Intermolecular Forces 1 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Intermolecular Forces Intermolecular forces are attractive forces between molecules. Intramolecular forces hold atoms together in a molecule. Intermolecular vs Intramolecular • 41 kJ to vaporize 1 mole of water (inter) • 930 kJ to break all O-H bonds in 1 mole of water (intra) “Measure” of intermolecular force Generally, intermolecular boiling point forces are much weaker melting point than intramolecular forces. ΔHvap ΔHfus 2 ΔHsub Intermolecular Forces Dipole-Dipole Forces Attractive forces between polar molecules Orientation of Polar Molecules in a Solid 3 Intermolecular Forces Ion-Dipole Forces Attractive forces between an ion and a polar molecule Ion-Dipole Interaction 4 Interaction Between Water and Cations in solution 5 Intermolecular Forces Dispersion Forces Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules ion-induced dipole interaction dipole-induced dipole interaction 6 Induced Dipoles Interacting With Each Other 7 Intermolecular Forces Dispersion Forces Continued Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted. Polarizability increases with: • greater number of electrons • more diffuse electron cloud Dispersion forces usually increase with molar mass. 8 What type(s) of intermolecular forces exist between each of the following molecules? HBr HBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules. CH4 CH4 is nonpolar: dispersion forces. S SO2 SO2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO2 molecules. 9 Intermolecular Forces Hydrogen Bond The hydrogen bond is a special dipole-dipole interaction between they hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. A H … B or A H … A A & B are N, O, or F 10 Hydrogen Bond HCOOH and water 11 Why is the hydrogen bond considered a “special” dipole-dipole interaction? Decreasing molar mass Decreasing boiling point 12 Gases 13 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Elements that exist as gases at 250C and 1 atmosphere 14 15 Physical Characteristics of Gases • Gases assume the volume and shape of their containers. • Gases are the most compressible state of matter. • Gases will mix evenly and completely when confined to the same container. • Gases have much lower densities than liquids and solids. 16 NO2 gas Force Pressure = Area (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa 17 Gases Very weak intermolecular forces: Van der Waals and London. Gases occupy all the volume of the container. Pressure: gas pressure is due to the particle collisions onto recipient surface. Pressure increases with T, because in this way increases the Ekin. of the particles, or decreasing the recipient volume (or by adding gas inside the recipient). Four parameters are necessary to characterize a gas: mass, volume, pressure and temperature. Ideal gases To characterize a gas we need a function, which depends on four parameters f(m, V, T, P) Model: Ideal gas The function describing the state of an ideal gas is called: ideal gas equation. Ideal Gases 1. A gas is composed of molecules that are separated from each other by distances far greater than their own dimensions. The molecules can be considered to be points; that is, they possess mass but have negligible volume. 2. Gas molecules are in constant motion in random directions, and they frequently collide with one another. Collisions among molecules are perfectly elastic. 3. Gas molecules exert neither attractive nor repulsive forces on one another. 4. The average kinetic energy of the molecules is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy KE = ½ mu2 20 Ideal Gases • The definition of ideal gas removes the chemical differences among gases, so allowing the definition of general laws. • Definition of experimental laws: they were derived from experimental observations • Boyle, Charles and Volta Gay-Lussac 21 Ideal Gases Real gases behave as ideal gases at low pressure (low concentration) and high temperature (high Ekin) Boyle Law – constant Temperature PV=cost PoVo = P1V1 Boyle’s Law P ∝ 1/V P x V = constant Constant temperature Constant amount of gas P1 x V1 = P2 x V2 23 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P x V 726 mmHg x 946 mL P = 1 1 = = 4460 mmHg 2 154 mL V2 24 Ideal gases Charles Law – constant Pressure V/T = cost Vo/To = V1/T1 Increasing T, increases V: Vt=V0(1+αt) with α=1/273,15 Ideal gases: Charles law Vt=V0(1+αt) Ideal gases: Charles law Repeating the experiment with different gases or different starting V, we can observe different linear curves, but the intercept with X axis is always at -273,15 °C, corresponding to V = 0: this is the minimum value of T. T(K)= t(°C) + 273,15 A sample of carbon monoxide gas occupies 3.20 L at 125 °C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? ° T1 = 125 ( C) + 273.15 (K) = 398.15 K V x T 1.54 L x 398.15 K T = 2 1 = = 192 K 2 3.20 L V1 28 Ideal gases: Avogadro law It is a consequence of the Charles law, although it was independently defined. The volume of a gas is always the same with equal number of moles (with the same conditions of T and P) In particur at 0 °C e 1 atm (normal conditions), the volume occupied by a mole of a gas is V0 = 22,414 liters. Vt/T=V0/T0 Avogadro’s Law V ∝ number of moles (n) Constant temperature Constant pressure V = constant x n V1 / n1 = V2 / n2 30 Ideal gases Volta Gay-Lussac Law – constant Volume P/T = cost Po/To = P1/T1 Increasing T, increases P: Pt=P0(1+βt) with β=1/273,15 Pt/T=P0/T0 Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 1 volume NH3 1 volume NO 32 Summary of Gas Laws Boyle’s Law 33 Charles Law 34 Avogadro’s Law 35 Ideal Gas Equation 1 Boyle’s law: P ∝ (at constant n and T) V Charles’ law: V ∝ T (at constant n and P) Avogadro’s law: V ∝ n (at constant P and T) nT V ∝ P nT nT V = constant x = R R is the gas constant P P PV = nRT 36 The conditions 0 °C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K) 37 What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 °C = 273.15 K P = 1 atm PV = nRT 1 mol HCl nRT n = 49.8 g x = 1.37 mol V = 36.45 g HCl P 1.37 mol x 0.0821 L • atm x 273.15 K mol•K V = 1 atm V = 30.7 L 38 Density (d) Calculations PM m is the mass of the gas in g d = m = V RT M is the molar mass of the gas Molar Mass (M ) of a Gaseous Substance dRT = d is the density of the gas in g/L M P Related density dA/dB = MA / MB 39 A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 °C. What is the molar mass of the gas? dRT m 4.65 g g = d = = = 2.21 M P V 2.10 L L g L•atm 2.21 x 0.0821 x 300.15 K L mol•K M = 1 atm M = 54.5 g/mol 40 Gas Stoichiometry 0 What is the volume of CO2 produced at 37 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) g C6H12O6 mol C6H12O6 mol CO2 V CO2 1 mol C6H12O6 6 mol CO2 5.60 g C6H12O6 x x = 0.187 mol CO2 180 g C6H12O6 1 mol C6H12O6 L•atm 0.187 mol x 0.0821 x 310.15 K nRT mol•K V = = = 4.76 L P 1.00 atm 41 Gas mixtures For ideal gases there is no differences in the application of gas laws for pure gases or mixtures of them, since there are no chemical differences. Each component of a mixture can access to the recipient volume and contributes to the total pressure by a partial pressure. The partial pressure of a gas is the pressure exerted by the gas present alone into the recipient: Dalton Law P = P1 + P2 + P3 + ……Pn = ∑Pi Gas mixture For the l’i-th component: PiV = niRT (1) For the mixture: ∑PiV = ∑niRT (2) con ∑ni = n1 + n2 + n3 + ……nn = n Dividing (1) with (2) PiV/ PV = niRT/ nRT Pi/ P = ni/ n = χi= molar fraction Pi = χi P Dalton’s Law of Partial Pressures V and T are constant P P = P + P P1 2 total 1 2 44 Consider a case in which two gases, A and B, are in a container of volume V.