Chapter 9 in Chang Text

Transcription

1 Section 8.0: CHEMICAL EQUILIBRIUM Chapter 9 in Chang Text

2 ..the direction of spontaneous change at constant T and P is towards lower values of Gibbs Energy (G). this also applies to chemical reactions

3 Gibbs energy of reaction (G) consider the reaction A B..if an infinitessimal amount of A turns into B then the change in the amount of A may be expressed as: d n A = - d.and the change in the amount of B present.is d n B = + d.where quantity is called the extent of the reaction (its units are moles) When extent of reaction changes by finite amount, the amount of A and B present may be expressed as n A - n B +

4 Gibbs Energy of Reaction ( R G) is defined as slope of plot of Gibbs Energy (G) versus the extent of the reaction ().. R G = [ G / ] P,T ( here signifies derivative of slope of G wrt )..so if a reaction progresses by d then the change in Gibbs energy can be written.. dg = A dn A + B dn B = - A d + B d = ( B - A ) d [ G / ] P,T = ( B - A ) so R G = ( B - A ).. R G is related to difference in chemical potentials of reactants and products at the specific composition of reaction mixture.

5 chemical potential varies with composition slope of G vs. changes as rxn. proceeds G vs A B R G = ( B - A ) Rxn. spontaneous in forward direction when A B Rxn. spontaneous in reverse direction when B A At Equilibrium: R G = 0 when A = B Figure from Atkins Text

6 At constant T and P: R G 0 frwd. rxn is spontaneous Rxn. is exergonic can be used to drive another reaction. R G 0 reverse. rxn is spontaneous Rxn. is endergonic R G = 0 at equilibrium Figure from Atkins Text

7 We can use a favourable rxn. to DRIVE an un-favourable one.exergonic rxns. can be used to drive endergonic rxns. An example of an exergonic rxn is ATP hydrolysis: ATP + H 2 O ADP + P i + H 3 O + ATP to ADP, with ΔG = 30 kj (at 37 C) ΔH = - 20 kj ; ΔS = + 34 J/K This exergonic rxn. can couple with endergonic reactions, and help drive processes or cascades of reactions. 30 kj is available for driving other reactions. ***since ΔS is large the.. ΔG is sensitive to T.

8 Biological Energy Conversion in cells energy released by the oxidation of foods is stored in ATP in the cell each ATP molecule can be used to drive endergonic reactions.where ΔG does not exceed 30 kj example. Biosynthesis of proteins is highly endergonic (due to enthalpic contribution as well as decrease in entropy that accompanies assembly of amino acids into a specific sequence). Formation of a peptide link is endergonic and will typically consume approx. 3 ATP molecules for each link. ex. In myoglobin..there are approx. 150 peptide links biosynthesis of this protein consumes 450 ATP molecules

9 8.2. CHEMICAL EQUILIBRIUM in GASEOUS SYSTEMS

10 Consider the reaction: aa (g) bb (g) Remember equation for chemical potential of x th component in a mixture: = x + RT ln [P x / P ] x P x is the partial pressure of component x in the mixture. x is the standard state chemical potential and P = 1 bar. So for A and B in rxn. above we can write: A B = A + RT ln [P A / P ] = B + RT ln [P B / P ]

11 Gibbs Energy change for the reaction can be written as: R G = b B -a A R G = b B - a A + b RT ln[p B / P ] - a RT ln [P A / P ] Standard Gibbs Energy of reaction is.. R G = b B -a A R G = R G + RT ln {[P B / P ] b /[P A / P ] a } At equilibrium R G = 0 0 = R G + RT ln {[P B / P ] b /[P A / P ] a } R G = - RT ln K P (where p denotes concn. are in pressures)

12 K p, the equilibrium constant: K p = {[P B / P ] b /[P A / P ] a } = (P Bb /P Aa ) (P ) a-b At a given temperature the value of R G is a constant that depends only on nature of reactants and products as well as temperature. If we know R G we can calculate K P and vice versa...remember we can calculate R G from Gibbs energy of formation of reactants and products ( f G )...so we can calculate K P also using values for f G (note..we are talking about standard state conditions i.e. 1 bar and 298 K)

13 Total Gibbs Energy vs. extent of reaction for the reaction: aa (g) bb (g). In this case R G 0 R G = b B -a A when there is no mixing Gibbs energy decreases linearly as rxn. progresses mix G is a neg. quantity Gibbs energy lower for mixing than non-mixing case equilibrium point is compromise between opposing tendencies for reactant and product to mix and conversion of reactants into products.

