthalene to be non-volatile for this problem. naphthalene 30
The Elevation of the Boiling Point Temperature Let s return to the phase diagram showing the vapour pressures of a solvent and a solution of the solvent. P pure solvent solvent containing non-volatile solute The solid line is for the pure solvent and the dotted line is for the solution. At any given temperature the vapour pressure of the solution is lower than that for the pure solvent. You ll remember that the liquid/vapour equilibrium line is also the boiling point temperature line. Therefore at any given pressure the boiling point temperature of a solvent increases upon addition of non-volatile solute. T P 1 atm T b solution T b pure solvent T Therefore the boiling point temperature of a solvent is elevated upon addition of a non-volatile solute. The degree of change in the boiling point temperature can also be quantified by the following equation. ΔT b = K b m where ΔT b = T b (solution) T b (solvent) m = molality K b = molal boiling point elevation constant K b is sometimes referred to as the ebullioscopic constant. Whatever we call it, it is important to remember that K b is a constant for the solvent involved. Question 2.17: If we measure ΔT b in Kelvin what units must K b have? The above equation again emphasises that the identity of the solute is incidental it is the number of solute particles that counts. When considering the number of solute particles, however, you must be careful. If, for instance, we add 1 mole of sodium chloride to 1 000 g of water, what is the resulting molality of the solution? 31
The answer is 1 m NaCl. However, sodium chloride is an ionic solid. As such, when it is placed in water, it dissociates into sodium ions and chloride ions. Therefore 1 mole of NaCl supplies 2 moles of particles to the solvent and thus the effective number of particles is 2 moles. To take into account the effective number of particles in the solution we add another factor into the equation. ΔT b = K b m i Here i is the van t Hoff factor (named after the 1 st recipient of the Nobel prize for Chemistry in 1901). For sodium chloride the van t Hoff factor is equal to two. Therefore for ionic solids which provide ions in solution, one must always be aware of how many moles of ions are added per mole of solute. Question 2.18 Determine the van t Hoff factor for each of the following species when they are added to water (a) magnesium bromide (b) potassium sulfate (c) ethanol, CH 3 CH 2 OH (d) aluminium oxalate Sometimes the van t Hoff factor can be non-integral. This arises when species associate in solution. When an ionic substance is dissolved in water the ions still tend to be attracted to each other due to electrostatic forces. Ion pairs can form in solution, where a cation and an anion have a brief encounter! The ion pairs form and quickly break up but the net result is that the effective number of particles in solution at any given time is less than would be expected if the ions move independently of each other. X - M + M + X - M+ X - ion pairs Organic substrates in aqueous solution give non-integral values as well but this time the van t Hoff factor is less than one. This again is due to the organic molecules liking each others company in solution i.e. the intermolecular forces between solute and solute are stronger than the solute-solvent intermolecular forces. association in solution 32
When association does take place in solution then the van't Hoff factor is concentration dependent. The more solute particles around, the more association in solution occurs, and the lower is the van't Hoff factor. For instance a 0.10 mol dm -3 KBr(aq) solution has an i value of 1.88 whereas a 1.0 mol dm -3 KBr solution has an i value of 1.77. Table 1: Properties of Common Solvents Solvent Normal boiling point temperature o C Normal freezing point temperature o C K b o C kg mol -1 K f o C kg mol -1 Water 100 0.00 0.512 1.86 Benzene 80.1 5.48 2.53 5.12 Acetic acid 118 16.6 3.07 3.90 Nitrobenzene 210.88 5.7 5.24 7.00 Phenol 182 43 3.56 7.40 Camphor 207.42 178.40 5.61 40.0 Carbon tetrachloride 76.6-23.0 5.03 31.8 Chloroform 61.7-63.5 3.63 4.68 Ethanol 78.5-117.30 1.22 1.99 Example 2.3 Question: What is the normal boiling point temperature of a 1.45 mol dm -3 aqueous solution of sucrose? Answer: 33
Example 2.4 Question: What mass of naphthalene, C 10 H 8, must be dissolved in 422 g of nitrobenzene to produce a solution which boils at 213.76 o C at 1.00 atm. Answer: Question 2.19 A solution contains 2.50 g of biphenyl, C 12 H 10, dissolved in 100.0 g of benzene. The boiling point temperature of the solution is 0.411 o C higher than that for pure benzene. Estimate K b for benzene. biphenyl Question 2.20 6.63 g of an electrolyte MX (molar mass = 124 g mol -1 ) is added to 54.2 g of water to give a solution which boils at 100.92 o C at 1.00 atm. Discuss the degree of dissociation of the electrolyte in solution. The Depression of the Freezing Point Temperature What happens to the freezing point temperature of a solvent when one adds a non-volatile solute to it? If we think about the last two sections carefully it will become evident that when one adds a solute to a solvent it becomes energetically more stable. It takes more energy to get the solution to boil that it would take for an equivalent amount of solvent (i.e. T b as the amount of solute ). It also takes more energy for the molecules to escape from the surface of the liquid when evaporating when a solute is added to a solvent (i.e. the vapour pressure as the amount of solute ). This increase in stability of the solution is a direct result of it becoming more disordered as compared with pure solvent and pure solute i.e. its entropy has increased. If the liquid phase has become more stable what will happen to the temperature it freezes at? It will decrease. The solution phase will have a larger temperature window in which it can exist. 34
The addition of a non-volatile solute will decrease or depress the freezing point temperature of a solvent. If you re not convinced by this argument consider the situation from the solid state s point of view. For a solid to form, the molecules must become ordered i.e. they lose their translational degree of freedom. The more disordered the solution is, the more difficult it becomes to order the molecules, and therefore T f decreases. Again this effect depends upon the effective number of solute molecules present. The effect is quantified by the equation T f = K f m i K f is the molal freezing point depression constant, or the cryoscopic constant, of the solvent. Values of K f for common solvents are given in Table 4.1. Note here that T f is always positive. T f = T f (solvent) T f (solution) Example 2.5 Question: Calculate the normal freezing point temperature of a 1.74 m aqueous solution of sucrose. Answer: The effect of a solute is normally bigger upon the freezing point temperature than the boiling point temperature (compare the values of K f with K b in Table 1). Some solutes have such a large K f value that the freezing point temperature depression of a solvent can be used to measure accurately the molar mass of an unknown solute. Example 2.6 Question: 1.20 g of an unknown organic compound was dissolved in 50.0 grams of benzene. The solution had a T f of 4.92 o C. Calculate the molar mass of the organic solute. Answer: 35
If we now plot the phase diagram for a pure solvent and a solution of the same solvent for the solid/liquid equilibrium line we see the following. P solution pure solvent T If we compare the complete phase diagram for a pure solvent, and a solution including that solvent, the following picture arises. P solvent solution T The boiling point temperature of the solvent increases and the freezing point temperature decreases when a non-volatile solute is added to the solvent. Question 2.21 Why is the sublimation equilibrium line unaffected by the addition of a solute? Question 2.22 Copper has a melting point of 1083 C. Its cryoscopic constant is 23 C kg mol -1. Calculate the melting point temperature of a brass which contains 88.2 mass% Cu and 11.8 mass% Zn. 36