Section 8.0: CHEMICAL EQUILIBRIUM Chapter 9 in Chang … 8 Fall 2010.pdf · Chapter 9 in Chang Text …………..the direction of spontaneous change at constant T and P is towards

Section 8.0:

CHEMICAL EQUILIBRIUM

Chapter 9 in Chang Text

…………..the direction of spontaneous change at constant T and P is towards lower values of GibbsEnergy (G)…………….

this also applies to chemical reactions

Gibbs energy of reaction (G)………………

……consider the reaction A B

…..if an infinitessimal amount of A turns into B then the changein the amount of A may be expressed as:

d nA = - d

….and the change in the amount of B present ….is d nB = + 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………

nA -

nB +

Gibbs Energy of Reaction (RG) is defined as slope of plot of Gibbs Energy (G) versus the extent of the reaction () ………..

RG = [ 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 = AdnA + BdnB = - A d + Bd = (B - A) d

[ G / ]P,T = (B - A)

so RG = (B - A)

….. RG is related to difference in chemical potentials of reactantsand products at the specific composition of reaction mixture.

A B

RG = (B - A)

Rxn. spontaneous in forward direction when A B

Rxn. spontaneous in reverse direction when B A

At Equilibrium:

RG = 0 when A = B

• chemical potential varies with composition

• slope of G vs. changes as rxn. proceeds

G vs

Figure from Atkins Text

At constant T and P:

RG 0 frwd. rxnis spontaneous

Rxn. is exergonic

…can be used to driveanother reaction.

RG 0 reverse. rxnis spontaneous

Rxn. is endergonic

RG = 0 at equilibrium

Figure from Atkins Text

An example of an exergonic rxn is ATP hydrolysis:

ATP + H2O ADP + Pi + H3O+

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.

We can use a favourable rxn. to ‘DRIVE’ an un-favourable one

…….exergonic rxns. can be used to drive endergonic rxns…….

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 endergonicreactions ….where ΔG° does not exceed 30 kJ

example. Biosynthesis of proteins is highly endergonic(due to enthalpic contribution as well as decreasein entropy that accompanies assembly of aminoacids into a specific sequence).

Formation of a peptide link is endergonic and willtypically consume approx. 3 ATP molecules for eachlink.

ex. In myoglobin …..there are approx. 150 peptide linksbiosynthesis of this protein consumes 450 ATP molecules

8.2. CHEMICAL EQUILIBRIUMin GASEOUS SYSTEMS

Consider the reaction:

aA (g) bB (g)

Remember equation for chemical potential of xth componentin a mixture:

x = x° + RT ln [Px/ P°]

Px 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 = A° + RT ln [PA/ P°]

B = B° + RT ln [PB/ P°]

Gibbs Energy change for the reaction can be written as:

RG = b B - a A

RG = bB° - aA° + b RT ln [PB/ P°] - a RT ln [PA/ P°]

Standard Gibbs Energy of reaction is ………..

RG° = b B° - a A°

RG = RG° + RT ln {[PB/ P°] b/[PA/ P°] a}

At equilibrium RG = 0

0 = RG° + RT ln {[PB/ P°] b/[PA/ P°] a }

RG° = - RT ln KP (where p denotes concn. are in pressures)

Kp, the equilibrium constant:

Kp = {[PB/ P°] b/[PA/ P°] a } = (PBb/ PA

a ) (P°) a-b

At a given temperature the value of RG° is a constant thatdepends only on nature of reactants and products as wellas temperature.

If we know RG° we can calculate KP and vice versa.

…..remember we can calculate RG° from Gibbs energy of formation of reactants and products (fG°).

…..so we can calculate KP also using values for fG°

(note…..we are talking about standard state conditionsi.e. 1 bar and 298 K)

Total Gibbs Energy vs. extent of reaction for the reaction: aA (g) bB (g).

In this case RG° 0

RG° = b B° - a A°

• when there is no mixingGibbs energy decreaseslinearly as rxn. progresses

• mixG is a neg. quantity

• Gibbs energy lower formixing than non-mixing case

• equilibrium point is “compromise” between opposingtendencies for reactant and product to mix and conversion of reactants into products.