14 R G is usually not equal to zero, but when it is zero then K = 1. K = 1 means products and reactants are equally favored at equilibrium. If R G 0 then K 1 If R G 0 then K 1 this does not mean the reaction does not occur. Example. R G = 10 kj, T = 298 K and K P = in this case products may still form if large quantities of reactants are used for the reaction. K is dimensionless, and dependent on Temperature.

15 Sample Problem: From thermodynamic data given calculate the equilibrium constant for the following reaction at 298 K: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) f G (NH 3 (g)) = kj/mol

16 Reactions are rarely carried out under standard state conditions: so usually we are interested in R G R G = R G + RT ln Q At a given temperature the value of R G is constant but we can change value of R G by changing partial pressure of gases. (i.e. concentration dependence).q is the reaction quotient Q = K P if R G = 0 **If R G is a large positive number (e.g. 50 kj) then sign of R G is primarily determined by R G unless reactants or products are present in large quantities to override effect of R G (i.e. equal in magnitude but opposite in sign)

17 Can determine direction of reaction using following expression: R G = RT ln (Q/K) If Q K P R G is negative and reaction proceeds in forward direction If Q K P R G is positive and reaction proceeds in reverse direction

18 Sample Problem 9.2 At 298 K the partial pressures of the gases in reaction below are P N2 = 190 torr, P H2 = 418 torr, P NH3 = 722 torr. Calculate the value of R G for the reaction. N 2 (g) + 3H 2 (g) 2NH 3 (g) R G = kj

19 The following equation applies only to an ideal gas: x = x + RT ln [P x / P ] For real gases we write as follows x = x + RT ln [f/ P ].where here f (fugacity) has replaced partial pressure. Fugacity has same units as pressure. At low pressure, gas behaves ideally and fugacity equals pressure. At high pressures, deviations occur fugacity coefficient: = f/p 1 indicates attractive intermolecular forces are dominant 1 indicates repulsive intermolecular forces are dominant

20 Fugacity versus pressure. lim = 1 (P 0) - fugacity can only be calculated for further discussion refer to Appendix 9.1 in Chang

21 Therefore for a reaction involving real gases: a A (g) b B (g) K f = {[f B / 1 bar] b /[f A / 1 bar] a } since = f/p K f = Bb {[P B / 1 bar] b / Aa [P A / 1 bar] a } = K K p where K = Bb / A a and K p = [P B / 1 bar] b / [P A / 1 bar] a K f K p called thermodynamic equilibrium constant called apparent equilibrium constant (value not constant, depends on pressure)

22 at low pressures K P approx. equal to K f significant deviations occur as total pressure inc. beyond 300 bar K f remains approx constant over entire pressure range

23 Sample Problem on Equilibrium in Gaseous Systems Chang text # 9.3 Gaseous nitrogen dioxide is actually a mixture of nitrogen dioxide (NO 2 ) and dinitrogen tetroxide (N 2 O 4 ). If the density of such a mixture is 2.3 g/l at 74 C and 1.3 atm, calculate the partial pressures of the gases and the value of K P for the dissociation of N 2 O 4.

24 The equilibrium constant for the dissociation reaction: N 2 O 4 (g) 2 NO 2 (g)

25 Chang Text # 9.5 Consider the reaction PCl 5 (g) PCl 3 (g) + Cl 2 (g) K P = 1.05 at 250ºC. A quantity of 2.50 g of PCl 5 is placed in an evacuated flask of volume L and heated to 250 C. (a) Calculate the pressure of PCl 5 if it did not dissociate. Solution

26 (b) Calculate the partial pressure of PCl 5 at equilibrium. The partial pressure of PCl 5 at equilibrium can be determined using The K P expression.

27 How to solve for x.. Quadratic Formula: x = (-b b 2 4 ac ) / 2 a where a, b and c are given within the quadratic equation: a x 2 + b x + c = 0

28 (C) What is the total pressure at equilibrium? (D) What is the degree of dissociation of PCl 5?

