Lecture #13. Chapter 17 Enthalpy and Entropy

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1 Lecture #13 Chapter 17 Enthalpy and Entropy

2 First Law of Thermodynamics Energy cannot be created or destroyed The total energy of the universe cannot change Energy can be transferred from one place to another ΔE universe = 0 = ΔE system + ΔE surroundings However, this does not tell us the direction of the change

3 First Law of Thermodynamics For an exothermic reaction, lost heat from the system goes into the surroundings. Energy is lost from the system as heat (q) or work (w) Energy conservation requires that the energy change of a system equal the heat lost plus the work done ΔE = q + w ΔE = ΔH + PΔV E is a state function

4 Spontaniety Spontaneous process A process that occurs without outside intervention. Nonspontaneous process A process that only occurs as long as energy is continually added to the system. Spontaneity depends on dispersion of energy that occurs during a process. Spontaneity is determined by comparing the free energy of a system before a reaction to that after a reaction. If the system contains less free energy than when it started, the process is spontaneous.

5 Spontaniety Any spontaneous process is also irreversible. A reversible process will proceed back and forth. If a process is spontaneous in one direction, it must be non spontaneous and irreversible in the other direction.

6 Thermodynamics Spontaneous vs. Fast Kinetics

7 Factors Affecting Spontaneity Enthalpy change (ΔH) a comparison of bond energy of reactants and products Entropy change (ΔS) a comparison of the randomness of reactants and products

8 Spontaneity and Enthalpy (H) Although many spontaneous processes are exothermic (i.e., combustion), this is not true for all spontaneous reactions: Endothermic (H > 0), but reaction is spontaneous.

9 A Spontaneous Endothermic Reaction A Spontaneous Endothermic Chemical Reaction. water Ba(OH). 2 8H 2 O(s) + 2NH 4 NO 3 (s) Ba 2+ (aq) + 2NO 3- (aq) + 2NH 3 (aq) + 10H 2 O(l) ΔH 0 rxn = kj

10 Thermodynamic Entropy A thermodynamic function which increases as the number of energy-equivalent ways of arranging the components increases.

11 The Concept of Entropy (The State of Order) A change in order is a change in the number of ways of arranging particles. Increasing Entropy, (ΔS>0) More order fewer particles less molecular motion Less Order more particles more molecular motion solid liquid gas solid + liquid ions in solution

12 Second Law of Thermodynamics The total entropy of the universe increases in any spontaneous process. For a reversible process: ΔS univ = 0 For a spontaneous process: ΔS univ = ΔS sys + ΔS surr > 0 If the entropy of the system decreases, the entropy of the surroundings must increase by a larger amount for a process to be spontaneous.

13 Spontaneity of Reaction ΔS univ = ΔS sys + ΔS surr as a Function of Δ S sys and Δ S surr ΔS univ > 0

14 Entropy and Changes of State When materials change state, the number of macro states it can have changes as well. solid liquid gas Increasing Degrees of Motion

15 Temperature Dependence of ΔS surroundings Exothermic processes add heat to the surroundings, increasing the entropy of the surroundings. Endothermic processes take heat from the surroundings, decreasing the entropy of the surroundings. The amount the entropy of the surroundings changes depends on the temperature it is at originally. The higher the original temperature, the less effect addition or removal of heat has. eat has

16 Calculations with Thermodynamic Values A Recollection

17 a) Magnesium carbonate decomposes to magnesium oxide and carbon dioxide; Hrxn = kj/mol kj + MgCO3 (s) MgO (s) + CO2 (g) Is the reaction exothermic or endothermic? What is the equation for the reverse reaction? MgO (s) + CO2 (g) MgCO3 (s) kj Hrxn = kj/mol What is the H when 35.5 g of CO2 reacts? 35.5 g CO2 x 1.00 mol CO g CO2 x kj 1.00 mol CO2 = kj

18 b) Given the following reactions, ½ N2 (g) + ½ O2 (g) NO (g) H = kj/mol NO (g) + ½ Cl2 (g) NOCl (g) H = kj/mol Calculate Hrxn for 2 NOCl (g) N2 (g) + Cl2 (g) + O2 (g) Hrxn= kj/mol NOCl (g) NO (g) + ½ Cl2 (g) H = -(-38.6 kj/mol) = kj/mol 2 NOCl (g) 2 NO (g) + Cl2 (g) H = (2)(38.6 kj/mol) = kj/mol * NO (g) ½ N2 (g) + ½ O2 (g) H = kj/mol 2 NO (g) N2 (g) + O2 (g) H = (2)(-90.3 kj/mol) = kj/mol *

19 1) Consider the combustion of propane gas: C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) ΔHrxn = kj (a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25ºC. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous? (a) T = = 298 K ΔSsurr = = -ΔHrxn T -(-2044) kj 298 K = kj/k = x 10 3 J/K

20 1) Consider the combustion of propane gas: C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) ΔHrxn = kj (a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25ºC. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous? (b) C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) 6 mol gas 7 mol gas ΔSsys is positive!!