RG° is usually not equal to zero, but when it is zero then K = 1.

K = 1 means products and reactants are equally favoredat equilibrium.

If RG° 0 then K 1

If RG° 0 then K 1 …………………this does not mean thereaction does not occur.

Example. RG° = 10 kJ , T = 298 K and KP = 0.018.

in this case products may still form if large quantities of reactants are used for the reaction.

K is dimensionless, and dependent on Temperature.

Sample Problem:From thermodynamic data given calculate the equilibrium

constant for the following reaction at 298 K:

N2 (g) + 3 H2(g) ↔ 2 NH3(g)

fG°(NH3(g)) = - 16.6 kJ/mol

Reactions are rarely carried out under standard state conditions:

so usually we are interested in RG

RG = RG° + RT ln Q

At a given temperature the value of RG° is constant but we canchange value of RG by changing partial pressure of gases.

(i.e. concentration dependence)

……………………….Q is the reaction quotient Q = KP if RG = 0

**If RG° is a large positive number (e.g. 50 kJ) then sign of RG is primarily determined by RG° unless reactants or productsare present in large quantities to override effect of RG°(i.e. equal in magnitude but opposite in sign)

Can determine direction of reaction using following expression:

RG = RT ln (Q/K)

If Q KP RG is negative and reaction proceeds in forward direction

If Q KP RG is positive and reaction proceeds in reverse direction

Sample Problem 9.2

At 298 K the partial pressures of the gases in reaction below are PN2 = 190 torr, PH2 = 418 torr, PNH3 = 722 torr. Calculate the value of RG for the reaction.

N2 (g) + 3H2(g) 2NH3(g) RG° = -33.2 kJ

The following equation applies only to an ideal gas:

x = x° + RT ln [Px/ 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

Fugacity versus pressure.

lim = 1 (P 0)

- fugacity can only be calculatedfor further discussion refer to Appendix 9.1 in Chang

Therefore for a reaction involving real gases:

a A (g) b B (g)

Kf = {[fB/ 1 bar] b/[fA/ 1 bar] a }

since = f/P

Kf = Bb {[PB/ 1 bar] b/ A

a [PA/ 1 bar] a } = K Kp

where K = Bb / A

a

and Kp = [PB/ 1 bar] b/ [PA/ 1 bar] a

Kf called thermodynamic equilibrium constant

Kp called apparent equilibrium constant(value not constant, depends on pressure)

• at low pressures KP approx. equal to Kf

• significant deviations occur as total pressure inc. beyond 300 bar• Kf remains approx constant over entire pressure range

Sample Problem on Equilibrium in Gaseous Systems Chang text # 9.3

Gaseous nitrogen dioxide is actually a mixture of nitrogen dioxide(NO2) and dinitrogen tetroxide (N2O4). If the density of such a mixture is 2.3 g/L at 74°C and 1.3 atm, calculate the partial pressuresof the gases and the value of KP for the dissociation of N2O4.

The equilibrium constant for the dissociation reaction:

N2O4 (g) 2 NO2 (g)

Chang Text # 9.5

Consider the reaction PCl5 (g) PCl3 (g) + Cl2 (g) KP = 1.05 at 250ºC.

A quantity of 2.50 g of PCl5 is placed in an evacuated flask of volume0.500 L and heated to 250°C. (a) Calculate the pressure of PCl5 if it did not dissociate.

Solution

(b) Calculate the partial pressure of PCl5 at equilibrium.

The partial pressure of PCl5 at equilibrium can be determined using The KP expression.

How to solve for x …………..

Quadratic Formula:

x = (-b b2 – 4 ac ) / 2 a

where a , b and c are given within the quadratic equation:

a x2 + b x + c = 0

(C) What is the total pressure at equilibrium ?

(D) What is the degree of dissociation of PCl5 ?