29 8.3. Chemical Equilibrium: Reactions in Solution

30 Reactions in Solution.here we express concentrations in molarity or molality aa bb A = A + RT ln [m A / m ] * assuming ideal behavior and expression conc. in molality where m = 1 mol solute/kg solvent R G = - RT ln K m where K m = {[m B / m ] b /[m A / m ] a } With concentrations in molarity Kc = {([B]/ 1M ) b /([A]/ 1 M) a }

31 R G = - RT ln K c for equilibrium reactions R G = R G + RT ln Q for non-equilibrium rxns. For non ideal solutions we replace concentration with activities. x = x + RT ln a x ( this is like substitution of fugacity for pressure with gases) K a = a Bb / a a A a = m so K a = Bb / a A {[m B / m ] b /[m A / m ] a } = K K m where K m is apparent equilibrium constant and K a is thermodynamic equilibrium constant

32 8.4. Heterogeneous Equilibrium.to this point we have only considered homogeneous equilibrium meaning reactions where we only have one phase..heterogeneous equilibrium refers to reactions where we have more than one phase.. Example: CaCO 3 (s) CaO (s) + CO 2 (g)..the equilibrium constant would be K c = [CO 2 ] / 1 M or K P = P CO2 / 1 bar *** since we don t include solids in equilibrium constant expression

33 The thermodynamic equilibrium constant, K a, for this reaction would be: K a = [a CaO a CO2 ] / a CaCO3 = [1 a CO2 ] / 1 = a CO2..by convention activities of pure solids (and pure liquids) in their standard states ( at 1 bar) are 1. At low pressures /moderate pressures we can assume that activities of solids don t change much. K f = f CO2 / 1 bar or if we assume ideal behavior. K P = P CO2 / 1 bar

34 Plot of equilibrium pressure of CO 2 over CaO and CaCO 3 as a function of T. Try example 9.3 in Chang text. Chang text, page 311

35 Partial Molal Free Eenergy Consider the General Reaction a A + b B c C + d D The Gibbs Free Energy Functional should now be written as: G = G(T,P,n A,n B,n C,n D ) dg = G T P,nA,nB,nC,nD dt + G P T,nA,nB,nC,nD dp + G n A T,P,nB,nC,nD dn A + G n B T,P,nA,nC,nD dn B + G n C T,P,nA,nB,nD dn C + G n D T,P,nA,nB,nC dn D

36 = S dt + V dp + μ A dn A + μ B dn B + μ C dn C + μ D dn D From the stoichiometry of the reaction: dn A = a dn B = b dn C c = dn D d dξ dξ = measure of the extent of the reaction Therefore: dn A = a dξ dn B = b dξ dn C = + c dξ dn D = + d dξ

37 Then: dg = S dt + V dp + (c μ C + d μ D a μ A b μ B ) dξ = S dt + V dp + [ c (μ C o + RT ln a C ) + d (μ D o + RT ln a D ) a (μ A o + RT ln a A ) b (μ B o + RT ln a B ) ] dξ = S dt + V dp + [ (c μ C o + d μ D o a μ A o b μ B o ) + RT (c ln a C + d ln a D a ln a A b ln a B ) ] dξ

38 = S dt + V dp c d + [ (c μ o C + d μ o D a μ o A b μ o a a C D B ) + RT ln a b a a A B ] dξ If temperature and pressure are constant then dt = 0 and dp = 0 Define: Δμ o (c μ C o + d μ D o a μ A o b μ B o ) Then: c d dg = [ Δμ o a a C D + RT ln ] dξ a b a a A B

39 If system is at equilibrium, then dg = 0 and: Δμ o + RT ln eq b B a A d D c C a a a a = 0 or: ΔG o Δμ o = RT ln eq b B a A d D c C a a a a

40 UNITS UNITS UNITS Recall: dg = S dt + V dp + [ c (μ o C + RT ln a C ) + d (μ o D + RT ln a D ) a (μ o A + RT ln a A ) b (μ o B + RT ln a B ) ] dξ Students (undergraduate AND graduate) usually have trouble with the units for the terms included in the coefficient of dξ. Let s examine one of those terms in detail (the others will be the same) by inserting all the units of each component EXPLICITLY. For instance: c (μ o C + RT ln a C )

41 c (μ C o + RT ln a C ) Now include the units explicitly in addition to the numerical value; Note that c, μ C o, R, T and a C from this point on are simply numbers. (c (mol))[ μ C o (kj mol 1 ) + (R (kj mol 1 K 1 ))(T (K)) ln a C ] After multiplication of the numerical values and cancellation of the units we are left with the following: c μ C o (kj) + c R T (kj) ln a C = c μ C o (kj) + R T (kj) ln a C c All the terms in the coefficient of dξ are in the same unit (kj). c, μ C o, R, T and a C are simply numbers.

