Chapter 17 Free Energy and Thermodynamics Lesson 1

2 First Law of Thermodynamics First Law of Thermodynamics: Energy cannot be created or destroyed the total energy of the universe cannot change it can be transfered from one place to another   E universe = 0 =  E system +  surroundings system = reactants & products surroundings = everything else (the transfer of energy from one to the other does not change the energy of the universe)

3 First Law of Thermodynamics For an exothermic reaction, heat from the system goes into the surroundings two ways energy can be “lost” from a system, converted to heat, q used to do work, w Energy conservation requires that the energy change in the system = heat exchanged + work done on the system.  E = q + w (  E = internal energy change)  E =  H – P  V (at const. P, q p =  H, enthalpy change)

4 Enthalpy, H related to (includes) the internal energy  H generally kJ/mol stronger bonds = more stable molecules if products more stable than reactants, energy released; exothermic  H = negative if reactants more stable than products, energy absorbed; endothermic  H = positive The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions. Hess’ Law:  H° rxn =  (  H f ° prod ) -  (  H f ° react )

5 Thermodynamics and Spontaneity thermodynamics predicts whether a process will proceed (occur) under the given conditions spontaneous process  nonspontaneous process does not occur under specific conditions. spontaneity is determined by comparing the free energy (G) of the system before the reaction with the free energy of the system after reaction. if the system after reaction has less free energy than before the reaction, the reaction is thermodynamically favorable. spontaneity ≠ fast or slow (rate); this is kinetics

6 Spontaneous Nonspontaneous ice 25 o C water 25 o C 2Na(s) + 2H 2 O(l)  H 2 (g) + 2NaOH(aq)  H 2 (g) + 2NaOH(aq) 2Na(s) + 2H 2 O(l) ball rolls downhill ball rolls uphill water -10 o C ice -10 o C

Copyright McGraw- Hill 2009 Entropy Changes in a System Qualitative S solid < S liquid

Copyright McGraw- Hill 2009 Entropy Changes in a System Qualitative S liquid < S vapor

Copyright McGraw- Hill 2009 Entropy Changes in a System Qualitative S pure < S aqueous

Copyright McGraw- Hill 2009 Entropy Changes in a System Qualitative S lower temp < S higher temp

11 Diamond → Graphite Graphite is thermodynamically more stable than diamond, so the conversion of diamond into graphite is spontaneous – but it’s kinetically too slow (inert) it will never happen in many, many generations. kinetics Spontaneity: direction & extent kinetics: how fast

12 Factors Affecting Whether a Reaction Is Spontaneous The two factors that determine the thermodynamic favorability are the enthalpy and the entropy. The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond.  H The entropy factors relate to the randomness/orderliness of a system  S The enthalpy factor is generally more important than the entropy factor

14 Entropy, S Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases.  S generally in J/K (joules/K) S = k lnW k = Boltzmann Constant (R/N A ) = 1.38  J/K W is the number of energetically equivalent ways, (microstates). It is unitless. Entropy is usually described as a measure of the randomness or disorder; the greater the disorder of a system, the greater its S. The greater the order  the smaller its S.

15 Entropy & Microstates, W Energetically Equivalent States for the Expansion of a Gas (4 gas molecules) 1 microstate 6 microstates (most probable distribution) S = k ln W  S = k ln W f - k ln W i if W f > W i,  S > 0 & entropy increases.

16 Changes in Entropy,  S entropy change is favorable when the result is a more random system (State C: higher entropy ).  S is positive (  S > 0) Some changes that increase the entropy are: rxns where products are in a more disordered state. (solid > liquid > gas) less order (solid< liquid < gas)  larger S (disorder) reactions which have larger numbers of product molecules than reactant molecules. increase in temperature (more movement) solids dissociating into ions upon dissolving

Copyright McGraw- Hill 2009 Entropy Changes in a System Qualitative S fewer moles < S more moles

18 Changes in Entropy in a System (melting) Particles fixed in space Particles can occupy many positions

19 Changes in Entropy in a System (vaporization) Particles occupy more space (larger volume)

20 Changes in Entropy in a System (solution process) Structure of solute and solvent disrupted (also more solute particles)

Strategy To determine the entropy change in each case, we examine whether the number of microstates of the system increases or decreases. The sign of ΔS will be positive if there is an increase in the number of microstates and negative if the number of microstates decreases. Solution (a)Upon freezing, the ethanol molecules are held rigid in position. This phase transition reduces the number of microstates and therefore the entropy decreases; that is, ΔS < 0.

(b) Evaporating bromine increases the number of microstates because the Br 2 molecules can occupy many more positions in nearly empty space. Therefore, ΔS > 0. (c) Glucose is a nonelectrolyte. The solution process leads to a greater dispersal of matter due to the mixing of glucose and water molecules so we expect ΔS > 0. (d) The cooling process decreases various molecular motions. This leads to a decrease in microstates and so ΔS < 0.

23 The 2nd Law of Thermodynamics The entropy of the universe increases in a spontaneous process.  S universe =  S system +  S surroundings > 0  S universe =  S system +  S surroundings = 0 (equilibrium) If  S system >> 0,  S surroundings 0! If  S system > 0 for  S universe > 0! the increase in  S surroundings often comes from the heat released in an exothermic reaction,  system < 0.

