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Second Year Chemistry

Second Year Chemistry. 1 st semester: Organic 1 st semester: Physical (2005-2006) December exams 2 nd : Analytical & Environmental 2 nd : Inorganic Summer exams Physical: 3 lecturers Þ ~ 8 topics Dónal Leech: four topics Thermodynamics Gases, Laws & Phases , Equilibrium.

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Second Year Chemistry

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  1. Second Year Chemistry • 1st semester: Organic • 1st semester: Physical (2005-2006) • December exams • 2nd: Analytical & Environmental • 2nd: Inorganic • Summer exams • Physical: 3 lecturers Þ ~8 topics • Dónal Leech: four topics • Thermodynamics • Gases, Laws& Phases, Equilibrium

  2. Introduction Energetics and Equilibria What makes reactions “go”! This area of science is called THERMODYNAMICS Thermodynamics is expressed in a mathematical language BUT Don’t, initially anyway, get bogged down in the detail of the equations: try to picture the physical principle expressed in the equations We will develop ideas leading to one important Law, and explore practical applications along the way The Second Law of Thermodynamics

  3. Lecture Resources 12 lectures leading to four exam questions (section A, you must answer two from this section) • Main Text: “Elements of Physical Chemistry” Atkins & de Paula, 4th Edition (Desk reserve) http://www.oup.com/uk/booksites/content/0199271836/ OTHERS. “Physical Chemistry” Atkins & de Paula, 7th Edition or any other PChem textbook These notes available on NUI Galway web pages at http://www.nuigalway.ie/chem/degrees.htm See also excellent lecture notes from James Keeler, Cambridge, although topics are treated in a different running order than here.

  4. Thermodynamics: the 1st law The internal energy of an isolated system is constant Energy can neither be created nor destroyed only inter-converted Energy: capacity to do work Work: motion against an opposing force System: part of the universe in which we are interested Surroundings: where we make our observations (the universe) Boundary: separates above two

  5. System and Surroundings Systems • Open: energy and matter exchanged • Closed: energy exchanged • Isolated: no exchange • Diathermic wall: heat transfer permitted • Adiabatic wall: no heat transfer

  6. Work and Heat • Work (w): transfer of energy that changes motions of atoms in the surroundings in a uniform manner • Heat (q): transfer of energy that changes motions of atoms in the surroundings in a chaotic manner • Endothermic: absorbs heat • Exothermic: releases heat

  7. Work • Mechanical work can generally be described by dw = -F.dz • Gravitational work (mg.dh) • Electrical work (.dq) • Extension work (f.dl) • Surface expansion work (.d) As chemists we will concentrate on EXPANSION WORK (many chemical reactions produce gases) Expansion against constant external pressure dw = -F.dz but pex = F/A therefore w = -pex.DV

  8. Expansion Work Expansion against zero external pressure (free expansion) w = -pex.DV = 0 (external pressure = 0) Reversible isothermal expansion • In thermodynamics “reversible” means a process that can be reversed by an infinitesimal change of a variable. • A system does maximum expansion work when the external pressure is equal to that of the system at every stage of the expansion

  9. Isothermal reversible expansion Exercise: Calculate the work done when 1.0 mol Ar(g) confined in a cylinder of volume 1.0 dm3 at 25°C expands isothermally and reversibly to 2.0 dm3.

  10. 1st Law of Thermodynamics The internal energy of an isolated system is constant Energy can neither be created nor destroyed only inter-converted U = q+w Exercise: A car battery is charged by supplying 250 kJ of energy to it as electrical work, but in the process it loses 25kJ of energy as heat to the surroundings. What is the change in internal energy of the battery? INTERNAL ENERGY is a State Function Use calorimetry. If we enclose our system in a constant volume container (no expansion), provided no other kind of work can be done, then w = 0. U = qV How do we measure heat?

  11. Bomb calorimetry • By measuring the change in Temperature of the water surrounding the bomb, and knowing the calorimeter heat capacity, C, we can determine the heat, and hence DU. Heat Capacity Amount of energy required to raise the temperature of a substance by 1°C (extensive property) For 1 mol of substance: molar heat capacity (intensive property) For 1g of substance: specific heat capacity (intensive property)

  12. Calorimeter calibration Can calibrate the calorimeter, if its heat capacity is unknown, by passing a known electrical current for a given time to give rise to a measured temperature change. Amperes.Volts.Sec = Coulombs.Volts = Joules Exercise: In an experiment to measure the heat released by the combustion of a fuel, the compound was burned in an oxygen atmosphere inside a calorimeter and the temperature rose by 2.78°C. When a current of 1.12 A from an 11.5 V source was passed through a heater in the same calorimeter for 162 s, the temperature rose by 5.11°C. What is the heat released by the combustion reaction?

  13. Enthalpy Most reactions we investigate occur under conditions of constant PRESSURE (not Volume) ENTHALPY: Heat of reaction at constant pressure! Use a “coffee-cup” calorimeter to measure it Heat capacity Exercise: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffee-cup calorimeter, the temperature increases from 21°C to 27.5°C. What is the enthalpy change, if the density is 1g/mL and specific heat 4.18 J/g.K?

