CHE140 Lecture Notes 11.2 - 11.13
CHE140 Lecture Notes 11.2 - 11.13 CHE140
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Date Created: 11/13/15
CHE140 Lecture Notes 11.2.15 – 11.13.15 11.2.15 Hydrogen Bonds Strongest Dipole – Dipole Forces London Dispersion Forces Weakest Hydrogen Bonds Any H bonded directly to O, H, or N are called bond donors O, H, and N are bond acceptors Dispersion Forces (London Dispersion Forces) Weakest intermolecular force The only intermolecular force present in non-polar molecules Always present – in all molecules Caused by a temporary dipole o A temporary dipole is caused by a temporary upset due to an uneven distribution of electrons (happens very quickly) Can we predict the strength of dispersion forces? Strength is determined by size and number of electrons Polarizability – the tendency of the electrons in an atom or molecule to be distorted, creating a temporary dipole o High polarizability stronger dispersion forces Larger molecules with more electrons o Low polarizability weaker dispersion forces Smaller molecules with less electrons Hydrocarbons – molecules with C and H only o Non-polar How can we measure the strength of intermolecular forces? Boiling (and melting) points o o Molecule Melting Point ( C) Boiling Point ( C) CH 4 -187 161 CF 4 -152.2 -82.1 CCl 4 -22.9 +76.7 CBr 4 +91 +189.5 CI 4 +171 - ** Different structures will change the boiling point Molecule Boiling Point (oC) H 3CH CH2CH 2 3 -0.5 H 3CH CH2CH 2H CH2 2 3 +69 H 3 – CH – CH – 2H – CH2 3 +60 | CH 3 H 3 – CH – CH – CH 3 +57.9 | | CH C3 3 Now we can look at structures and determine what kind of intermolecular force (IMF) it has. London Dispersion and Dipole – Dipole: London Dispersion because it is present in every molecule, and Dipole-Dipole because the O with two lone pairs creates a slight negative charge, or dipole H 3 C = O (O having two lone pairs) H 3 London Dispersion and Dipole – Dipole: London Dispersion because it is present in every molecule, and Dipole-Dipole because the O with two lone pairs creates a slight negative charge, or dipole H 3CH CH2OH 2 London Dispersion, Dipole-Dipole, and Hydrogen: London Dispersion because it is present in every molecule, Dipole-Dipole because the OH bond is polar, and Hydrogen bond because the H is bonded directly to an O H 3CH CH2CH 2 3 London Dispersion only: because London Dispersion is present in every molecule 11.4.15 Lecture Polarity, IMFs and Solubility Predict solubility using the “like dissolves like” rule o Polar compounds are soluble in polar solvents o Non-polar compounds are soluble in non-polar solvents Miscible Two liquids are miscible if they can form a homogenous solution in all proportions o AKA solubility of liquids Water, H O is a polar solvent; Hexane, C H is a non-polar solvent 2 6 14 o Sugar, C H 12is22o11r, and would be soluble in water, but not hexane o Naphthalene, C H i6 n10-polar (only carbon and hydrogen), and would be soluble in hexane, but not water Solubility and Structure H C – (CH ) – OH 3 2 x #C Structure Solubility in H 2 1 H 3 – OH Fully miscible 2 H 3 – CH – 2H Fully miscible 3 H 3 – CH CH2– O2 Fully miscible 4 H 3 – CH CH2CH 2 OH 2 Miscible ≤ 9.1mL/100mL H 2 6 H 3 – (CH ) 2 5H Miscible ≤ 0.59mL/100mL H 2 8 H 3 – (CH ) 2 7H Insoluble The more CH groups present, the less miscible the substance is going to be 2 Solution Process Ions will start dissolving from the corners, exposing new corners, until the solute is completely dissolved Dissolving increases the entropy of the system o Entropy – measure of the amount of disorder Most solids are more soluble as the temperature increases (energy input) All gases are less soluble as the temperature increases o The solubility of gases increases with increasing pressure Henry’s Law C gas K H P gas K is the constant and P is the pressure Hydrophillic – water loving/dipole-dipole bonding and hydrogen bonding Hydrophobic – water fearing/London dispersion Phase Diagram 11.6.15 Surface Tension Surface tension – a liquid’s resistance to increasing its surface area o The energy required to separate the molecules at a liquids surface The stronger the IMFs, the greater the surface tension Meniscus – curve in the surface of water due to surface tension The downward meniscus shown with water is called a cohesive force The upward meniscus shown with mercury is called an adhesive force Capillary Action Capillary action is due to IMFs Capillary action – the ability of a liquid to flow in narrow spaces without