PART 10 – Chemical EquilibriumPART 10 Chemical Equilibrium

Reference: Chapter 14 17 in textbookReference: Chapter 14, 17 in textbook

1

Basic Properties1. Dynamic Equilibrium

e.g. N2O4 (g) 2 NO2 (g)

At ilib iAt equilibrium, vforward = vbackward

showing the system reaches an equilibrium state.g y q

Q: How to draw the graph by v ~ t?

2

Basic Properties

2. System reaches an equilibrium state Sys e eac es a equ b u s a espontaneously

All th ilib i t t d t f ditiAll the equilibrium states are under a set of conditions,

and can be affected when conditions change.

3. It does NOT matter whether the system starts from reactant side or product side.

e g N O (g) 2 NO (g)e.g. N2O4 (g) 2 NO2 (g)

Case 1: Starting from 1 mol N2O4 (g)

Case 2: Starting from 2 mol NO2 (g)3

Basic Properties

4 Dri ing Force4. Driving ForceThe coexistence of two driving forces leads to an

equilibrium.

e.g. N2O4 (g) 2 NO2 (g)

Reduced energy (enthalpy): 2 NO2 (g) N2O4 (g)Reduced energy (enthalpy): 2 NO2 (g) N2O4 (g)

Increased entropy: N2O4 (g) 2 NO2 (g)

At Equilibrium: ΔG = 0, (i.e. ΔH = T • ΔS)

4

Equilibrium Constant

Equilibrium Constant (K)q ( )Revisit: Kw, Ka, Kb, Ksp, ……

Formally defined – Mass Law

For a reaction: a A + b B c C + d DFor a reaction: a A b B c C d D

At equilibrium:

K = (CCc • CD

d) / (CAa • CB

b)

K (P c P d) / (P a P b) (if A B C D )K = (PCc • PD

d) / (PAa • PB

b), (if A, B, C, D are gases)

Equilibrium Constant K is only the functionEquilibrium Constant K is only the function of T 5

a. The representation of K should be consistent with thea. The representation of K should be consistent with the chemical reaction, and designated with the temperature.

Example: N2O4 (g) 2 NO2 (g) K0373K = 11.01

b. In a heterogeneous (multi-phase) equilibrium, K only l t t th d l ti t tirelates to the gas pressures and solution concentrations.

Example: CaCO3 (s) CaO (s) + CO2 (g)

K = PCO2

Example: Zn (s) + 2 H+ (aq) Zn2+ (aq) + H (g)Example: Zn (s) + 2 H+ (aq) Zn2+ (aq) + H2 (g)

K = [Zn2+] • PH2 / [H+]2

6

I (dil t d) t l ti th [ t ] i 1 d h dc. In (diluted) water solution, the [water] is 1 and unchanged.

Example: Cr2O72- (aq) + H2O (l) 2 CrO4

2- (aq) + 2 H+ (aq) p 2 7 ( q) 2 ( ) 4 ( q) ( q)

K = ([CrO42-]2 • [H+]2) / [Cr2O7

2-]

d. For different chemical reactions, K has different values.

N2 (g) + 3 H2 (g) 2 NH3 (g) KTN2 (g) + 3 H2 (g) 2 NH3 (g) KT

½ N2 (g) + 3/2 H2 (g) NH3 (g) KT’ = (KT)1/2

2 NH3 (g) N2 (g) + 3 H2 (g) KT’’ = 1 / KT

7

PracticeQ: For a reaction: COCl2 (g) CO (g) + Cl2 (g)

In a fixed volume container at 900 K there was someIn a fixed volume container at 900 K, there was some COCl2 (g) with initial pressure of 101.3 kPa. When this dissociation reaction reached equilibrium, the total pressure of q , pthe container is 189.6 kPa. Calculate the equilibrium constant. Solution:

COCl2 (g) CO (g) + Cl2 (g)