  14. Perfect gas enthalpy • Use intensive property of molar enthalpy and internal energy • At 25°C, RT = 2.5 kJ/mol Thermicity-Revision Endothermic reaction (q>0) results in an increase in enthalpy (DH>0) Exothermic reaction (q<0) results in an increase in enthalpy (DH<0) NB: Internal energy and Enthalpy are STATE FUNCTIONS

  15. Temperature variation of enthalpy Convenient empirical expression to use for heat capacity is: Exercise: What is the change in molar enthalpy of N2 when it is heated from 25°C to 100 °C, given that:

  16. Relation between heat capacities

  17. Thermochemistry Chemists report data for a set of standard conditions: The standard state of a substance (°) is the pure substance at exactly 1 bar It is conventional (though not obligatory) to report data for a T of 298.15K Standard enthalpies of phase transition Energy that must be supplied (or is evolved) as heat, at constant pressure, per mole of molecules that undergo the phase transition under standard conditions (pure phases), denoted DH° Note: the enthalpy change of a reverse transition is the negative of the enthalpy change of the forward transition

  18. DH°

  19. Sublimation • Direct conversion of a solid to a vapour The enthalpy change of an overall process is the sum of the enthalpy changes for the steps into which it may be divided

  20. Enthalpies of ionisation (kJ/mol) DionH°(T)= Ionisation energy(0) + (5/2)RT (see Atkins & de Paula, Table 3.2)

  21. Problems • Ethanol is brought to the boil at 1 atm. When the electric current of 0.682 A from a 12.0 V supply is passed for 500 s through a heating coil immersed in the boiling liquid, it is found that the temperature remains constant but 4.33 g of ethanol is vapourised. What is the enthalpy of vapourisation of ethanol at its boiling point at 1 atm? • Calculate the standard enthalpy of sublimation of ice at 0°C given that DfusH° is 6.01 kJ/mol and DvapH°is 45.07 kJ/mol, both at 0°C. • DsubH° for Mg at 25°C is 148 kJ/mol. How much energy as heat must be supplied to 1.00 g of solid magnesium metal to produce a gas composed of Mg2+ ions and electrons?

  22. Bond enthalpies (kJ/mol)

  23. Problem • Estimate the standard reaction enthalpy for the formation of liquid methanol from its elements as 25°C

  24. Enthalpies of combustion Enthalpies (heats) of combustion: complete reaction of compounds with oxygen. Measure using a bomb calorimeter. Most chemical reactions used for the production of heat are combustion reactions. The energy released when 1g of material is combusted is its Fuel Value. Since all heats of combustion are exothermic, fuel values are reported as positive. • Most of the energy our body needs comes from fats and carbohydrates. • Carbohydrates are broken down in the intestines to glucose. Glucose is transported in the blood to cells where it is oxidized to produce CO2, H2O and energy: • C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(l) DcH°=-2816 kJ • The breakdown of fats also produces CO2 and H2O • Any excess energy in the body is stored as fats

  25. Heats of formation If one mole of the compound is formed under standard conditions from its elements in their reference state then the resulting enthalpy change is said to be the standard molar enthalpy (Heat) of formation, fH° where the subscript indicates this. The reference state is the most stable form under the prevailing conditions.

  26. Hess’s Law To evaluate unknown heats of reaction The standard enthalpy of a reaction is the sum of the standard enthalpies for the reactions into which the overall reaction may be divided rxnHo = nDfHom(products) - nDfHom(reactants)

  27. Variation of DrH° with T DrH°(T2) = DrH°(T1) + DrCp°(T2-T1) Kirchoff’s Law DrCp° = S nCp,m°(products) - S nCp,m°(reactants) If heat capacity is temperature dependent, we need to integrate over the temperature range

  28. Thermodynamics: the 2nd law Deals with the direction of spontaneous change (no work required to bring it about) Kelvin Statement No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work Impossible!

  29. Entropy The apparent driving force for spontaneous change is the dispersal of energy A thermodynamic state function, Entropy, S, is a measure of the dispersal of energy (molecular disorder) of a system 2nd Law: The Entropy of an isolated system increases in the course of spontaneous change DStot>0

  30. Thermodynamic definition of S • Concentrates on the change in entropy: DS = qrev/T Can use this equation to quantify entropy changes. We will see later (3rd & 4th year) a statistical description of entropy S = k lnW (Boltzmann formula)

  31. Heat Engines All heat engines have a hot region “source” and a cold region “sink”: some energy must be discarded into the cold sink as heat and not used to do work

  32. Expansion entropy • Intuitively can guess that entropy increases with gas expansion. • Thermodynamic definition allows us to quantify this increase Recall that: w = -nRT ln (Vf/Vi) BUT qrev = -w (DU = 0 for isothermal processes) DS = nR ln (Vf/Vi) Note: independent of T Also: Because S is a state function, get the same value for an irreversible expansion

  33. Heating Entropy

  34. Entropy of phase transition • Entropy of fusion • Entropy of vapourisation Trouton’s rule The entropy of vapourisation is approximately the same (85 J/K.mol) for all non-polar liquids

  35. Phase transitions • To evaluate entropies of transition at T other than the transition temperature Entropy of vapourisation of water at 25°C? • Sum of DS for heating from 25°C to 100°C, DS for vapourisation at 100°C, and DS for cooling vapour from 100°C to 25°C. Try it! (+118 J/K.mol).

