Lecture #14. Chapter 17 Free Energy and Equilibrium Constants

Transcription

1 Lecture #14 Chapter 17 Free Energy and Equilibrium Constants

2 Josiah W. Gibbs

3 Gibbs Free Energy (G) The maximum energy released by a system occurring at constant temperature and pressure that is available to do useful work. ΔSuniverse = ΔSsystem + ΔSsurroundings ΔSuniv = ΔSsys + -ΔHsys/T -TΔSuniv = -TΔSsys + ΔHsys -TΔSuniv = ΔHsys - TΔSsys ΔG = -TΔS univ

4 Gibbs Free Energy (G) The maximum amount of energy from the system available to do work on the surroundings (the system s chemical potential ) ΔG = ΔH - TΔS = -TΔSuniv ΔG is proportional to the negative of ΔSuniv and is thus another criterion for spontaneity. ΔG < 0 for a spontaneous process ΔG > 0 for a nonspontaneous process ΔG = 0 for a process at equilibrium

5 Calculating ΔGº at 25º: ΔGº reaction = ΣnΔGº f(products) - ΣnΔGº f(products) at temperatures other than 25º: ΔGº reaction = ΔHº reaction - TΔSº reaction

6 3) Ozone in the lower atmosphere is a pollutant that can be formed by the following reaction involving the oxidation of unburned hydrocarbons: CH4 (g) + 8 O2 (g) CO2 (g) + 2 H2O (g) + 4 O3 (g) Use the standard free energies of formation to determine Gºrxn for this reaction at 25 ºC. Reactant/Product CH4 (g) O2 (g) CO2 (g) H2O (g) O3 (g) ΔGfº, (kj/mol )

7 CH4 (g) + 8 O2 (g) CO2 (g) + 2 H2O (g) + 4 O3 (g) ΔGº reaction = ΣnΔGº f(products) - ΣnΔGº f(products) ΔGº rxn = [ΔG f º (CO2(g) + 2(ΔG f º (H2O(g) ) + 4(ΔG f ºO2(g))] -[ΔG f ºCH4(g)) + 8(ΔG f ºO2(g))] = [ kj + 2( kj) +4(163.2 kj)] -[-50.5 kj) + 8(0.0 kj)] = [ kj)] +[50.5 kj)] = kj Reactant/Product CH4 (g) O2 (g) CO2 (g) H2O (g) O3 (g) ΔGfº, (kj/mol )

8 4) One of the possible initial steps in the formation of acid rain is the oxidation of the pollutant SO2 to SO3 by the reaction: SO2 (g) + ½ O2 (g) SO3 (g) a) Calculate Gºrxn at 25 ºC and determine whether the reaction is spontaneous. Reactant/Product SO2 (g) O2 (g) ΔHfº, (kj/mol ) Sº, (J/mol ᐧK) SO3 (g)

9 4a) SO2 (g) + ½ O2 (g) SO3 (g) ΔHº rxn = Σn p ΔH f º (products) Σn r ΔH f º (reactants) ΔHº rxn = [ΔH f º,SO3(g)] [ΔH f º,SO2(g) + ½(ΔH f º,O2(g))] = [ kj] [ kj kj] = kj Reactant/Product SO2 (g) O2 (g) ΔHfº, (kj/mol ) Sº, (J/mol ᐧK) SO3 (g)

10 4a) SO2 (g) + ½ O2 (g) SO3 (g) ΔSº rxn = Σn p Sº (products) Σn r Sº (reactants) = [SºSO3(g)] -[(SºSO2(g)) + ½(SºO2(g))] = [256.8 J/K] -[248.2 J/K + ½(205.2 J/K)] = J/K) T = 298 K ΔGº reaction = ΔHº reaction - TΔSº reaction = x 10 3 J K (-94.0 J/K) = x 10 3 J = kj The reaction is spontaneous at this temperature.

11 4b) For the same reaction, estimate the value of Gºrxn at 125 ºC. Does the reaction become more or less spontaneous at this elevated temperature; that is, does the value of Gºrxn become more negative (more spontaneous) or more positive (less spontaneous)? SO2 (g) + ½ O2 (g) SO3 (g) T = 398 K ΔGº reaction = ΔHº reaction - TΔSº reaction = x 10 3 J K (-94.0 J/K) = x 10 3 J = kj The reaction is less spontaneous at this temperature than at 25ºC.

