Week lectures--tentative 0.7 Kinetic-Molecular Theory 40 Application to the Gas Laws 0.8 Molecular Effusion and Diffusion 43 Graham's Law of Effusion Diffusion and Mean Free Path 0.9 Real Gases: Deviations 47 The van der Waals Equation Chapter Intermolecular Forces, Liquids, and Solids. A Molecular Comparison of Liquids and Solids 444. Intermolecular Forces 445 Ion-Dipole Forces Dipole-Dipole Forces London Dispersion Forces Hydrogen Bonding Comparing Intermolecular Forces.3 Some Properties of Liquids 453 Viscosity Surface Tension.4 Phase Changes 455 Energy Changes Accompanying Phase Changes Heating Curves Critical Temperature and Pressure.5 Vapor Pressure 459 Explaining Vapor Pressure on the Molecular Level Volatility, Vapor Pressure, and Temperature Vapor Pressure and Boiling Point 0.7 Theory developed to explain gas behavior. Theory based on properties at the molecular level. Kinetic molecular theory gives us a model for understanding pressure and temperature at the molecular level. Pressure of a gas results from the number of collisions per unit time on the walls of container. 0.7 There is a spread of individual energies of gas molecules in any sample of gas. As the temperature increases, the average kinetic energy of the gas molecules increases. Assumptions: Gases consist of a large number of molecules in constant random motion. Volume of individual molecules negligible compared to volume of container. Intermolecular forces (attractive or repulsive forces between gas molecules) are negligible. Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature. Average kinetic energy of molecules is proportional to temperature. Magnitude of pressure given by how often and how hard the molecules strike. Gas molecules have an average kinetic energy. Each molecule may have a different energy. As kinetic energy increases, the velocity of the gas molecules increases. Root mean square speed, u, is the speed of a gas molecule having average kinetic energy. Average kinetic energy, ε, is related to root mean square speed: ε = mu
Do you remember how to calculate v xy from v x and v y? xy ( v v ) v = + And how about v from all three components? x y The Maxwell-Boltzmann Distribution of Velocities Note that the mean value of velocity is zero! [ v + v v ] v = + x y z Remember these equations!! They ll pop up again in Chap.. The Maxwell Distribution of Speeds Most Probale Speed = v mp RT = M = u mp Average Speed = 8RT v = π M = < u > = u rms Speed = v rms 3RT = M = u rms = < u > 8 And, v mp : v : v rms = : : 3 π = :.8 :.5 u mp <u> The Maxwell-Boltzmann Distribution of Velocities u rms This is also the form of a Gaussian (normal) distribution, where u mp = <u> = u rms.
Application to Gas Laws As volume increases at constant temperature, the average kinetic of the gas remains constant. Therefore, u is constant. However, as the volume increases the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases. If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases. Molecular Effusion and Diffusion As kinetic energy increases, the velocity of the gas molecules increases. Average kinetic energy of a gas is related to its mass: ε = mu Consider two gases at the same temperature: the lighter gas has a higher velocity than the heavier gas. Mathematically: u = 3RT M Molecular Effusion and Diffusion The lower the molar mass, M, the higher the u rms. 3RT M = u rms SAMPLE EXERCISE 0.4 Calculating a Root-Mean-Square Speed Calculate the rms speed, u, of an N molecule at 5 C. Solution Analyze: We are given the identity of the gas and the temperature, the two quantities we need to calculate the rms speed. Plan: We will calculate the rms speed using Equation 0.. M Solve: In using Equation 0., we should convert each quantity to SI units so that all the units are compatible. We will also use R in units of J/mol-K (Table 0.) in order to make the units cancel correctly. (These units follow from the fact that J = kg-m /s ) 3RT u rms = Comment: This corresponds to a speed of 50 mi/hr. Because the average molecular weight of air molecules is slightly greater than that of N, the rms speed of air molecules is a little slower than that for N. The speed at which sound propagates through air is about 350 m/s, a value about two-thirds the average rms speed for air molecules. Graham s Law of Effusion As kinetic energy increases, the velocity of the gas molecules increases. Effusion is the escape of a gas through a tiny hole. The rate of effusion can be quantified. Graham s Law of Effusion Consider two gases with molar masses M and M, the relative rate of effusion is given by: r M = r M Only those molecules that hit the small hole will escape through it. Therefore, the higher the u rms the greater the likelihood of a gas molecule hitting the hole. 3