I iti l 101 3 0 0Initial: 101.3 0 0

Change: –x x x

Equilibrium: 101.3 – x x x

(101 3 – x) + x + x = 189 6 x = 88 3 kPa

8

(101.3 – x) + x + x = 189.6 x = 88.3 kPa

K = (88.3) * (88.3) / (101.3 – 88.3) = 600

Equilibrium Constant & ΔG

For an arbitrary gas reaction:For an arbitrary gas reaction:

a A (g) + b B (g) c C (g) + d D (g)

ΔG = c ΔGC + d ΔGD – (a ΔGA + b ΔGB)

ΔG ΔGo + d ΔGo ( ΔGo + b ΔGo )ΔG = c ΔGoC + d ΔGo

D – (a ΔGoa + b ΔGo

B)

+ RT [(c ln PC + d ln PD) – (a ln PA+ b ln PB)][( C D) ( A B)]

since: c ΔGoC + d ΔGo

D – (a ΔGoa + b ΔGo

B) = ΔGo

Therefore:

ΔG = ΔGo + RT ln [ (PCc • PD

d) / (Paa • PB

b) ]

9

ΔG ΔG RT ln [ (PC PD ) / (Pa PB ) ]

Equilibrium Constant & ΔG

Or in a solution reaction:

ΔG = ΔGo + RT ln [ (CCc • CD

d) / (Caa • CB

b) ]

dC dC

PbB

aA

dD

CC Q

PPPP

=⋅⋅

CbB

aA

dD

CC Q

CCCC

=⋅⋅

ΔGT = ΔGT° + RT ln Q

BA BA

GT GT Q

ΔGT < 0 Reactant side Product side

ΔGT = 0 No net reaction (System in equilibrium)

ΔG > 0 Product side Reactant side

10

ΔGT > 0 Product side Reactant side

When ΔGT = 0:T

0 = ΔGT = ΔGT° + RT ln Q

At this time, all species’ concentrations do not change with

ti i th t h h d ilib itime any more, i.e., the system has reached equilibrium.

Therefore:Therefore:

Q = K

0 = ΔGT° + RT ln K

ΔGT° = – RTln K

11ln K = – ΔGT° / RT or: log K = – ΔGT° / 2.303RT

ΔGT = ΔGT° + RT ln QT T

ΔGT = – RT ln K + RT ln Q

Q < K, ΔG < 0 Reactant side Product side

Q = K, ΔG = 0 No net reaction (in equilibrium)

Q > K ΔG > 0 Product side Reactant sideQ > K, ΔG > 0 Product side Reactant side

Therefore:

ΔG determines the direction of reaction;

ΔG° determines the level of reaction12

ΔG determines the level of reaction.

PracticeQ1: For a reaction

H ( ) + I ( ) 2 HI ( )H2 (g) + I2 (g) 2 HI (g)

At T = 700 K, K = 55.3. When PHI = 0.7 atm, PH2 = PI2, ,= 0.2 atm, what is the direction of reaction?

Solution:

Q = (PHI)2 / [ (PH2) • (PI2) ] = (0.7)2 / (0.2)*(0.2)

= 12.25 < K

So the net reaction is to grow more HI product13

So the net reaction is to grow more HI product.

PracticeQ2: For a reaction

BaCl H O (s) BaCl (s) + H O (g)BaCl2·H2O (s) BaCl2 (s) + H2O (g)

If ΔG°298K of this reaction is 19.50 kJ·mol-1 . What is 298Kthe vapor pressure of water at 25°C ?

Solution:

ΔG ° RT l K l K ΔG °/ RTΔGT° = – RT lnK lnK = – ΔGT°/ RT

lnK = – 19.50 x 1000 / (8.314 x 298) = – 7.87lnK 19.50 x 1000 / (8.314 x 298) 7.87

K = 3.82 x 10-4

14K = PH2O = 3.82 x 10-4 atm

A Few Notes of ΔGoT and KT

Units for R and ΔG sho ld be consistentUnits for R and ΔG should be consistent.

K is a function of temperature T, i.e. K (T). (From handbooks ΔGo is given for 298 K(From handbooks, ΔGo

f is given for 298 K.

From K’s at different T’s, we can calculate corresponding ΔGo

T then we cancorresponding ΔG T , then we can calculate ΔHo and ΔSo . (Assume ΔHo and ΔSo are temperature independent )ΔSo are temperature independent.)

