Chapter_11_Part_1

# Chapter_11_Part_1 - Chapter 11 States of Matter Liquids and...

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Unformatted text preview: Chapter 11: States of Matter: Liquids and Solids States of Matter We began our study of states of matter at the beginning of the course, when we studied gases in Chapter 5. We then moved inside atoms to study their electronic structure. Now we return to states of matter to better understand liquids and solids. 1. Comparison of Gases, Liquids and Solids When we studied gases, we compared the three states of matter in several ways • gases and liquids take the shape of the container; solids retain their own shape • gases expand to fill a container; gases can be compressed; liquids and solids have a definite volume We can explain these properties by comparing how the units (molecules, atoms or ions) exist in each phase: • in a gas the units are far apart and in constant motion • in a liquid the units are close together and can tumble over each other • in a solid the units are close together and in fixed positions relative to each other Changes of State 2. Phase Transitions A phase transition is the change from one state of matter to another. Each transition has a specific name: • solid –––> liquid melting • solid –––> gas sublimation • liquid –––> gas vaporization (evaporation) These three transitions all require energy, so are endothermic. • gas –––> liquid condensation • gas –––> solid deposition (condensation) • liquid –––> solid freezing These three transitions all release energy, so are exothermic. Chapter 11 Part 1 1 of 7 In the diagram below a change from lower to higher is endothermic; a change from higher to lower is exothermic. gas boiling sublimation condensation liquid freezing melting condensation or deposition solid a. Vapor Pressure Vapor pressure is the partial pressure of a vapor over the liquid measured at equilibrium at a given temperature. Equilibrium in this case means that the rate of vaporization is equal to the rate of condensation. The graph to the right represent the rates of vaporization and condensation after a liquid has been put into a closed vessel: Vapor pressure depends on temperature. The higher the temperature, the higher the vapor pressure. This is because as temperature is increased a greater proportion of the molecules have enough energy to change from the liquid to the gaseous state. The graph below shows the fraction of molecules that have enough energy to vaporize at two temperatures. Chapter 11 Part 1 2 of 7 The graph to the below shows the change in vapor pressure with increasing temperature for four different liquids: b. Boiling Point and Melting Point The boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to atmospheric pressure. The normal boiling point of a liquid is the temperature at which the vapor pressure of the liquid is equal to 1 atmosphere (760 mmHg). Boiling differs from evaporation because when a liquid boils, bubble of vapor (gas) form within the body of the liquids, whereas evaporation occurs only at the surface of the liquid. See the diagram to the right. The freezing point of a liquid is the temperature at which the liquid changes to a crystalline solid. The melting point of a solid is the temperature at which a crystalline solid changes to a liquid. The value of the freezing point is identical to the value of the melting point for a given substance. c. Chapter 11 Heat of Phase Transition Heat is required to raise the temperature of a substance. The quantity of heat required is related to the specific heat capacity of the substance in each physical state. The specific heat of ice differs from that of liquid water and from that of water vapor. Heat is also required to change a substance from solid to liquid and from liquid to gas. The quantity of heat required is related to the heat of the phase transition. Part 1 3 of 7 The graph below illustrates the change in temperature of a substance as heat is added at a constant rate. Note that when only one phase is present, the temperature increases as heat is added. However, when a phase change is occurring, that is, when two phases are in equilibrium, there is no temperature change. All of the energy being added is used for the phase change. The heat of fusion , DHfus, is the energy required to change from solid to liquid. The heat of vaporization, DHvap, is the energy required to change from liquid to gas. Example: How much heat is required to vaporize 1.00 kg of ammonia. DHvap = 23.4 kJ/mol. Given: mass = 1.00 kg ammonia; DHvap = 23.4 kJ/mol Process: We begin this problem with the amount of ammonia given, convert kg to grams, convert grams to moles, and finally convert moles to kJ using DHvap. Molar mass ammonia (NH3) = 1(14.00674) + 3(1.00794) = 17.03056 g 1.00 kg d. Chapter 11 3 10 g 1 kg 1 mol 23.4 kJ 17.03056 g 3 mol = 1374.00062 kJ = 1.37 x 10 kJ Clausius–Clapeyron