SAMPLE EXERCISE 0.5 Applying Graham s Law An unknown gas composed of homonuclear diatomic molecules effuses at a rate that is only 0.355 times that of O at the same temperature. Calculate the molar mass of the unknown, and identify it. Solution Analyze: We are given the rate of effusion of an unknown gas relative to that of O, and we are asked to find the molar mass and identity of the unknown. Thus, we need to connect relative rates of effusion to relative molar masses. Plan: We can use Graham s law of effusion, Equation 0.3, to determine the molar mass of the unknown gas. If we let r x and represent the rate of effusion and molar mass of the unknown gas, Equation 0.3 can be written as follows: Solve: From the information given, SAMPLE EXERCISE 0.5 continued We now solve for the unknown molar mass, Because we are told that the unknown gas is composed of homonuclear diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I. PRACTICE EXERCISE Calculate the ratio of the effusion rates of Thus, Graham s Law of Effusion Consider two gases with molar masses M and M, the relative rate of effusion is given by: 3RT r u M M = = = r u 3RT M M Only those molecules that hit the small hole will escape through it. Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole. Diffusion and Mean Free Path Diffusion of a gas is the spread of the gas through space. Diffusion is faster for light gas molecules. Diffusion is significantly slower than rms speed (consider someone opening a perfume bottle: it takes while to detect the odor but rms speed at 5 C is about 50 mi/hr). Diffusion is slowed by gas molecules colliding with each other. Diffusion and Mean Free Path Average distance of a gas molecule between collisions is called mean free path. At sea level, mean free path is about 6 0-6 cm. Real Gases: Deviations From the ideal gas equation, we have PV = n RT or PV nrt For mol of gas, PV/nRT = for all pressures. In a real gas, PV/nRT varies from significantly and is called Z. Z = PV nrt The higher the pressure the more the deviation from ideal behavior. = 4
Real Gases: Deviations From the ideal gas equation, we have PV = n RT For mol of gas, PV/RT = for all temperatures. As temperature increases, the gases behave more ideally. The assumptions in kinetic molecular theory show where ideal gas behavior breaks down: the molecules of a gas have finite volume; molecules of a gas do attract each other. Real Gases: Deviations As the pressure on a gas increases, the molecules are forced closer together. As the molecules get closer together, the volume of the container gets smaller. The smaller the container, the more space the gas molecules begin to occupy. Therefore, the higher the pressure, the less the gas resembles an ideal gas. Real Gases: Deviations As the gas molecules get closer together, the smaller the intermolecular distance. Real Gases: Deviations The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules. Therefore, the less the gas resembles and ideal gas. As temperature increases, the gas molecules move faster and further apart. Also, higher temperatures mean more energy available to break intermolecular forces. 5
Real Gases: Deviations Therefore, the higher the temperature, the more ideal the gas. Real Gases: Deviations The van der Waals Equation We add two terms to the ideal gas equation one to correct for volume of molecules and the other to correct for intermolecular attractions The correction terms generate the van der Waals equation: nrt n a P = V nb V where a and b are empirical constants characteristic of each gas. Real Gases: Deviations The van der Waals Equation nrt n a P = V nb V Corrects for molecular volume Corrects for molecular attraction General form of the van der Waals equation: n a P + ( V nb) = nrt V SAMPLE EXERCISE 0.6 Using the van der Waals Equation If.000 mol of an ideal gas were confined to.4 L at 0.0 C, it would exert a pressure of.000 atm. Use the van der Waals equation and the constants in Table 0.3 to estimate the pressure exerted by.000 mol of Cl (g) in.4 L at 0.0 C. Solution Analyze: The quantity we need to solve for is pressure. Because we will use the van der Waals equation, we must identify the appropriate values for the constants that appear there. Plan: Using Equation 0.6, we have SAMPLE EXERCISE 0.6 continued Comment: Notice that the first term,.003 atm, is the pressure corrected for molecular volume. This value is higher than the ideal value,.000 atm, because the volume in which the molecules are free to move is smaller than the container volume,.4 L. Thus, the molecules must collide more frequently with the container walls. The second factor, 0.03 atm, corrects for intermolecular forces. The intermolecular attractions between molecules reduce the pressure to 0.990 atm. We can conclude, therefore, that the intermolecular attractions are the main cause of the slight deviation of Cl (g) from ideal behavior under the stated experimental conditions. Solve: Substituting n =.000 mol, R = 0.0806 L-atm/mol-K, T = 73. K, V =.4 L, a = 6.49 L - atm/mol, and b = 0.056 l/mol: Check: We expect a pressure not far from.000 atm, which would be the value for an ideal gas, so our answer seems very reasonable. 6