42 Equilibrium Constant Calculations ΔG o Δμ o = RT ln a a c C a A a a d D b B eq ΔG o and Δμ o have the same unit: kj Therefore the RHS of the equation must also have kj as the unit. However, R normally has units of (kj mol 1 K 1 ) and T has the unit (K). The product, RT, then should have the unit (kj mol 1 ). WHAT IS WRONG?? It is important to remember that the mol unit has been cancelled, as shown previously. In effect we are substituting an R which has the same numerical value but appears to have a different unit: (kj K 1 )

43 Jacobus Henricus van t Hoff ( ) st Nobel Prize in Chemistry

44

45 8.5. Influence of Temperature, and Catalysts on Equilibrium Constant Influence of Temperature: van t Hoff Equation: ln (K 2 /K 1 ) = ( R H /R) [( 1/T 1 ) (1/T 2 )] = ( R H /R) [(T 2 -T 1 ) /(T 1 T 2 )] Equation allows calculation of K at one temperature if we know K at another temperature..here we assume R H is temperature independent.

46 The van t Hoff Equation ΔG o = ΔH o T ΔS o = RT ln K eq If ΔH o and ΔS o are independent of temperature then for temperatures T 1 and T 2 we may write: ln K 2 = H RT o 2 + S o R (A) ln K 1 = H RT o 1 + S o R (B) Subtract (B) from (A):

47 ln K 2 ln K 1 = ln 1 2 K K = 1 2 o T 1 T 1 R H = o T 1 T 1 R H The van't Hoff Equation

48 ln K = - ( R G / RT ) and R G = R H - T R S so ln K = - R H /RT + R S /R A plot of ln K versus 1 / T gives a straight line with Slope = - R H /R y-intercept = R S /R..this plot only yields a straight line if R S and R H are independent of Temperature. for narrow temperature ranges this method works (i.e. 50 K or less)

49 a) R H 0 (b) R H 0 (c) R H = 0

50 ln (K 2 /K 1 ) = ( R H /R) [(T 2 -T 1 ) /(T 1 T 2 )].if R H is positive (endothermic) and T 2 T 1 then K 2 K 1.if R H is negative (exothermic) and T 2 T 1 then K 2 K 1 For Endothermic reaction increase in T.shifts reaction to right.. favors formation of products. For Exothermic reaction increase in T.shifts reaction to left.. favors formation of reactants.

51 LeChatelier s Principle: states that if an external stress is applied to a system at equilibrium, the system will adjust itself in such a way to partially offset the stress as it tries to re-establish equilibrium. (Chang page 314) so if T is raised the system at equilbrium will shift in the endothermic direction.then energy is absorbed as heat and rise in temperature is opposed Exothermic rxn.. Temp..favors the reactants Endothermic rxn. Temp..favors the products

52 Endothermic Exothermic (a) Increase in Temperature, Boltzmann distribution adjusts and population changes increased population of higher energy state at expense of population in lower energy state. Population of B becomes more abundant in equilibrium mixture. Atkins text. page 237

53 Effect of a Catalyst a catalyst can speed up the rate of a reaction without itself being consumed. a catalyst cannot shift the position of equilibrium for a reaction not at equilibrium a catalyst will increase both the forward and reverse rates so equilibrium will be reached sooner

54 Q: what happens if we add reactant A to a system at equilibrium? Equilibrium re-establishes A: the system responds by driving forward the reaction to use up reactant A and produce more product B add more A add more B If we add more A, then drive A B If we add more B then we drive A B pure A fraction of A in mix pure B 0 1

55 Chang Text Problem # 9.10 The vapor pressure of dry ice (solid CO 2 ) is torr at C and 1486 torr at C. Calculate the molar heat of sublimation of CO 2. The van t Hoff equation ln (K 2 /K 1 ) = ( R H/R) [ T 1-1 T 2 1 ] CO 2 (s) CO 2 (g)

56 Problem 9.22 in Chang Text: Photosynthesis can be represented by 6CO 2 (g) + 6H 2 O (l) C 6 H 12 O 6 (s) + 6O 2 (g) R H = 2801 kj How would equilibrium be affected by the following changes : (a) Partial pressure of CO 2 is increased (b) O 2 is removed from reaction mixture (c) Sucrose is removed from reaction mixture (d) More water is added (e) A catalyst is added (f) Temperature is decreased

57 8.6. Principles of Coupled Reactions Let s revisit concept of one reaction driving another.. We call the favourable reaction (with ΔG negative) EXERGONIC And the un-favourable reaction (with ΔG positive) ENDERGONIC ΔG TOTAL = ΔG EXERGONIC + ΔG ENDERGONIC *** we can make small weight move upwards by coupling to falling of large weight.