12 Gibbs Free Energy Change (ΔG) ΔG = ΔH - (TΔS) A process will be spontaneous when ΔG is negative. ΔH is negative and TΔS is positive ΔH is negative and large and TΔS is negative and small ΔH is positive and small and TΔS is positive and large A process will be nonspontaneous when ΔG is positive. ΔH is positive and TΔS is negative

13 Temperature and Spontaneity ΔG = ΔH - TΔS ΔG ΔG

14 Temperature and Spontaneity ΔG = ΔH - TΔS H = positive, S = positive G = H T S = negative above a certain temperature Spontaneous, ΔG<0 ΔH=TΔS ΔG=0 TΔS ΔG ΔH Nonspontaneous, ΔG>0 ΔG=ΔH-TΔS Temperature, K

15 Temperature and Spontaneity ΔG = ΔH - TΔS H = negative, S = negative G = H T S = negative below a certain temperature Temperature, K ΔH=TΔS ΔG=0 Nonspontaneous, ΔG>0 ΔG=ΔH-TΔS ΔG ΔH Spontaneous, ΔG<0 TΔS

16 5) Consider the reaction for the decomposition of carbon tetrachloride gas: CCl4 (g) C (s, graphite) + 2 Cl2 (g) ΔH = kj; ΔS = J/K (a) Calculate G at 25 ºC and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25 ºC, determine at what temperature (if any) the reaction becomes spontaneous. (a) T = = 298 K ΔG = ΔH - (TΔS) ΔG = 95.7 x 10 3 J - (298 K)(142.2 J/K) ΔG = 95.7 x 10 3 J - (42.2 x 10 3 J) ΔG = x 10 3 J The reaction is not spontaneous!!

17 5) Consider the reaction for the decomposition of carbon tetrachloride gas: CCl4 (g) C (s, graphite) + 2 Cl2 (g) ΔH = kj; ΔS = J/K (a) Calculate G at 25 ºC and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25 ºC, determine at what temperature (if any) the reaction becomes spontaneous. ΔG = ΔH - (TΔS) 0 = ΔH - (TΔS) 0 = 95.7 x 10 3 J - (T)(142.2 J/K) T = (95.7 x 10 3 J) /(142.2 J/K) T = 673 K

18 ΔG Relationships If a reaction can be expressed as a series of reactions, the sum of the ΔG values for the individual reactions is the ΔG for the total reaction. If a reaction is reversed, the sign of its ΔG reverses. If the amount of material involved in a reaction is multiplied by a factor, the value of ΔG is multiplied by the same factor.

19 Free Energy and Chemical Equilibrium

20 Free Energy and Chemical Equilibrium G < 0; K >> 1

21 Free Energy and Chemical Equilibrium G slightly negative

22 Free Energy and Chemical Equilibrium G > 0; K << 1

23 The Relationship Between ΔG 0 and K at 25 0 C ΔG 0 (kj) K Significance x x10-18 Essentially no forward reaction; reverse reaction goes to completion x10-9 2x10-2 7x x10 1 6x10 8 3x x10 35 Forward and reverse reactions proceed to same extent FORWARD REACTION Forward reaction goes to completion; essentially no reverse reaction REVERSE REACTION

24 A(g) B(g), K < 1 A(g) B(g), K > 1 G G Pure A Pure B Pure A Pure B ΔG = 0 A(g) B(g), K = 1 ΔG = 0 G G = H-TS ΔG = ΔH-TΔS spontaniety Pure A Pure B ΔG = 0

25 Free energy change for a reaction: ΔGº reaction = Σn prod ΔGº products Σn react ΔGº reactants Depends on reaction conditions relative to standard states: ΔG rxn = ΔGº rxn + RT ln Q When reaction is at equilibrium: ΔG rxn = 0, and Q = K ΔGº rxn = -RT ln K (K = e ΔGºrxn/RT )