1) Consider the combustion of propane gas: C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) ΔHrxn = -2044 kj (a) Calculate the entropy change in the surroundings associated with this reaction occurring at

21 1) Consider the combustion of propane gas: C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g) ΔHrxn = kj (a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25ºC. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous? (c) ΔSuniverse is positive and the reaction is spontaneous!!

22 Absolute Entropy and the Third Law of Thermodynamics

23 Entropy and Temperature Entropy increases as temperature increases. Increases in kinetic energy increase the number of accessible molecular motions. Decreasing temperature decreases entropy. At what temperature does all molecular motion cease??

24 The absolute entropy of a substance is the energy a substance has due to dispersal of energy through its particles Third Law of Thermodynamics: The absolute entropy of a perfect crystal is zero at absolute zero. Every substance which is not a perfect crystal has an absolute entropy greater than 0. Standard molar entropy (Sº): The absolute entropy of 1 mole of a substance in its standard state at 298 K and 1 bar of pressure. (Calculated from measurement of molar heat capacities as a function of temperature)

25 Standard States of Pure Substances and Solutions Physical State Solid Liquid Gas Solution Standard State Pure solid, most stable allotrope of an element Pure liquid Pure gas 1 M Pressure 1 bar 1 bar 1 bar 1 bar Standard molar entropy (Sº): Calculated for a particular state, a particular allotrope, a particular molecular complexity, a particular molar mass, and a particular degree of dissolution.

26 Selected Standard Molar Entropy Values Standard molar entropy (Sº): Calculated for a particular state, a particular allotrope, a particular molecular complexity, a particular molar mass, and a particular degree of dissolution.

27 Relative Standard Molar Entropy Values Gas > Liquid > Solid Larger molecule > Smaller molecule Less constrained allotrope > More constrained allotrope More complex molecule > Less complex molecule Dissolved substance > Solid crystal

Substance S, (J/mol K) H 2 O (g) 70.0 H 2 O (l) 188.8 Substance He (g) Ne (g) Sº, (J/mol ᐧK) 126.2 146.1 Substance C (diamond) (s) Sº, (J/mol ᐧK) 2.4 Ar (g) Kr (g) 154.

28 Substance S, (J/mol K) H 2 O (g) 70.0 H 2 O (l) Substance He (g) Ne (g) Sº, (J/mol ᐧK) Substance C (diamond) (s) Sº, (J/mol ᐧK) 2.4 Ar (g) Kr (g) C (graphite) (s) 5.7 Xe (g) Molar Substance Mass S, (J/mol K) Ar (g) NO (g) Substance S, (J/mol K) KClO 3 (s) KClO 3 (aq) 265.7

29 Entropy and Structure Entropy increases as the complexity of molecular structure increases. More bonds, more opportunities for internal motion (more internal vibrational motion).

30 Entropy and Vibrational Motion NO NO 2 N 2 O 4

31 Predict the Sign of the Entropy Changes H 2 O (g) H 2 O (l) CO 2 (s) CO 2 (g) N 2 O (g) 2 N 2 (g) + O 2 (g) + Mg (s) + Cl 2 (g) MgCl 2 (s) 2 H 2 S(g) + 3 O 2 (g) 2 H 2 O(g) + 2 SO 2 (g) O 3 (g) 3 O 2 (g) + NH 3 (g) + HCl (g) NH 4 Cl (s) -

32 Calculating Entropy Changes

33 2) Calculate Sºrxn for the balanced chemical equation 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g) ΔS rxn = ΔS sys = S final S initial ΔSº rxn = Σn products Sº products Σn reactants Sº reactants Reactant or Product NH3 (g) O2 (g) NO (g) Sº, (J/mol ᐧK) H2O (g) ΔSº rxn = Σn p Sº (products) Σn r Sº (reactants) = [4(SºNO(g)) + 6(SºH2O(g))] -[4(SºNH3(g)) + 5(SºO2(g))] = [4(210.8 J/K) + 6(188.8 J/K)] -[4(192.8 J/K) + 5(205.2 J/K)] = J/K J/K = J/K

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax.

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro First Law of Thermodynamics Chapter 17 Free Energy and Thermodynamics You can t win! First Law of Thermodynamics: Energy cannot be created or destroyed