58 many chemical and biological reactions are endergonic and not spontaneous under standard state conditions some of these reactions can be carried out by coupling them to exergonic reactions Example: Consider the extraction of copper from its ore (Cu 2 S)..heating the ore will not yield much copper because process is associated with large positive Δ R G. Cu 2 S (s) 2Cu (s) + S (s) Δ R G = 86.2 kj if the thermal decomposition is coupled to the oxidation of sulfur then overall process is spontaneous

59 Cu 2 S (s) 2Cu (s) + S (s) Δ R G = 86.2 kj S (s) + O 2 (g) SO 2 (g) Δ R G = kj Overall: Cu 2 S + O 2 (g) 2Cu (s) + SO 2 (g) Δ R G = kj Gibbs energy for the overall process is the sum of the Gibbs Energies of each reaction. Overall process has large neg. value for Δ R G and thus is spontaneous.

60 Characteristics of Coupled Biological Reactions: 1- endergonic reaction is coupled with an exergonic reaction..combined overall reaction is exergonic 2- exergonic reaction is most often hydrolysis of ATP 3- coupled reaction is catalyzed by an enzyme

61 ATP Currency of Energy in Biological Systems ATP hydrolysis is used to drive many biological processes including protein synthesis, ion transport, muscle contraction, electrical activity in nerve cells. Hydrolysis Reaction of ATP at ph = 7: ATP 4- + H 2 O ADP 3- + H + + HPO 2-4 The exact value of Δ R G associated with ATP hydrolysis is dependent on Temperature, ph and presence of metal ions. Usually ranges between Δ R G = - 25 to - 40 kj At ph = 7, T = 310 K and in presence of Mg 2+ ions Δ R G = kj

62 ATP Note: 4 negative charges in molecule Upon hydrolysis ATP ADP and loses terminal phosphate group ADP can be further hydrolyzed to produce AMP

63 Why is hydrolysis of ATP to form ADP associated with negative value for Δ R G? Two factors must be taken into account: 1- electrostatic repulsion 2- resonance stabilization 1 - electrostatic repulsion - 4 negative charges in molecule - proximity of charges causes repulsion - repulsion reduced upon hydrolysis reduces no. of negative charges from 4 to resonance stabilization - ADP and HPO 4 2- possess more resonance structures than ATP see page 329 in Chang text.

64 Examples of Exergonic Hydrolysis Reactions

65 Example: alanine + glycine alanylglycine Δ R G = 17.2 kj K = 1 x 10 3 at 298 K Reaction is said to be thermodynamically controlled. Positive value for Δ R G signifies that reaction is not spontaneous If coupled to ATP hydrolysis this reaction will occur spontaneously.

66 reactions may also be kinetically controlled a kinetically controlled reaction has a negative value for Δ R G but the rate is very slow, negligible, in absence of a catalyst Example: phosphorylation of glucose to form glucose 6-phosphate coupled to ATP hydrolysis is an exergonic process but proceeds very slowly in absence of enzyme hexokinase (catalyst).

67 8.7. Equilibrium in Biological Processes: Consider Binding of Oxygen to Myoglobin MYOGLOBIN can bind (uptake) OXYGEN in the body to form an oxygenated complex Mb (aq) + O 2 (g) MbO 2 (aq) K = [MbO 2 ] / [O 2 ] [Mb] = [MbO 2 ] / P O2 [Mb]..P represents the partial pressure of the O 2 gas. We are interested in concentration of oxygenated complex: [MbO 2 ] = K P O2 [Mb] Reference: Atkins Text (The Elements of Phys Chem with Applications in Biology)

68 Mb (aq) + O 2 (g) MbO 2 (aq) Y = fractional saturation Y = [MbO 2 ] / ([Mb] + [MbO 2 ]) [MbO 2 ] = K P O2 [Mb] Y = K P O2 [Mb] / ([Mb] + K P O2 [Mb]) so Y = K P O2 / (1 + K P O2 ) (divide all by [Mb]) Therefore, as we increase P of oxygen we increase Y.

69 Oxygen saturation curve for Myoglobin.

70 9.34. A polypeptide can exist in either the helical or random coil forms. The equilibrium constant for equilibrium reaction of the helix to the random coil transition is 0.86 at 40 C and 0.35 at 60 C. Calculate the values of R H and R S for the reaction.

71 9.36 At 720 K the equilibrium partial pressures are P NH3 = atm, P N2 = 69.6 atm, P H2 = atm, respectively. (a) Calculate the value of K P for the reaction given below. (b) Calculate the thermodynamic equilibrium constant if NH3 = 0.782, N2 = and H2 = The reaction is N 2 (g) + 3 H 2 (g) 2 NH 3 (g)