15

PracticeQ3: For a reaction,

CO (g) + H2O (g) H2 (g) + CO2 (g), ΔHo298K= –9.85 kJ/mol(g) 2 (g) 2 (g) 2 (g) 298K

At 1.00 atm and 800 K, mix equal moles of CO (g) and H2O (g). When the reaction reaches equilibrium, the pressure of CO2 (g) is e t e eact o eac es equ b u , t e p essu e o CO2 (g) smeasured as 0.498 atm. Calculate ΔSo

298K of this reaction.

Solution： PCO2 = PH2 = 0.498 atm

PCO = PH2O =( 1.00 – 2 x 0.498 ) / 2 = 0.002 atm CO H2O

K = (PCO2 · PH2 ) / (PCO · PH2O) = 0.4982 / 0.0022 = 6.2 x 104

4ΔGo800K = – RT lnK = – (8.314 x 800 x ln 6.2x104) = –73390 J/mol

ΔGo800K = ΔHo

298K – T ΔSo298K

16

800K 298K 298K

–73390 = –9850 – 800 x ΔSo298K ΔSo

298K = 79.4 J/mol·K

Hess Law for ΔG & KAs ΔG°T is an extensive property and can be added up. Both ΔG°T and K relate to certain chemical reactions.

Example:

(1) CO2 (g) ½ O2 (g) + CO (g) ΔG°1( ) CO2 (g) ½ O2 (g) CO (g) G 1

(2) H2 (g) + ½ O2 (g) H2O (g) ΔG°2

(1) + (2) ：

(3) H2 (g) + CO2 (g) H2O (g) + CO (g) ΔG°32 2 2 3

ΔG°3 = ΔG°1 + ΔG°2

– RT ln K°3 = – RT ln K°1 + (– RT ln K°2)

17

3 1 ( 2)

So: K°3 = K°1• K°2

Example:

N2 (g) + 3 H2 (g) 2 NH3 (g)

ΔG° RT l KoΔG° = – RT ln Ko

½ N2 (g) + 3/2 H2 (g) NH3 (g)

ΔG°’ = ½ ΔG° = – ½ RT ln K = – RT ln Ko ’

Ko ’ = (Ko)1/2

2 NH3 (g) N2 (g) + 3 H2 (g)2 NH3 (g) N2 (g) + 3 H2 (g)

ΔG°’’ = – ΔG° = RT ln Ko = – RT ln Ko’’

18K0 ’’ = 1 / Ko

Multiple Ionic Equilibria in SolutionExample 1:

H2S + H2O H3O+ + HS- Ka1

HS- + H2O H3O+ + S2- Ka2

H S + 2 H O 2 H O+ + S2- K = K • KH2S + 2 H2O 2 H3O + S2 Ka = Ka1 • Ka2

Example 2:Example 2:

AgCl (s) = Ag+ (aq) + Cl- (aq) Ksp

Ag+ (aq) + 2 NH3 (aq) = Ag(NH3)2+ Kf

19

AgCl(s) + 2 NH3 (aq) = Ag(NH3)2+ + Cl- (aq) K= Ksp • Kf

Shift of Chemical EquilibriaLe Châtelier’s Principle

– Qualitative prediction of the shift of an equilibrium

Eff t f T t h ?Effect of Temperature changes ?

Effect of Concentration changes ?g

Effect of Pressure changes ?

Effect of Adding inert gas ?

Effect of Catalyst ?Effect of Catalyst ?