8.3. Chemical Equilibrium:

Reactions in Solution

Reactions in Solution

….here we express concentrations in molarity or molality

aA ↔ bB

A = A° + RT ln [mA/ m°]

* assuming ideal behavior and expression conc. in molality

where m° = 1 mol solute/kg solvent

RG° = - RT ln Km

where Km = {[mB/ m°] b/[mA/ m°] a }

With concentrations in molarity

Kc = {([B]/ 1M ) b/([A]/ 1 M) a }

RG° = - RT ln Kc for equilibrium reactions

RG = RG° + RT ln Q for non-equilibrium rxns.

For non – ideal solutions we replace concentration with activities.

x = x° + RT ln ax

(…this is like substitution of fugacity for pressure with gases)Ka = aB

b / aAa

a = m

so Ka = Bb / A

a {[mB/ m°] b/[mA/ m°] a }

= K K mwhere Km is apparent equilibrium constant

and Ka is thermodynamic equilibrium constant

8.4. Heterogeneous Equilibrium

….to this point we have only considered homogeneous equilibriummeaning reactions where we only have one phase………

…..heterogeneous equilibrium refers to reactions where we havemore than one phase………..

Example:CaCO3 (s) ↔ CaO (s) + CO2 (g)

…..the equilibrium constant would be

Kc = [CO2] / 1 M or KP = PCO2 / 1 bar

*** since we don’t include solids in equilibrium constantexpression

The thermodynamic equilibrium constant, Ka, for this reaction would be:

Ka = [aCaO aCO2] / aCaCO3 = [1 aCO2] / 1 = aCO2

………..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.

Kf = fCO2 / 1 bar

or if we assume ideal behavior……………….

KP = PCO2 / 1 bar

Plot of equilibrium pressure of CO2 over CaO and CaCO3 as a function of T.

Try example 9.3 in Changtext.

Chang text, page 311

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,nA,nB,nC,nD)

dG = nD,nC,nB,nA,PT

G

dT +

nD,nC,nB,nA,TPG

dP

+ nD,nC,nB,P,TAn

G

dnA +

nD,nC,nA,P,TBnG

dnB

+ nD,nB,nA,P,TCn

G

dnC +

nC,nB,nA,P,TDnG

dnD

= – S dT + V dP + μA dnA + μB dnB + μC dnC + μD dnD

From the stoichiometry of the reaction:

adnA =

bdnB =

cdnC =

ddnD ≡ – dξ

dξ = measure of the extent of the reaction

Therefore:

dnA = – a dξ dnB = – b dξ

dnC = + c dξ dnD = + d dξ

Then:

dG = – S dT + V dP + (c μC + d μD – a μA – b μB) dξ

= – S dT + V dP

+ [ c (μCo + RT ln aC) + d (μD

o + RT ln aD)

– a (μAo + RT ln aA) – b (μB

o + RT ln aB) ] dξ

= – S dT + V dP

+ [ (c μCo + d μD

o – a μAo – b μB

o)

+ RT (c ln aC + d ln aD – a ln aA – b ln aB ) ] dξ

= – S dT + V dP

+ [ (c μCo + d μD

o – a μAo – b μB

o) + RT ln

b

B

a

A

d

D

c

C

aaaa ] dξ

If temperature and pressure are constant

then dT = 0 and dP = 0

Define:

Δμo ≡ (c μCo + d μD

o – a μAo – b μB

o)

Then:

dG = [ Δμo + RT ln

b

B

a

A

d

D

c

C

aaaa ] dξ

If system is at equilibrium, then dG = 0 and:

Δμo + RT ln eq

b

B

a

A

d

D

c

C

aaaa

= 0

or:

ΔGo ≡ Δμo = – RT ln eq

b

B

a

A

d

D

c

C

aaaa

UNITS – UNITS – UNITS

Recall:

dG = – S dT + V dP

+ [ c (μCo + RT ln aC) + d (μD

o + RT ln aD)

– a (μAo + RT ln aA) – b (μB

o + RT ln aB) ] 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 (μCo + RT ln aC)

c (μCo + RT ln aC)

Now include the units explicitly in addition to the numerical value;

Note that c , μC

o , R , T and aC from this point on are simply “numbers”.

