Second Law of Thermodynamics

Transcription

1 Second Law of Thermodynamics

2 First Law: the total energy of the universe is a constant Second Law: The entropy of the universe increases in a spontaneous process, and remains unchanged in a process at equilibrium

3 Entropy and the second law A process is spontaneous if ΔS univ is positive ΔS univ = ΔS system + ΔS surr > 0 A process is at equilibrium if: ΔS univ = ΔS system + ΔS surr = 0 Therefore we need to consider how the entropy of the system and the surroundings change during a process

4 Entropy changes in the System

5 Entropy changes in the System ΔS system is + if disorder increases ΔS system is - if products are more ordered than reactants Can be calculated from tables of thermodynamic values ΔS rxn = ΣnS (products) Σm S (reactants) We can often make good guesses as to the sign of ΔS system

6 Entropy changes in the System Calculate the standard entropy change for: 2CO(g ) + O (g ) 2CO (g ) 2 2 qualitative prediction: 2 moles of gas on the right, 3 on the left ; the products are more ordered than reactants; the sign of ΔS is -

7 Entropy changes in the System If a reaction produces excess gas ΔS system is + If a reaction produces no net change in gas molecules ΔS system may be ( +) or ( - ) but the change will have a small value General Rule: a reaction that increases the total number of molecules or ions will increase ΔS system

8 Example: Calculate the standard entropy change for: 2CO(g ) + O 2 (g ) 2CO 2 (g ) S (reactants) 2 mol(197.9 J/K mol) + 1mol(205 J/K mol) = 600,8 J/K S (products) 2 mol(213.6 J/K mol) = J/K ΔS rxn = J/K J/K ΔS rxn = J/K

9 Third Law of Thermodynamic and Absolute Entropy the entropy of a perfect crystalline substance is zero at the absolute zero of temperature : 0 K ( -273º C ) important in the determination of absolute entropy values

10 Entropy changes in the surroundings

11 Entropy changes in the surroundings How are surroundings affected by heating and cooling exothermic reactions increases entropy of surroundings endothermic reactions decrease entropy of surroundings ΔH system ΔS surr = q T ΔS surr = T

12 Surroundings Surroundings System Heat System

13 Surroundings Surroundings System Heat Entropy System Heat

14 Surroundings Surroundings System Heat Entropy System Heat Entropy

15 What are the possibilities? ΔS sys ΔS surr + = ΔS univ spontaneous? ?? yes no

16 Spontaneity and temperature a reaction may be spontaneous at one temperature and nonspontaneous at a different temperature ΔS = ΔS + ΔS univ system surr ΔH system T

17 Gibbs Free Energy ΔS universe = ΔS surroundings + ΔS system we study the system; therefore reference surroundings in of terms the system ΔS universe = ΔH system / T + ΔS system TΔS universe = ΔH system + TΔS system TΔS universe = ΔH system TΔS system ΔG = ΔH system TΔS system

18 Gibbs Free Energy Free Energy (ΔG) is a measure energy available to do work. a release of free energy during a chemical reaction is spontaneous

19 Criterion for spontaneity is the Gibbs free energy change at constant temperature and pressure, if ΔG system is negative,the reaction is spontaneous positive, the reaction is not spontaneous zero, the system is at equilibrium

20 Standard Free-Energy Changes

21 Standard Free Energies of Formation the change in free energy that accompanies the formation of 1 mole of a substance from its constituent elements at standard conditions ΔGº = kj ΔGº = kj/mol f

22 From Standard Free Energies of Formation ΔG rxn = ΣnΔG f (products) Σm ΔG f (reactants) the standard free energies of formation of any element in its stable form equals zero

23 Example: Calculate the standard free-energy changes for the following reaction at 25 C : 2C 2 H 6 (g ) + 7O 2 (g ) 2mol( 32.9 kj/mol) + 7mol(0 kj/mol) 4CO 2 (g ) + 6H 2 O(l ) 4mol( kj/mol) + 6mol( kj/mol) 66 kj 3001 kj ΔG = 3001 ( 66) = 2935 kj

24 Applications of ΔG = ΔH TΔS

25 ΔG = ΔH TΔS ΔH negative and ΔS positive; reaction spontaneous at all temperatures ΔH positive and ΔS negative; reaction can t be spontaneous at any temperature ΔH positive and ΔS positive; increasing temperature favors spontaneity ΔH negative and ΔS negative; increasing temperature works against spontaneity

26 ΔG = ΔH TΔS ΔH negative and ΔS positive; reaction spontaneous at all temperatures Click for example ΔH positive and ΔS negative; reaction can t be spontaneous at any temperature ΔH positive and ΔS positive; increasing temperature favors spontaneity ΔH negative and ΔS negative; increasing temperature works against spontaneity

27 Endothermic dissolution: at 25ºC ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔH f kj/mol kj/mol kj/mol S 94.6 J/mol K J/mol K 56.5 J/mol K

28 Endothermic dissolution: at 25 º C ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔH rxn ΔS = = [ kj/mol + ( kj/mol) ] - = 15.4 kj/mol [112.8 J/mol K J/mol K ] - ( kj/mol) (94.6 J/mol K) = 74.7 J/mol K.0747 kj/mol K =

29 Endothermic dissolution: at 25 º C ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔH rxn = 15.4 kj/mol ΔS =.0747 kj/mol K Τ = 298 K ΔG = ΔH TΔS ΔG = 15.4 kj/mol 298 K (.0747 kj/mol K) ΔG = -6.86

30 Endothermic dissolution: at 25 º C ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔH f kj/mol kj/mol kj/mol S 94.6 J/mol K J/mol K 56.5 J/mol K ΔG f kj/mol kj/mol kj/mol ΔG rxn = -6.8 kj/mol

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32 Example ΔH negative and ΔS positive; reaction spontaneous at all temperatures 2H 2 O 2 (l ) 2H 2 O(l ) + O 2 (g )

33 Example ΔH positive and ΔS negative; reaction can t be spontaneous at any temperature 3O 2 (g) 2O 3 (g)

34 Example ΔH positive and ΔS positive; increasing temperatures favors spontaneity H 2 (g ) + I 2 (g ) 2HI(g )

35 Example ΔH negative and ΔS negative; increasing temperatures works against spontaneity NH 3 (g ) + HCl(g ) NH 4 Cl(s )