24 Entropy Changes in the Surroundings (  S surr ) Exothermic Process  S surr > 0 Endothermic Process  S surr < 0

25 The 3 rd Law of Thermodynamics S = k ln W S = k ln W = k ln 1 = 0  S = S f – S i ; where S i = 0 K the absolute entropy of a substance is always (+) positive at the new T - allows determination of entropy of substances. (W = 1, there is only one way to arrange the particles to form a perfect crystal) the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K

26 Third Law of Thermodynamics The entropy of a perfect crystalline substance is zero at the absolute zero of temperature. S = k ln W W = 1 S = 0

27 Standard Entropies S° entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution Values can be used to calculate the standard entropy change for a reaction,  S o rxn (=  S o sys )

29 Trends: Standard Entropies Molar Mass For monatomic species, the larger the molar mass, the larger the entropy available energy states more closely spaced, allowing more dispersal of energy through the states

30 Trends: Standard Entropies States the standard entropy of a substance in the gas phase is greater than the standard entropy of the same substance in the solid or liquid phase at a particular temperature Substance S°, (J/mol∙K) H 2 O (l)70.0 H 2 O (g)188.8

31 Trends: Standard Entropies Allotropes the more highly ordered form has the smaller entropy -different forms of an element

32 Trends: Standard Entropies Molecular Complexity (inc # of atoms) larger, more complex molecules generally have larger entropy more available energy states, allowing more dispersal of energy through the states Substance Molar Mass S°, (J/mol∙K) Ar (g) NO (g)

33 Trends: Standard Entropies Dissolution dissolved solids generally have larger entropy distributing particles throughout the mixture Substance S°, (J/mol∙K) KClO 3 (s)143.1 KClO 3 (aq)265.7

34 Q. Arrange the following in order of increasing 25 o C! (lowest to highest) Ne(g), SO 2 (g), Na(s), NaCl(s) and H 2 (g) Na(s) < NaCl(s) < H 2 (g) < Ne(g) < SO 2 (g)

Calculate  S  for the reaction 4 NH 3(g) + 5 O 2(g)  4 NO (g) + 6 H 2 O (g)  S is +, as you would expect for a reaction with more gas product molecules than reactant molecules standard entropies from Appendix IIB  S, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SS S  NH3, S  O2, S  NO, S  H2O, Substance S , J/mol  K NH 3 (g)192.8 O2(g)O2(g)205.2 NO(g)210.8 H 2 O(g)188.8

Consider 2 H 2 (g) + O 2 (g)  2 H 2 25 o C ∆S o = 2 S o (H 2 O) - [2 S o (H 2 ) + S o (O 2 )] ∆S o = 2 mol (69.9 J/Kmol) - [2 mol (130.6 J/Kmol) + 1 mol (205.0 J/Kmol)] ∆S o = J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid. Calculating ∆S o for a Reaction ∆S o =  S o (products) -  S o (reactants)

The more negative  H syst and the lower the temperature the higher (more positive)  S surr 37 Temperature Dependence of  S surroundings  system 0 )  system > 0 (endothermic), it takes heat from the surroundings, decreasing the entropy of the surroundings (  S surroundings < 0 )

The reaction C 3 H 8(g) + 5 O 2(g)  3 CO 2(g) + 4 H 2 O (g) has  H rxn = kJ at 25°C. Calculate the entropy change of the surroundings. combustion is largely exothermic, so the entropy of the surroundings should increase (inc in # gas mol)  H system = kJ, T = 298 K  S surroundings, J/K Check: Solution: Concept Plan: Relationships: Given: Find: SS T,  H

2 H 2 (g) + O 2 (g)  2 H 2 25 o C ∆S o system = J/K Calculating ∆S o for the surroundings ∆S o surroundings = J/K Can calculate ∆H o system = ∆H o rxn = kJ (2 mol H 2 0 (l) - 0) (2 mol H 2 0 (l) - 0)

2 H 2 (g) + O 2 (g)  2 H 2 25 o C ∆S o system = J/K ∆S o surroundings = J/K ∆S o universe = J/K The entropy of the universe is increasing, so the reaction is spontaneous ( product- favored).  S universe =  S system +  S surroundings Given S o surr, S o sys and T, determine S o univ and predict if the reaction will be spontaneous. Given  S o surr,  S o sys and T, determine  S o univ and predict if the reaction will be spontaneous.

Spontaneous or Not? originally: originally:  S universe =  S system +  S surroundingsbut but  S universe =  S system –  system /T  system  S system Spontaneous? Exothermic  H sys < 0 Less order  S sys > 0 Spontaneous under all conditions;  S univ > 0 Exothermic  H sys < 0 More order  S sys < 0 Favorable at low T Endothermic  H sys > 0 Less order  S sys > 0 Favorable at high T Endothermic  H sys > 0 More order  S sys < 0 Not spontaneous under any conditions  S univ < 0

42 Without doing any calculations, determine the sign of S sys and S surr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous. 2CO(g) + O 2 (g)  2CO 2 (g)  rxn = kJ 2NO 2 (g)  O 2 (g) + 2NO(g)  rxn = kJ -/T)  S universe =  S system + ( -  sys /T)  S system = ( -); 3 mol gas form 2 mol gas  S surr = (+); low T  S system = (+); 2 mol gas form 3 mol gas  S surr = ( -) ; high T -/T)  S surr = ( -  sys /T)

43 Without doing any calculations, determine the sign of  S sys and  S surr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous. 2H 2 (g) + O 2 (g)  2H 2 O(g)  rxn = kJ CO 2 (g)  C(s) + O 2 (g)  rxn = kJ -/T)  S universe =  S system + ( -  sys /T)  S system = ( -) ; 3 mol gas form 2 mol gas  S surr = (+) ; low T  S system = ( -) ; complicated gas forms a solid & gas  S surr = ( -) ; all T -/T)  S surr = ( -  sys /T)

44 At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings, if  H o rxn = -127 kJ and  S o rxn = 314 J/K. Plan: set  S o rxn =  S o surr and solve for T; convert kJ to J rxn implies system!!! Ans: T = +404 K