26 R = (L atm)/(mol K) 1 atm = kpa R = (J)/(mol K) = (L kpa/mol K)

27 6. Determine the equilibrium constant for the following reaction at 298 K. Cl(g) + O 3 (g) ClO(g) + O 2 (g) ΔG = kj A) B) C) D) E) ln K = ΔGº/RT K = e ΔGºrxn/RT ΔGºrxn/RT = -(-34.5kJ/(8.314 x 10-3 kj/mol K)(298)) = e = 1.01 x 10 6

28 Influence of Temperature on Equilibrium Constants

29 Temperature Dependence of K ΔGº = ΔH rxn º TΔS rxn º ln K = ΔGº/RT For an exothermic reaction, ΔH rxn º/RT >0, increasing the temperature decreases the value of the equilibrium constant. For an endothermic reaction, ΔH rxn º/RT <0, increasing the temperature increases the value of the equilibrium constant.

30 From previous sections: ΔGº = ΔHº TΔSº ln K = ΔGº/RT Assuming Sº is independent of T, K will vary with temperature as: Combining and rearranging: y = ln K; x = 1/T; slope = ΔHº/R; intercept = ΔSº/R y = mx +b

31 6b. Determine the equilibrium constant for the following reaction at 500 K. Cl(g) + O 3 (g) ClO(g) + O 2 (g) ΔH = kj K2 ln ( 1.0 x 10 6 ) = -162,200 J/mol ( ) ( 1-1 ) J/(mol K) K2 1.0 x 10 6 = (-26.4) e K2 = 3.4 x 10-6

32 Driving the Human Engine: Coupled Reactions Living systems drive nonspontaneous reactions by coupling them to spontaneous reactions.

33 Free Energy Changes in Coupled Reactions Biochemical reactions can often be described in terms of free energy changes: Reaction #1: Glucose + P i Glucose-6-phosphate + H 2 O Reaction #2: ΔGº = kj/mol, endergonic ATP + H 2 O ADP + P i ΔGº = kj/mol, exergonic To couple the reactions we add the reactants and products together and add the ΔGº values Glucose + ATP Glucose-6-phosphate + ADP ΔGº = kj/mol, exergonic

34 Mechanical Model ΔG>0 ΔG<0 Downward motion of an object releases potential energy (an exergonic process) that can be used to do mechanical work, moving another object upward. (an endergonic process) A coupling mechanism is required to enable the exergonic process to drive the endergonic process.

35 Chemical Model + Reaction #1 endergonic Reaction #2 exergonic Coupled Reaction exergonic

36 Chemical Model The coupled reaction is exergonic; it will go spontaneously (forward, left to right) in the cell, but will it proceed at a rate consistent with cellular needs? There is no information about rates in the value of a ΔG. Most biological reactions would proceed at a very slow rate if they are not catalyzed. the catalyst is usually an enzyme.

37 Coupled Reactions Gº = kj/mol (i.e ) CH 2 O H ATP ADP CH 2 O PO 3 2- O OH OH HO OH D-Glucose Hexokinase O OH OH HO OH D-Glucose-6-phosphate An enzyme

38 Free Energy Changes for Glycolysis Reactions Steps* Glycolytic Reaction Step ΔG(kJ/mol) 1 glucose + ATP glucose-6-phosphate + ADP glucose-6-phosphate fructose-6-phosphate 0 to fructose-6-phosphate + ATP fructose-1,6-bisphosphate + ADP fructose-1,6-bisphosphate DHP + G3P 0 to -6 dihydroxyacetone phosphate glyceraldehyde-3-phosphate 0 to 4 5 glyceraldehyde-3-phosphate 1,3-bisphosphoglycerate -2 to 2 6 1,3-bisphosphoglycerate + ADP 3-phosphoglycerate + ATP 0 to phosphoglycerate 2-phosphoglycerate 0 to phosphoglycerate phosphoenolpyruvate 0 to phosphoenolpyruvate + ADP pyruvate + ATP -17 *ΔGº s are positive and negative. Under cellular conditions, however, the values of ΔG are negative or near zero. Three steps, however are irreversible.

39 Free Energy Changes for Glycolysis Reactions Steps