Equation: Relating Vapor Pressure and Liquid Temperature The graph showing the increase in vapor pressure as temperature increases was not linear. However, when the natural log of vapor pressure (ln P) is plotted 1 against the reciprocal of temperature ( /T) a linear graph results. The equation describing this graph is called the Clausius–Clapeyron equation. Part 1 4 of 7 P2 ln P1 = 1 DHvap R T1 1 – T2 Example: Carbon disulfide, CS2, has a normal boiling point of 46oC (vapor pressure = 760 mmHg) and a heat of vaporization of 26.8 kJ/mol. What is the vapor pressure of carbon disulfide at 35oC? Given: P1 = 760 mmHg P2 = ? T1 = 46ºC Process: Given: T2 = 35ºC DHvap = 26.8 kJ/mol P2 = ? We begin this problem by converting temperature to K and then use the Clausius Clapeyron equation to solve for P2. Note that R uses J while DHvap uses kJ. This requires that we convert to J. P1 = 760 mmHg P2 = ? T1 = 46ºC + 273 = 319 K T2 = 35ºC + 273 = 308 K R = 8.314 J/(mol•K) 3 DHvap = 26.8 x 10 J/mol P2 = 3 1 P2 26.8 x 10 J/mol ln 760 mmHg = 8.314 J/(mol•K) 319 K – To solve this begin with the subtraction inside the parenthesis: 1 308 K 3 P2 26.8 x 10 J/mol ln 760 mmHg = 8.314 J/(mol•K) We can now multiply by DHvap and divide by R: –0.0011957/K P2 760 mmHg = –0.360891006 Note that there are no units on the right side of the equation; they have canceled. The next step is to take the antilog of both sides. This is either the INV LN key or eX key on your calculator. P2 760 mmHg = 0.697054968 Solve now for P2: 529.7617763 mmHg = 530. mmHg P2 = ln Note that the decimal point is required in order to indicate the three significant figures. Example: Carbon disulfide, CS2, has a normal boiling point of 46.5oC (vapor pressure = 760.0 mmHg) and a vapor pressure of of 400.0 mmHg at 28.0oC? Given: P1 = 760.0 mmHg P2 = 400.0 mmHg T1 = 46.5ºC T2 = 28.0ºC DHvap =? Chapter 11 Part 1 5 of 7 Process: Given: We begin this problem by converting temperature to K and then use the Clausius Clapeyron equation to solve for P2. Because a decimal place is given for temperature, we need to use 273.15 for the conversion. P1 = 760 mmHg P2 = 400.0 mmHg T1 = 319.65 K T2 = 301.15 K R = 8.314 J/(mol•K) DHvap = ? 400.0 mmHg 1 1 DHvap ln 760.0 mmHg = 8.314 J/(mol•K) 319.65 K – 301.15 K To solve this begin with with the left side (where there are no variables). The natural log of the ratio: 1 1 DHvap –0.641853886 = 8.314 J/(mol•K) – 319.65 K 301.15 K Save this result in your memory. Now work with the parenthesis: DHvap –0.641853886 = 8.314 J/(mol•K) –0.000192182 Divide this result by R: –0.641853886 = (–0.000023115)mol/J DHvap Now solve for DHvap: = 27767.1971 J/mol DHvap 3. 4 = 2.777 x 10 J/mol or 27.77 kJ/mol Phase Diagrams A phase diagram is a graph that shows the temperature and pressure conditions that shows to the different various sets of conditions under which each state of matter is stable. A phase diagram is made of up regions that are separated by lines. Each region corresponds to a different state of matter. In the phase diagram for water that is shown to the right, we can see that the region for gas is at the highest temperatures and the lowest pressures. The region for the solid is at the lowest temperatures and highest pressures. The lines represent the conditions under which the two phases are in equilibrium. There are also two specific points: Triple point: all three phases exist in equilibrium at these conditions (point A) Critical point: temperature above which liquid cannot exist. Chapter 11 Part 1 6 of 7 a. Melting–Point Curve On the diagram above, the melting point curve is the from A to B. Crossing this line from left to right is melting. Crossing this line from right to left is freezing. When the solid is more dense than the liquid (the usual case) this line slants to the right. When the solid is less dense than the liquid (as is the case for water and is shown above) this line slants to the left. b. Vapor–Pressure Curves for the Liquid and the Solid On the diagram above, the line from A to C represents the equilibrium between liquid and gas. Crossing this line from left to right is boiling. Crossing this line from right to left is condensation. On the diagram above, the line from A to D represents the equilibrium between solid and gas. Crossing this line from left to right is sublimation. Crossing this line from right to left is deposition or condensation. c. Chapter 11 Critical Temperature and Pressure On the graph above, point C is the critical point. The temperature at this point is called the critical temperature. The pressure corresponding to this temperature is called the critical pressure. Part 1 7 of 7 ...
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## This note was uploaded on 04/02/2008 for the course CHM 1220 taught by Professor Barber during the Fall '07 term at Wayne State University.

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