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1 Third Year Chemistry 2 nd semester: Physical (2007-2008) May exams Physical: 4 lecturers 8 topics Dónal Leech: one topic Thermodynamics Mixtures and phase diagrams
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2 Phase Equilibria Phase transitions Changes in phase without a change in chemical composition Gibbs Energy is at the centre of the discussion of transitions Molar Gibbs energy G m = G/n Depends on the phase of the substance A substance has a spontaneous tendency to change into a phase with the lowest molar Gibbs energy
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3 Variation of G with pressure We can derive (see derivation 5.1 in textbook) that G m = V m p Therefore G m >0 when p>0 Can usually ignore pressure dependence of G for condensed states Can derive that, for a gas: G m = RT ln(p f /p i )
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4 To be presented in Lecture
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5 Variation of G with temperature G m = -S m T Can help us to understand why transitions occur The transition temperature is the temperature when the molar Gibbs energy of the two phases are equal. The two phases are in EQUILIBIRIUM at this temperature
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6 Phase diagrams Map showing conditions of T and p at which various phases are thermodynamically stable At any point on the phase boundaries, the phases are in dynamic equilibrium
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7 Location of phase boundaries Clapeyron equation (see derivation 5.4) Clausius-Clapeyron equation (derivation 5.5) Constant is vap S/R
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8 Derivations dG m = V m dp – S m dT dG m (1) = dG m (2) V m (1)dp – S m (1)dT = V m (2)dp – S m (2)dT {V m (2) – V m (1)}dp = {S m (2) – S m (1)}dT trs V dp = trs S dT T trs V dp = trs H dT dp/dT = trs H/(T trs V)
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9 Derivations: liquid-vapour transitions l To be presented in lecture
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10 Characteristic points When vapour pressure is equal to external pressure bubbles form: boiling point Normal bp: 1 atm, Standard bp: 1 bar When a liquid is heated in a closed vessel the liquid density eventually becomes equal to the vapour density: a supercritical fluid is formed.
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11 Using the C-C equation The vapour pressure of mercury is 160 mPa at 20°C. What is its vapour pressure at 50°C given that its enthalpy of vapourisation is 59.3 kJ/mol? The vapour pressure of pyridine is 50.0 kPa at 365.7 K and the normal boiling point is 388.4 K. What is the enthalpy of vapourisation of pyridine? Estimate the normal and standard boiling point of benzene given that its vapour pressure is 20.0kPa at 35°C and 50.0kPa at 58.8°C. Remember: BP: temperature at which the vapour pressure of the liquid is equal to the prevailing atmospheric pressure. At 1atm pressure: Normal Boiling Point (100°C for water) At 1bar pressure: Standard Boiling Point (99.6°C for water; 1bar=0.987atm, 1atm = 1.01325bar)
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12 Phase Rule Can more than 3 phases co-exist (for a single substance)? Gibbs energies are equal: G m (1)= G m (2) G m (2)= G m (3) G m (3)= G m (4) All a function of p and T. Need to solve 3 equations for 2 unknowns: impossible! F = C-P+2 Phase rule
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13 CO 2 Dry ice fog-special effects Supercritical fluids Caffeine extraction from coffee beans Dry-cleaning Polymerisations Chromatography
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14 Water Ice I structure Solid-liquid boundary slopes to the left with increasing pressure volume decreases when ice melts, liquid is denser that solid at 273 K
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15 Introduction to mixtures Homogeneous mixtures of a solvent (major component) and solute (minor component). Introduce partial molar property: contribution that a substance makes to overall property. V = n A V A + n B V B Note: can be negative, if adding solute to solvent results in decrease in total volume (eg MgSO 4 in water)
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16 The chemical potential, We can extend the concept of partial molar properties to state functions, such as Gibbs energy, G. This is so important that it is given a special name and symbol, the chemical potential, . G = n A G A + n B G B G = n A A + n B B
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17 The chemical potential of perfect gases in a mixture Recall that G m (p f ) = G m (p i ) + RT ln (p f /p i ) At standard pressure G m (p) = G m ° + RT ln (p/p°) Therefore, for a mixture of gases J = J ° + RT ln (p J /p°) More simply (at p° = 1 bar) J = J ° + RT ln p J System is at equilibrium when for each substance has the same value in every phase