(c (mol))[ μCo (kJ mol–1) + (R (kJ mol–1 K–1))(T (K)) ln aC ]

After multiplication of the numerical values and cancellation

of the units we are left with the following:

c μCo (kJ) + c R T (kJ) ln aC = c μC

o (kJ) + R T (kJ) ln aCc

All the terms in the coefficient of dξ are in the same unit (kJ).

c , μCo , R , T and aC are simply “numbers”.

Equilibrium Constant Calculations

ΔGo ≡ Δμo = – RT ln eq

b

B

a

A

d

D

c

C

aaaa

ΔGo and Δμo 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)

Jacobus Henricus van’tHoff

(1852-1911)

1901 – 1st Nobel Prize in Chemistry

8.5. Influence of Temperature, and Catalystson Equilibrium Constant

Influence of Temperature:

van’t Hoff Equation:

ln (K2/K1) = (RH° /R) [( 1/T1) – (1/T2)]

= (RH° /R) [(T2 - T1) /(T1 T2)]

Equation allows calculation of K at one temperature if we know K at another temperature.

….here we assume RH° is temperature independent.

The van’t Hoff Equation

ΔGo = ΔHo – T ΔSo = – RT ln Keq

If ΔHo and ΔSo are independent of temperature

then for temperatures T1 and T2 we may write:

ln K2 = – 2

o

RTH +

RSo (A)

ln K1 = – 1

o

RTH +

RSo (B)

Subtract (B) from (A):

ln K2 – ln K1 = ln

1

2

KK

= –

12

o

T1

T1

RH

= +

21

o

T1

T1

RH

The van't Hoff Equation

ln K = - (RG° / RT )

and RG° = RH° - TRS°

so ln K = - RH°/RT + RS° /R

A plot of ln K versus 1 / T gives a straight line with

Slope = - RH°/R

y-intercept = RS° /R

…..this plot only yields a straight line if RS° and RH° are independent of Temperature.

……for narrow temperature ranges this method works (i.e. 50 K or less)

a) RH ° 0 (b) RH ° 0 (c) RH ° = 0

ln (K2/K1) = (RH° /R) [(T2 - T1) /(T1 T2)]

….if RH° is positive (endothermic) and T2 T1 then K2 K1

….if RH° is negative (exothermic) and T2 T1 then K2 K1

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.

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 triesto 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

Endothermic Exothermic

(a) Increase in Temperature, Boltzmann distribution adjusts and population changes………increased population of higher energystate at expense of population in lower energy state. Population of B becomes more abundant in equilibrium mixture.

Atkins text. page 237

Effect of a Catalyst

• a catalyst can speed up the rate of a reaction without itself beingconsumed.

• a catalyst cannot shift the position of equilibrium

• for a reaction not at equilibrium a catalyst will increase boththe forward and reverse rates so equilibrium will be reachedsooner

Q: what happens if we add reactant “A” to a system at equilibrium ?

A: the system responds by driving forward the reaction to use up reactant “A” and produce more product “B”

add more “A”

Equilibriumre-establishes

add more “B”

pure ‘A’ fraction of ‘A’ in mix pure ‘B’

0 1

If we add more A , then drive A B

If we add more B then we drive A B

Chang Text Problem # 9.10

The vapor pressure of dry ice (solid CO2) is 672.2 torrat – 80.00°C and 1486 torr at –70.00°C. Calculate the molar heat of sublimation of CO2.

The van’t Hoff equation

ln (K2/K1) = (RH/R) [ T1-1 – T2

–1]

CO2 (s) CO2 (g)

Problem 9.22 in Chang Text:

Photosynthesis can be represented by 6CO2(g) + 6H2O (l) C6H12O6 (s) + 6O2 (g) RH = 2801 kJ

How would equilibrium be affected by the following changes :

(a) Partial pressure of CO2 is increased

(b) O2 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

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 upwardsby coupling to falling of large weight.