the assistance of external forces like gravity Viscosity – a liquid’s resistance to flow o The stronger the IMFs, the higher the viscosity Water Liquid at room temperature (unlike H S2 H S2, etc. which are gases at room temperature) Maximum density at +4 C (1.00g/mL) Ice is less dense (0.92g/mL) than liquid water Enthalpy of Solution, ∆H sol ∆H solH ion-ion∆H solvent-solvention-dipole ∆H ion-ionnergy in) is the lattice energy (u) o u = K x Q1 x Q2 d ∆H solvent-solvent ion-dipoleergy out) is ∆H hydrationydration enthalpy) ∆H sol= ∆H hyd– u Calculate u using the Born-Haber Cycle Na(g) + ½ Cl (g) NaCl(s) ∆H of + - Na (aq) + Cl (aq) NaCl(s) ∆H = u Whether using the standard enthalpy of formation or the lattice energy, you get NaCl 11.9.15 Lattice Energy Calculated using the Born-Haber Cycle u = K x Q1 x Q2 d ∆H =f2(∆sub) + 2(IE) + ½ (BDE – Enthalpy of dissociation) + (EA1 + EA2) + u -416 = 2(107) + 2(496) + ½ (499) + 603 + u -416 = 2058.5 + u u = -2475 kJ/mol Colligative Properties Properties of solutions that depend on the amount of solute particles, not the identity of the state 0.3M NaCl 0.6M ions o Multiply the 3 times 2 ions per unit NaCl 0.2M CaCl 2.6M ions o Multiply the 2 times 3 ions per unit CaCl2 0.6M C H12 22.11 molecules o Molecular structure, not ionic, so the M stays the same Most colligative properties depend on the molality (cursive m) molality = mole solute kg solvent Boiling Point Elevation ∆Tb = Kb x molality Kb = 0.52 C/molality Freezing Point Depression ∆Tf = Kf x molality Kf = 1.86oC/molality What are the melting and boiling points of a solution of 45.6g sugar (C12 22, 11 = 342.34 g/mol) in 250.g H2O? 45.6g C 12O 22 11 x 1000g sugar x 1mol sugar 250.g solvent 1kg solvent 342.34g = 0.533 molality o ∆Tb = (0.52)(0.533) = 0.28 C ∆Tb = 100.00 + 0.28 = 100.28 Co ∆Tf = (1.86)(0.533) = 0.99 C o ∆Tf = 0.00 – 0.99 = -0.99 C 11.11.15 Lecture Vapor Pressure The amount of a substance in the gas phase above a liquid Stronger IMFs, lower vapor pressure in pure liquids Solutions will have a lower vapor pressure than the pure solvent Vant Hoff Factor (i) A measure of the number of particles in solution per molecule/formula unit MgCl2 Mg 2+ + 2Cl- Ideal I = 3 (ions) ∆Tb = i x Kb x molality ∆Tf = i x Kf x molality The more ions in solution the further from the Vant Hoff Factor it will be Salt Ideal i Experimental i AlCl 3 4 3.2 MgCl 2 3 2.7 Ca(NO )3 2 3 2.5 Osmosis The flow of a fluid through a semipermeable membrane in order to balance the concentration of solute on both sides Solvent flows from the dilute solution to the more concentrated solution Dilute Concentrated More concentrated More Dilute 11.13.15 Lecture Osmosis – the flow of solvent through a semipermeable membrane to equalize the concentration of two solutions Osmotic pressure (∏) – the amount of pressure needed to stop osmotic flow ∏ = i x M x R x T o i is the Vant Hoff Factor o M is molarity o R is the ideal gas constant 0.0821 L x atm/mol x K ∆∏ = D(i x M) x R x T o D(i x M) is the difference of the two concentrations o i x1M – 1 x M2 2 What is the osmotic pressure across a semipermeable membrane separating a 2.04M CaCl 2 solution and a 2.21M NaCl solution at 25oC? iNaCl = 1.9 iCaCl2 = 2.4 T = 25 + 273 = 298K ∆∏ = [(2.4)(2.04M) – (1.9)(2.21M)] x 0.0821 x 298 ∆∏ = 0.697 x 0.0821 x 298 ∆∏ = 17.1 atm Reverse Osmosis Applying a pressure greater than ∏ to reverse the flow Determining the Molar Mass using Colligative Properties A solution of 19.4mg of a PCB in 1.00g CCl ha4 its freezing point lowered by 1.60 C. What is the molar mass of the PCB? o Kf = 29.8 C/molality ∆Tf = Kf x molality Molality = ∆Tf/Kf = 1.60 C/29.6 C/molality = 0.053691molality Molality = moles PCB Kg CCl 4 Moles PCB = molality x kg CCl4 = (0.053691)(0.00100kg) Moles PCB = 5.3691 x 10 mol PCB MM = g/mol = 0.0194g PCB -5 5.3691 x 10 mol PCB = 361 g/mol Fractional Distillation Mixture of liquids can be distilled (boiled off) The composition of the vapor above a mixture of liquids has a different composition o Each solvent has a different vapor pressure The composition of the vapor above a mixture of liquids is determined by Raoult’s law o o P solutionX solventP solvent o X solvents the mole fraction o P solven is the vapor pressure Vapor Pressure and Absolute Temperature Vapor pressure boiling point = atmospheric pressure Clausis-Clapeyron Equation Top left of the line – higher temp/pressure Bottom right of the line – lower temp/pressure
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