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18 Spontaneous mixing to be presented in lecture
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19 Gas mixtures Compare G mix = nRT {x A ln x A + x B ln x B } G = H – T S Therefore H = S mix = − nR {x A ln x A + x B ln x B } Perfect gases mix spontaneously in all proportions
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20 Ideal Liquid Solutions p J = x J p J * Due to effect of solute on entropy of solution Raoult’s Law
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21 Real Solutions
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22 Chemical potential of a solvent At equilibrium A (g) = A (l) A (l)= A °(g) + RT ln p A A (l)= A °(g) + RT ln x A p A * A (l)= A °(g) + RT ln p A * + RT ln x A └────────────────┘ A * A (l)= A *+ RT ln x A
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23 Is solution formation spontaneous? G = n A A + n B B Can show that G mix = nRT {x A ln x A + x B ln x B } and H = S mix = −nR {x A ln x A + x B ln x B }
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24 Ideal-dilute solutions Raoult’s law generally describes well solvent vapour pressure when solution is dilute, but not the solute vapour pressure Experimentally found (by Henry) that vp of solute is proportional to its mole fraction, but proportionality constant is not the vp of pure solute. Henry’s Law p B = x B K B
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25 Gas solubility Henry’s law constants for gases dissolved in water at 25°C Concentration of 4 mg/L of oxygen is required to support aquatic life, what partial pressure of oxygen can achieve this?
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26 Application-diving Table 1 Increasing severity of nitrogen narcosis symptoms with depth in feet and pressures in atmospheres. DepthP TotalP N 2 Symptoms 1004.03.0Reasoning measurably slowed. 1505.54.3Joviality; reflexes slowed; idea fixation. 2007.15.5 Euphoria; impaired concentration; drowsiness. 2508.36.4 Mental confusion; inaccurate observations. 30010.7.9Stupefaction; loss of perceptual faculties. Gas narcosis caused by nitrogen in normal air dissolving into nervous tissue during dives of more than 120 feet [35 m] Pain due to expanding or contracting trapped gases, potentially leading to Barotrauma. Can occur either during ascent or descent, but are potentially most severe when gases are expanding. Decompression sickness due to evolution of inert gas bubbles.
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27 Real Solutions-Activities J = J ° + RT ln a J
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28 Colligative properties Properties of solutions that are a result of changes in the disorder of the solvent, and rely only on the number of solute particles present Lowering of vp of pure liquid is one colligative property Freezing point depression Boiling point elevation Osmotic pressure
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29 Colligative properties Chemical potential of a solution (but not vapour or solid) decreases by a factor ( RT ln x A ) in the presence of solute Molecular interpretation is based on an enhanced molecular randomness of the solution Get empirical relationship for FP and BP (related to enthalpies of transition)
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30 Cryoscopic and ebullioscopic constants
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31 Osmotic pressure Van’t Hoff equation
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32 Phase diagrams of mixtures We will focus on two- component systems ( F = 4 ─ P ), at constant pressure of 1 atm ( F’ = 3 ─ P ), depicted as temperature- composition diagrams.
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33 Fractional Distillation-volatile liquids Important in oil refiningoil refining
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34 Exceptions-azeotropes Azeotrope: boiling without changing High-boiling and Low-boiling Favourable interactions between components reduce vp of mixture Trichloromethane/propanone HCl/water (max at 80% water, 108.6°C) Unfavourable interactions between components increase vp of mixture Ethanol/water (min at 4% water, 78°C)
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35 Liquid-Liquid (partially miscible) Hexane/nitrobenzene as example Relative abundances in 2 phases given by Lever Rule n’l’ = n’’l’’ Upper critical Temperature is limit at which phase separation occurs. In thermodynamic terms the Gibbs energy of mixing becomes negative above this temperature
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36 Other examples Water/triethylamine Weak complex at low temperature disrupted at higher T. Nicotine/water Weak complex at low temperature disrupted at higher T. Thermal motion homogenizes mixture again at higher T.
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37 Liquid-solid phase diagrams
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