1 CHEM 212 Chapter 3 By Dr. A. Al-Saadi

2 Introduction to The Second Law of Thermodynamics The two different pressures will be equalized upon removing the partition. This is a natural (usually irreversible) process or called “spontaneous”.

3 Introduction to The Second Law of Thermodynamics The irreversible natural “spontaneous” process takes place by the heat passing from the hot object to the cold one until we have T 1 = T 2. The process that is another way around is called “nonspontaneous” and has to be done reversibly. Hot solid Cold solid T1T1 T2T2 Where T 1 > T 2

4 Introduction to The Second Law of Thermodynamics 2H 2 (g) + O 2 (g)  2H 2 O(l) Is reaction spontaneous? It is spontaneous in going from left to right. For the reaction to go from right to left, work has to be done on H 2 O(l) to get the gases, and it is nonspontaneous.

5 Entropy (S) A measurement of spontaneity is the entropy. If a process occurs reversible from state A to state B with infinitesimal amounts of heat (dq rev ) absorbed at each stage, the “entropy change” is defined as:

6 Defining Entropy on the Basis of Steam Engine It is impossible for an engine to perform work by cooling a portion of matter to a temperature below that of the coldest part of the surroundings.

7 Historical Development of Steam Engines Newcomen steam engine: One-cylinder steam engine. Efficiency of converting heat into work is only 1%

8 Historical Development of Steam Engines Watt’s steam engine: Two-cylinder steam engine, mainly a cylinder (kept at T h ) and a condenser (kept at T c ). Efficiency of converting heat into work is only 10%

9 Historical Development of Steam Engines Watt’s steam engine: Two-cylinder steam engine, mainly a cylinder (kept at T h ) and a condenser (kept at T c ). Efficiency of converting heat into work is only 10%

10 The Carnot Engine It is a theoretical engine explaining the process in terms of T h and T c. It assumes: -An ideal engine. -An ideal gas, and -A reversible process. Carnot showed that such an engine would have the maximum possible efficiency for any engine working between T h and T c.

11 The Carnot Engine Four processes being done reversibly: 1)Isothermal expansion at T h. 2)Adiabatic expansion. 3)Isothermal compression at T c. 4)Adiabatic compression.

12 The Carnot Cycle Pressure-volume diagram for the Carnot cycle. AB and CD are isotherms. BC and DA are adiabatics

13 The Carnot Cycle The Carnot cycle for 1 mole of an ideal gas that starts at 10.0 bar and K with the value of C P /C V = γ = 1.40 (used for N 2 gas)

14 The Carnot Cycle

15 The Net Work from Carnot Cycle

16 Example 3.1

17 The Concept of Entropy q = 0 T c =T a +dT q b = q c + dq T c is unchanged

18 The Concept of Entropy Entropy is a measure of the disorder of the system. The change in the entropy (ΔS) indicates whether a change in the disorder of the system takes place or not. The sign of ΔS also indicated whether the disorder of the system increases or decreases.

19 The Concept of Entropy Inequality of Clausius: The change in entropy is generally given by: Any change that occurs irreversibly in nature is spontaneous and therefore is accompanied with a net increase in entropy. The energy of the universe is constant. The entropy of the universe tends always towards maximum.

20 Molecular Interpretation of Entropy When a spontaneous change takes place, the disorder of the system increases and its entropy, as a result, increases. Entropy is a measure of disorder. Example 1: Mixing solute and solvent Mixing process is a spontaneous process that increases the disorder of the system. Thus, entropy is expected to increase and ΔS is +ve.

21 Molecular Interpretation of Entropy Example 2: Phase change -Melting of solid. (ΔS is +ve) -Freezing of a liquid. (ΔS is -ve) -Evaporation of liquid. (ΔS is +ve) Example 3: Chemical reactions -H 2 (g)  2H(g) (ΔS is +ve) -2NH 2 (g)  3H 2 (g) + N 2 (g) (ΔS is +ve) Generally, in gaseous reactions, as the number of molecules increases (less pairing), entropy would increase. That is because less pairing involves more disorder and less restrictions.

22 Entropy of Mixing The mixing of ideal gases at equal pressures and temperatures

23 Calculation of Entropy Change

24 Calculation of Entropy Change for Solids and Liquids

25 Calculation of Entropy Change

26 Calculation of Entropy Change

27 The Third Law of Thermodynamics Nernst’s heat theorem states that: ΔS = 0 at the absolute zero provided that the states of the system are in thermodynamic equilibrium. Equilibrium is slowly and barely achieved at extremely low temperatures. The 3 rd law of thermodynamics: The entropies (S a, S b, S c, …) of all perfectly crystalline substances (thermodynamic equilibrium) must be the same at the absolute zero (S a = S b = S c = …).

