A"er todays lecture you should be able to do problems 1, 2, 3, 4, 5, 6, 7, 8, 9, 11 & & & & 18 In exam 3, Fall 2012

Transcription

1 A"er todays lecture you should be able to do problems 1, 2, 3, 4, 5, 6, 7, 8, 9, 11 & & & & 18 In exam 3, Fall

2 Third hourly exam Monday November 17 6:45-7:45 pm Rooms to be announced Alternate exam Monday morning 6:45 am Room 138 Chemistry ReservaMons required Room 185 Chemistry Before Friday at 5:00 pm 2

3 Exam will cover material in lecture notes inclusive Harrison review session Sunday November 16 3:00 4:00 pm Room 138 Chemistry 3

4 LRC mock exams Tuesday November 11 6:00 8:30 pm Room 108, Bessey Wednesday November 12 6:00 8:30 pm Room E100, VMC ReservaMons at lrc.msu.edu 4

5 Lecture notes 30 IntroducMon to free energy and entropy 2 nd law of thermodynamics 5

6 First Law of Thermodynamics Conserva;on of energy ΔE universe = ΔE system + ΔE surroundings = 0 ΔE system = q + w Constant T and P ΔH system = q p

7 Second Law of Thermodynamics Why do some process happen spontaneously while others do not? Gases expand into a lower pressure region. Heat flows from a hoier to a colder body.

8 vacuum 1 mole 1atm T E = 3 2 RT spontaneous Never happen T T 0.5 atm 0.5 atm E = 3 2 RT Spontaneous change at constant energy 8

9 In mechanical systems the equilibrium state is the lowest energy state available to the system In a chemical system this would mean (at constant T and P) the state of lowest enthalpy NH 4 Cl(s) H 2 O NH + 4 (aq) + Cl (aq) ΔH solution = +14.8kJ / mol H 2 O(l) H 2 O(v) ΔH vap = +40kJ / mol Enthalpy is not a reliable indicator of spontaneity 9

10 Spontaneous Process are irreversible Spontaneous does not necessarily mean instantaneous Spontaneous processes occur of their own accord Second Law of Thermodynamics states that for any spontaneous process the entropy change of the universe must be posi;ve ΔS universe > 0

11 ΔS universe = ΔS system + ΔS surroundings 0 Consider first the entropy change for the system ΔS system = S final S initial Entropy is a measure of the disorder in a system

12 The sign of the entropy change of the system will be posi;ve when: There is an increase in the temperature There is an expansion of a gas A solid is melted or a liquid is vaporized A solid or liquid (not a gas) is dissolved in a solvent There is an increase in the number of par;cles (atoms or molecules) The bonding becomes weaker A gas is produced in the reac;on

13 The more disordered a system the higher (more posi;ve) its entropy

14 Predict the sign of the entropy change in the following isothermal reac;ons CaCO 3 (s) CaO(s) + CO 2 (g) N 2 (g) + 3H 2 (g) 2NH 3 (g) HCl(g) + NH 3 (g) NH 4 Cl(s) ΔS > 0 ΔS < 0 ΔS < 0 NaCl(s) H 2 O Na + (aq) + Cl (aq) ΔS > 0 Cooling N 2 from 20 0 C to 50 0 C ΔS < 0

15 ΔS universe = ΔS system + ΔS surroundings Entropy is a state func;on (like the Energy) and can be calculated for any system So ΔS system is usually known Tables of absolute entropy exist for many substances What to do about ΔS Surroundings?

16 ΔS universe = ΔS system + ΔS surroundings What couples the system and the surroundings?

17 ΔS universe = ΔS system + ΔS surroundings What couples the system and the surroundings? When the system donates heat to the surroundings it increases the disorder in the surroundings system q system + q surroundings = 0 surroundings Conversely when the surroundings donate heat to the system it increases the disorder in the system

18 Rudolf Clausius 1865 At constant T and P ΔS surroundings = q surroundings T = q system T = ΔH system T 18

19 ΔS universe = ΔS system ΔH system T 0 ΔS system ΔH system T 0 ΔS ΔH T 0 T ΔS ΔH 0 ΔG = ΔH T ΔS 0 G is called the Gibbs Free Energy

20 The Gibbs free energy finesses the need to consider the surroundings when determining spontaneity! Second law of thermodynamics ΔS universe > 0 or ΔG system < 0 20

21 Consider the process H 2 O(s) 10 0 C H O(l) 2 Predict the signs of ΔH, ΔS, & ΔG ΔH = ΔS = ΔG =? Endothermic, posi;ve more disordered, posi;ve Depends on temperature ΔG = ΔH T ΔS

22 Assume ΔH and ΔS do not change with temperature ΔH & ΔS are both positive ΔG = ΔH T ΔS

23 Assume ΔH and ΔS do not change with temperature ΔG 0 T ΔH & ΔS are both positive ΔG = ΔH T ΔS

24 Assume ΔH and ΔS do not change with temperature not spontaneous ΔG Entropy driven 0 T equilibrium T Spontaneous at high T ΔH & ΔS are both positive ΔG = ΔH T ΔS

25 Note that at equilibrium (when ΔG = 0) ΔG = ΔH T ΔS = 0 ΔH T ΔS = 0 ΔS vap = ΔH vap T BP 25

26 So for the process H 2 O(s) 10 0 C H O(l) 2 Entropy driven ΔH = ΔS = ΔG = Endothermic, posi;ve more disordered, posi;ve Spontaneous at high temperature ΔG = ΔH T ΔS

27 Consider the process H 2 O(l) 10 0 C H O(s) 2 Predict the signs of ΔH, ΔS, & ΔG ΔH = ΔS = ΔG =? Exothermic, nega;ve More ordered, nega;ve Depends on temperature ΔG = ΔH T ΔS

28 Assume ΔH and ΔS do not change with temperature Enthalpy driven ΔG T equilibrium not spontaneous 0 T spontaneous at low T ΔH & ΔS are both negative ΔG = ΔH T ΔS

29 So for the process H 2 O(l) 10 0 C H O(s) 2 Enthalpy driven ΔH = ΔS = ΔG = Exothermic, nega;ve More ordered, nega;ve Spontaneous at low temperature ΔG = ΔH T ΔS

30 Assume ΔH and ΔS do not change with temperature ΔG 0 never spontaneous! T what if ΔH is positive & ΔS is negative? ΔG = ΔH T ΔS

31 Assume ΔH and ΔS do not change with temperature ΔG Always spontaneous! 0 T what if ΔH is negative & ΔS is positive? ΔG = ΔH T ΔS

32 Summary of ΔG and Temperature ΔH ΔS ΔG low T, - high T all T all T low T, + high T 32

33 Changes of State vapor liquid solid 33

34 ΔH + ΔS + evapora;on liquid mel;ng Changes of State vapor sublima;on solid ΔG = ΔH (+) T ΔS(+) 34

35 ΔH + ΔS + evapora;on Changes of State vapor condensa;on liquid sublima;on mel;ng freezing ΔH ΔS deposi;on solid ΔG = ΔH (+) T ΔS(+) ΔG = ΔH ( ) T ΔS( ) 35