SAMPLE INTEGRATIVE EXERCISE Putting Concepts Together Cyanogen, a highly toxic gas, is composed of 46.% C and 53.8% N by mass. At 5 C and 75 torr,.05 g of cyanogen occupies 0.500 L. (a) What is the molecular formula of cyanogen? (b) Predict its molecular structure. (c) Predict the polarity of the compound. Solution Analyze: First we need to determine the molecular formula of a compound from elemental analysis data and data on the properties of the gaseous substance. Thus, we have two separate calculations to do. (a) Plan: We can use the percentage composition of the compound to calculate its empirical formula. (Section 3.5) Then we can determine the molecular formula by comparing the mass of the empirical formula with the molar mass. (Section 3.5) Solve: To determine the empirical formula, we assume that we have a 00-g sample of the compound and then calculate the number of moles of each element in the sample: SAMPLE INTEGRATIVE EXERCISE continued The molar mass associated with the empirical formula, CN, is.0 + 4.0 = 6.0 g/mol. Dividing the molar mass of the compound by that of its empirical formula gives (5.0 g/mol)/(6.0 g/mol) =.00. Thus, the molecule has twice as many atoms of each element as the empirical formula, giving the molecular formula C N (b) Plan: To determine the molecular structure of the molecule, we must first determine its Lewis structure. (Section 8.5) We can then use the VSEPR model to predict the structure. (Section 9.) Solve: The molecule has (4) + (5) = 8 valence-shell electrons. By trial and error, we seek a Lewis structure with 8 valence electrons in which each atom has an octet and in which the formal charges are as low as possible. The following structure meets these criteria: (This structure has zero formal charges on each atom.) The Lewis structure shows that each atom has two electron domains. (Each nitrogen has a nonbonding pair of electrons and a triple bond, whereas each carbon has a triple bond and a single bond.) Thus the electrondomain geometry around each atom is linear, causing the overall molecule to be linear. Because the ratio of the moles of the two elements is essentially :, the empirical formula is CN. To determine the molar mass of the compound, we use Equation 0.. (c) Plan: To determine the polarity of the molecule, we must examine the polarity of the individual bonds and the overall geometry of the molecule. Solve: Because the molecule is linear, we expect the two dipoles created by the polarity in the carbon nitrogen bond to cancel each other, leaving the molecule with no dipole moment. Chapter -- Intermolecular Forces, Liquids, and Solids In many ways, this chapter is simply a continuation of our earlier discussion of real gases. Remember this nice, regular behavior described by the ideal gas equation. This plot for SO is a more representative one of real systems!!! This plot includes a realistic one for Volume as a function of Temperature! 7
Why do the boiling points vary? Is there anything systematic? What are the important Intermolecular Forces i.e, forces between molecules? What determines whether a substance exists as a gas, liquid, or solid? Two primary factors are involved: Kinetic Energy of the particles. Strength of attractions between the particles. Note that earlier chapters concentrated on Intramolecular Forces, those within the molecule. Important ones: ion-ion ion-dipole dipole-dipole dipole-induced dipole similar to atomic systems (review definition of dipoles) London Dispersion Forces: (induced dipole-induced dipole) related to polarizability van der Waals forces Hydrogen Bonding Relative Energies of Various Interactions How do you know the relative strengths of each? Virtually impossible experimentally!!! Most important though: Establish which are present. London Dispersion Forces: Always All others depend on defining property such as existing dipole for d-d. It has been possible to calculate the relative strengths in a few cases. d-d d-id disp Ar 0 0 50 N 0 0 58 C 6 H 6 0 0 086 C 3 H 8 0.0008 0.09 58 HCl 6 06 CH Cl 06 33 570 SO 4 0 05 H O 90 38 HCN 77 46 8
Primary factor here is London Dispersion Forces Ion-Dipole Interactions A fourth type of force, ion-dipole interactions are an important force in solutions of ions. The strength of these forces are what make it possible for ionic substances to dissolve in polar solvents. Dipole-Dipole Forces Let s take a closer look at these interactions: Ion-dipole interaction 9
Dipole-dipole interactions. This is the simple one. But we also have to consider other shapes. Review hybridization and molecular shapes. Dipole-dipole interactions Recall the discussion of sp, sp, and sp 3 hybridization? A Polarized He atom with an induced dipole molecule F Cl Br I CH 4 polarizability.3 4.6 6.7 0..6 molecular wt. 37 7 60 54 6 Molecular Weight predicts the trends in the boiling points of atoms or molecules without dipole moments because polarizability tends to increase with increasing mass. London dispersion forces or induced dipole-induced dipole 0
Water provides our best example of Hydrogen Bonding. But polarizability also depends on shape, as well as MW. Water is also unusual in the relative densities of the liquid and solid phases. These boiling points demonstrate the enormous contribution of hydrogen bonding. But hydrogen bonding is not limited to water: The crystal structure suggests a reason for the unusual high density of ice.