28 Achieving Very Low Temperatures 1- Expansion of the molecules to reduce the intermolecular forces (cooling to 1K). 2- Magnetization (entropy decrease and thus flow of heat from the sample to the surrounding) and then demagnetization (done after isolating the system “adiabatic step” causing cooling). (cooling to about 0.005K)

29 Absolute Entropies The absolute entropy of a substance can be calculated at another temperature by: cooling down that substance to nearly the absolute zero, and then increasing its temperature through reversible steps until the desired temperature is attained.

30 Absolute Entropies

31 Entropy Change of a Chemical Reaction Absolute entropies for substances are tabulated at standard conditions.tabulated

32 Conditions for Equilibrium The position of equilibrium must correspond to a state of maximum total entropy.

33 Molecular Interpretation of Gibbs Energy (G) In spontaneous processes at constant T and P, system moves towards a state of minimum Gibbs energy. (dG < 0) In the condition of equilibrium at constant T and P, there is no change in Gibbs energy (dG = 0)

34 Molecular Interpretation of Gibbs Energy (G) Consider the reaction H 2 ↔ 2H at T=300K H 2  2H (nonspontaneous) 2H  H 2 (spontaneous) For the spontaneous 2H  H 2 reaction: ΔG is –ve (The system approaches equilibrium as G goes to minimum). To show this in terms of enthalpy and entropy changes: ΔG = ΔH – TΔS ΔH is –ve (The reaction loses heat). ΔS is –ve, (More ordered arrangement as H 2 molecules are produced). ΔH – TΔS < 0 Thus, the affecting parameter (the weighting factor) is T: T is small, ΔH > TΔS, and ΔG will be –ve (spontaneous). T is too large, ΔH < TΔS, and ΔG will be +ve (nonspontaneous dissociation of H 2 molecules).

35 The Contribution to Spontaneous Change Enthalpy ChangeEntropy ChangeSpontaneous? ΔG = ΔH – TΔS ΔH 0 ΔH < 0 (Exothermic) ΔS < 0 ΔH < 0 (Exothermic) T = 0 ΔH > 0 (Endothermic) ΔS > 0 ΔH > 0 (Endothermic) ΔS < 0 ΔH > 0 (Endothermic) T = 0 Yes (ΔG < 0) Yes, if |TΔS| < |ΔH| Yes (ΔG < 0) Yes, if |TΔS| > |ΔH| No (ΔG > 0)

36 The vaporization of water at 100°C. (a) Liquid water at 100°C is in equilibrium with water vapor at 1 atm pressure. (b) Liquid water at 100°C is in contact with water vapor at 0.9 atm pressure, and there is spontaneous vaporization

37 Δ r Gº for Chemical Reactions Determine the Delta G under standard conditions using Gibbs Free Energies of Formation found in a suitable thermodynamic table for the following reaction: 4HCN(l) + 5O 2 (g) ---> 2H 2 O(l) + 4CO 2 (g) + 2N 2 (g) 2 moles ( kj/mole) = kj = Standard Free Energy of Formation for two moles H 2 O(l) 4 moles ( kj/mole) = kJ = Standard Free Energy of Formation for four moles CO 2 (g) 2 moles(0.00 kj/mole) = 0.00 kJ = Standard Free Energy of Formation for 2 moles of N 2 (g) Standard Free Energy for Products = ( kJ) + ( kJ) = kJ 4 moles ( 121 kj/mole) = 484 kJ = Standard Free Energy for 4 moles HCN(l) 5 moles (0.00 kJ/mole) = 0.00 kJ = Standard Free Energy of 5 moles of O 2 (g) Standard Free Energy for Reactants = (484) + (0.00) = 484 kJ Standard Free Energy Change for the Reaction = Sum of Free Energy of Products - Sum of Free Energy of Reactants = (-2052 kJ) - (484 kJ) = kJ

38 Maxwell Relations Important Relationships ; ; ; ;

39 Maxwell Relations

40 Maxwell Relations

41 Applications of Maxwell Relations Thermal expansion coefficient (α): During heat transfer, the energy that is stored in the intermolecular bonds between atoms changes. When the stored energy increases, so does the length of the molecular bond. As a result, solids typically expand in response to heating and contract on cooling. This response to temperature change is expressed as its coefficient of thermal expansion. coefficient of linear thermal expansion α α in /K at 20 °C material 60Mercury 42BCB 29Lead 23Aluminum 19Brass 17.3Stainless steel 17Copper 14Gold 13Nickel 12Concrete 11.1IronIron or SteelSteel 10.8Carbon steel 9Platinum 8.5Glass 5.8GaAs 4.6Indium Phosphide 4.5Tungsten 3.3Glass, PyrexPyrex 3Silicon 1.2Invar 1Diamond 0.59QuartzQuartz, fused

42 Applications of Maxwell Relations Prove that for 1 mole of a van der Walls gas, the internal pressure (∂U/∂V) T is equal to a/V 2 m.

43 Fugacity

44 Fugacity