• 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 (Cu2S)

…..heating the ore will not yield much copper because process isassociated with large positive ΔRG°.

Cu2S (s) 2Cu (s) + S (s) ΔRG° = 86.2 kJ

……if the thermal decomposition is coupled to the oxidation of sulfur then overall process is spontaneous…………

Cu2S (s) 2Cu (s) + S (s) ΔRG° = 86.2 kJ

S (s) + O2 (g) SO2 (g) ΔRG° = - 300 kJ

Overall: Cu2S + O2 (g) 2Cu (s) + SO2 (g) ΔRG° = - 213.9 kJ

Gibbs energy for the overall process is the sum of the Gibbs Energies of each reaction.

Overall process has large neg. value for ΔRG° and thus is spontaneous.

Characteristics of Coupled Biological Reactions: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

ATP ATP –– Currency of Energy in Biological SystemsCurrency of Energy in Biological Systems

ATP hydrolysis is used to drive many biological processesincluding protein synthesis, ion transport, muscle contraction,electrical activity in nerve cells.

Hydrolysis Reaction of ATP at pH = 7:

ATPATP44-- + H+ H22O O ADPADP33-- + H+ H++ + HPO+ HPO4422--

The exact value of ΔRG° associated with ATP hydrolysis is dependent on Temperature, pH and presence of metal ions. Usually ranges between ΔRG° = - 25 to - 40 kJ

At pH = 7, T = 310 K and in presence of Mg2+ ions ΔRG° = - 30.5 kJ

ATPATP

Upon hydrolysis ATP ADP and loses terminal phosphate group

ADP can be further hydrolyzed to produce AMP

Note:

4 negativecharges in molecule

Why is hydrolysis of ATP to form ADP associated withWhy is hydrolysis of ATP to form ADP associated withnegative value for negative value for ΔΔRRGG°° ??

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 3

2 - resonance stabilization- ADP and HPO4

2- possess more resonance structuresthan ATP see page 329 in Chang text.

Examples of Exergonic Hydrolysis Reactions

Example:

alanine + glycine alanylglycine ΔRG° = 17.2 kJ

K = 1 x 10 –3 at 298 K

Reaction is said to be thermodynamically controlled.

Positive value for ΔRG° signifies that reaction is not spontaneous

If coupled to ATP hydrolysis this reaction will occur spontaneously.

• reactions may also be kinetically controlled

• a kinetically controlled reaction has a negative value for ΔRG°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 butproceeds very slowly in absence of enzyme hexokinase(catalyst).

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) + O2 (g) MbO2 (aq)

K = [MbO2] / [O2] [Mb] = [MbO2] / PO2 [Mb]

…………..P represents the partial pressure of the O2 gas.

We are interested in concentration of oxygenated complex:

[MbO2] = K PO2 [Mb]

Reference: Atkins Text (The Elements of Phys Chem with Applications in Biology)

Mb (aq) + O2 (g) MbO2 (aq)

Y = fractional saturation

Y = [MbO2] / ([Mb] + [MbO2])

[MbO2] = K PO2 [Mb]

Y = K PO2 [Mb] / ([Mb] + K PO2 [Mb])

so Y = K PO2 / (1 + K PO2) (divide all by [Mb])

Therefore, as we increase P of oxygen we increase Y.

Oxygen saturation curve

for Myoglobin.

9.34. A polypeptide can exist in either the helical or random coilforms. The equilibrium constant for equilibrium reaction of the helixto the random coil transition is 0.86 at 40°C and 0.35 at 60°C.Calculate the values of RH° and RS° for the reaction.

9.36 At 720 K the equilibrium partial pressures are PNH3 = 321.6 atm,PN2 = 69.6 atm, PH2 = 208.8 atm, respectively.(a) Calculate the value of KP for the reaction given below.(b) Calculate the thermodynamic equilibrium constant if

NH3 = 0.782, N2 = 1.266 and H2 = 1.243

The reaction is N2 (g) + 3 H2 (g) 2 NH3(g)