  36. Entropy changes in the surroundings DStot= DSsys +DSsur DStot= DSsys –q/T Example: Water freezing to ice. Entropy change of system is -22 J/K.mol, and heat evolved is -6.01 kJ/mol. Entropy change in surroundings must be positive for this process to occur spontaneously. Check this for different temperatures. Note that DStot= 0 at equilibrium

  37. Spontaneity of water freezing DStot = DSsys - qsys/T • At 5°C: DStot = -22 JK-1mol-1 – (6,010Jmol-1/278K) • = -0.38 JK-1mol-1 • At -5°C: DStot = -22 JK-1mol-1 – (6,010Jmol-1/268K) • = +0.43 JK-1mol-1 • At 0°C: DStot = -22 JK-1mol-1 – (6,010Jmol-1/273K) • = 0.01 JK-1mol-1 • To find transition temperature, set DStot = 0 and solve for T. 273.18 K (slight error because of rounding of entropy and heats).

  38. Problem • Typical person heats the surroundings at a rate of 100W (=J/s). Estimate entropy change in one day at 20°C. • qsur = 86,400 s × 100 J/s • DSsur = qsur/T = (86,400 × 100 J)/293 K = 2.95 × 104 J/K

  39. 3rd Law • Entropy of sulfur phase transition is 1.09 J/K.mol. • Consider plot at left. Subtract entropy for phase transition (to give plot at right) • T=0 intercept is the same. Entropies of all perfectly crystalline substances are the same at T=0.

  40. Absolute and standard molar entropies (S and S0m) Absolute entropies can be determined by integration of areas under heat capacity/T as a function of T, and including entropies of phase transitions. Standard molar entropies are the molar entropies of substances at 1bar pressure (and usually 298 K)

  41. Standard molar entropies

  42. Standard reaction entropies • Difference in molar entropy between products and reactants in their standard states is called the standard reaction entropy and can be expressed (like enthalpy) as: • Note: absolute entropies, S, and standard molar entropies, S0m, are discussed in section 4.7 of the textbook rxnSo = nSom(products) - nSom(reactants)

  43. Spontaneity of reactions Consider the reaction: 2H2(g) + O2(g)  2H2O (l) DrS0 = 2(70 J/K.mol) –[2(131 J/K.mol) + (205 J/K.mol)] = -327 J/K.mol But this reaction is spontaneous (explosive even!) When considering the implications of entropy, we must always consider the total change of the system and its surroundings DrH0 = -572 kJ/mol. Therefore DrSsur = +1920 J/K.mol DrStot is positive +1590 J/K.mol (spontaneous reaction!).

  44. Gibbs Energy • Introduced by J.W. Gibbs to combine the calculations of 2 entropies, into one. • Because DStot = DS – DH/T (constant T and P) • Introduce G = H – TS (Gibbs “free” energy) • Then DG = DH – TDS (constant T) • So that DG = – TDStot(constant T and P) DG = DH – TDS In a spontaneous change at constant temperature and pressure, the Gibbs energy decreases

  45. Maximum non-expansion work • Can derive (see box 4.5 in textbook) that DG = w’max • Example: formation of water: enthalpy -286kJ, free energy -237kJ • Example: suppose a small bird has a mass of 30 g. What is the minimum mass of glucose that it must consume to fly to a branch 10 m above the ground? (DG for oxidation of glucose to carbon dioxide and water is -2828 kJ at 25°C) Exercise: A human brain operates at about 25 W (J/s). What mass of glucose must be consumed to sustain that power for 1 hour?

  46. Problems solved w’ = (30 × 10-3 kg)× (9.81 m s-2)× 10 m Note 1J = 1kg m2 s-2 n = 2.943 J/(2828× 103 J/mol) m = nM = (1.04× 10-6 mol) × 180 g/mol = 1.9× 10-4 g Answer 2: 5.7g

  47. 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 Gm = 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

  48. Variation of G with pressure • We can derive (see derivation 5.1 in textbook) that DGm = VmDp • Therefore DGm>0 when Dp>0 • Can usually ignore pressure dependence of G for condensed states • Can derive that, for a gas: DGm = RT ln(pf/pi)

  49. Proof-go back to fundamental definitions G = H –TS; H = U + pV; dU = dq + dw For an infinitesimal change in G: G + dG = H + dH – (T + dT)(S + dS) = H + dH – TS – TdS – SdT – dTdS dG = dH – TdS – SdT Also can write: dH = dU + pdV + Vdp dU = TdS – pdV(dS = dqrev/T and dw = -pdV) dG = Vdp – SdT Master Equations

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