But water isn t the only substance to show hydrogen bonding!.3 Some Properties of Liquids Viscosity Resistance of a liquid Viscosity the resistance to flow of a liquid, such as oil, water, gasoline, molasses, (glass!!!) Surface Tension tendency to minimize the surface area compare water, mercury Cohesive forces bind similar molecules together Adhesive forces bind a substance to a surface Capillary action results when these two are not equal to flow is called viscosity. It is related to the ease with which molecules can move past each other. Viscosity increases with stronger intermolecular forces and decreases with higher temperature. Soap reduces the surface tension, permitting one material to wet another more easily The SI unit is kg/m-s. Many tables still use the older unit of viscosity, the poise, which is g/cm-s, with typical values listed as cp = 0.0 P. Surface Tension Surface tension results from the net inward force experienced by the molecules on the surface of a liquid. Rationale for Surface Tension
Surface Tension Surface molecules are only attracted inwards towards the bulk molecules. Therefore, surface molecules are packed more closely than bulk molecules. Surface tension is the amount of energy required to increase the surface area of a liquid, in J/m. Cohesive forces bind molecules to each other. Adhesive forces bind molecules to a surface. Surface Tension Meniscus is the shape of the liquid surface. If adhesive forces are greater than cohesive forces, the liquid surface is attracted to its container more than the bulk molecules. Therefore, the meniscus is U-shaped (e.g. water in glass). If cohesive forces are greater than adhesive forces, the meniscus is curved downwards. Capillary Action: When a narrow glass tube is placed in water, the meniscus pulls the water up the tube. Remember that surface molecules are only attracted inwards towards the bulk molecules. Phase Changes Sublimation: solid gas. Vaporization: liquid gas. Melting or fusion: solid liquid. Deposition: gas solid. Condensation: gas liquid. Freezing: liquid solid. also called FUSION H vap : 40,670 J/mol SAMPLE EXERCISE.4 Calculating H for Temperature and Phase Changes Calculate the enthalpy change upon converting.00 mol of ice at 5 C to water vapor (steam) at 5 C under a constant pressure of atm. The specific heats of ice, water, and steam are.09 J/g-K, 4.8 J/g-K and.84 J/g- K, respectively. For H O, H fus = 6.0 kj/mol and H vap = 40.67 kj/mol. H fus : 6,00 J/mol C p (s): 37.6 J/mol-K C p (l): 7.4 J/mol-K C p (g): 33. J/mol-K Solution Analyze: Our goal is to calculate the total heat required to convert mol of ice at 5 C to steam at 5 C. Plan: We can calculate the enthalpy change for each segment and then sum them to get the total enthalpy change (Hess s law, Section 5.6). Solve: For segment AB in Figure.9, we are adding enough heat to ice to increase its temperature by 5 C. A temperature change of 5 C is the same as a temperature change of 5 K, so we can use the specific heat of ice to calculate the enthalpy change during this process: For segment BC in Figure.9, in which we convert ice to water at 0 C, we can use the molar enthalpy of fusion directly: The enthalpy changes for segments CD, DE, and EF can be calculated in similar fashion: 3
SAMPLE EXERCISE.4 continued The total enthalpy change is the sum of the changes of the individual steps: Check: The components of the total energy change are reasonable in comparison with the lengths of the horizontal segments of the lines in Figure.9. Notice that the largest component is the heat of vaporization..5 Vapor Pressure Explaining Vapor Pressure on the Molecular Level, Volatility, Vapor Pressure, and Temperature; Vapor Pressure and Boiling Point.6 Phase Diagrams The Phase Diagrams of H O and CO.7 Structures of Solids Unit Cells The Crystal Structure of Sodium Chloride Close Packing of Spheres.8 Bonding in Solids Molecular Solids Covalent-Network Solids Ionic Solids Metallic Solids Chapter 3 Properties of Solutions 3. The Solution Process Energy Changes and Solution Formation Solution Formation, Spontaneity, and Disorder Solution Formation and Chemical Reactions The observation of Vapor Pressure: And, at higher temperatures, the vp is higher. The same picture is useful to rationalize Vapor Pressure With the beaker covered, equilibrium is soon established. 4
Normal BP defined Contrast that with an uncovered beaker! But recall the definition of ANY Boiling Point. But, recall, we always prefer straight lines! ln P vap Hvap = + C R T (Add slide of C-C eq and plot.) Temp. increases Using the Clausius-Clapeyron Equation: Tabulate P in atm and T in K; Calculate lnp and /T Plot lnp vs /T The slope is = - H vap /R Or, with P the vapor pressure at T and P the vapor pressure at T H ln P = R vap subtract to give H vap + Cand ln P = + C T R T H vap ln P ln P = R T T or P H vap ln = P R T T The slope! 5