Q: Plot graphs of Concentration vs Time in the casesQ: Plot graphs of Concentration vs. Time, in the cases

when an old equilibrium is shifted to a new one? 20

Effect of Temperature Changeg

Temperature Affects Equilibrium Constantsp qK is a function of Temperature

ΔGoT = – RT ln K, or ln K = – ΔGo

T / RT

Because at a constant T:Because at a constant T:

ΔGoT = ΔHo – T·ΔSo

ln K = (– ΔHo + T·ΔSo) / RT

( ΔHo / RT) (ΔSo / R)= (– ΔHo / RT) + (ΔSo / R)

Plot K to 1/T: slope = – ΔHo / R,p

intercept = ΔSo / R 21

Wh T T l K ( ΔHo / RT ) + (ΔSo / R)When T = T1: ln K1 = (– ΔHo / RT1) + (ΔSo / R)

When T = T2: ln K2 = (– ΔHo / RT2) + (ΔSo / R)

So: ln (K2 / K1) = (– ΔHo / R) (1/T2 – 1/T1)

van’t Hoff Equation

Example： H2O (l) H2O (g)

A T K P A T K PAt T1: K1 = P1, H2O ; At T2: K2 = P2, H2O

Then: ln (P2 H2O / P1 H2O) = (– ΔHo / R) (1/T2 – 1/T1)Then: ln (P2, H2O / P1,H2O) = (– ΔH / R) (1/T2 – 1/T1)

Equation of how vapor pressure changes with temperature.

22

Effect of Concentration Changeg

Quantitative Prediction:

At Equilibrium: Q = K = ( [C]c · [D]d ) / ( [A]a · [B]b )

When the concentrations of products decrease to ½ :Q = ( [C/2]c · [D/2]d ) / ( [A]a · [B]b ) = (½)c+d · K Q < K Equilibrium moves to product side.

When the concentrations of reactants increase to twice :Q = ( [C]c · [D]d ) / ( [2A]a · [2B]b ) = (½)a+b · K Q < K Equilibrium moves to product side.Q q p

23

Effect of Pressure Changeg

No Effect for equilibrium of only solids & liquids.No Effect for equilibrium of only solids & liquids.

No Effect for reactions with same numbers of gas molecules before and after.

i.e. ( Δngas = ( c + d ) – ( a + b ) = 0 )

b K ( P P d ) / ( P P b )because: K = ( PCc · PD

d ) / ( PAa · PB

b )

When Ptotal doubles, each gas’ partial pressure doubles.When Ptotal doubles, each gas partial pressure doubles.

so: Q = [ (2PC) c · (2PD) d ] / [ (2PA) a · (2PB) b ] = 2(c+d-a-b) · K

24As: (c+d) – (a+b) = 0 Q = (2)0 · K = K

Effect of Pressure ChangegFor Δngas ≠ 0:

When Ptotal doubles, each gas’ partial pressure doubles.

so: Q = [ (2PC) c · (2PD) d ] / [ (2PA) a · (2PB) b ]

= 2(c+d-a-b) · K = 2 Δn · K= 2( ) K = 2 K

where: Δngas = ( c + d ) – ( a + b )

Δngas > 0 Q > K Equilibrium moves to reactants.

Δngas < 0 Q < K Equilibrium moves to products.

i e Increase total pressure shifts equilibrium to direction

25

i.e. Increase total pressure shifts equilibrium to direction of fewer gas molecules.

Effect of Adding Inert Gasg

If V of the system is not changed so allIf Vtotal of the system is not changed, so all partial pressures are not changed, the equilibrium does not shiftequilibrium does not shift.

If Ptotal of the system is not changed, which means the Vtotal is increased. So all partial pressures are reduced, the equilibrium shifts in the direction of increase gas molecules.

26

Practice

Q1: For a reaction:Q1: For a reaction:

2 NO2 (g) N2O4 (g) , At 25ºC, K = 8.81.2 (g) 2 4 (g)

If we put 3.0 mol N2O4 gas into a 4 liter container, when

the reaction reaches equilibrium:

(a) How many moles are there for each substance?

(b) What is the conversion yield of N2O4 ?

27

Solution:Solution:

2 NO2 (g) N2O4 (g)

2x mol (3.0 – x) mol

K = PN2O4 / PNO22 = 8.81N2O4 / NO2 8 8

PN2O4 = (3.0 – x) RT / V PNO2 = (2 x) RT / V

K = 8.81 = (3.0 – x) RT V2 / V ( 2 x RT)2

= (3.0 – x) 4 / (2 x)2 0.082 · 298 x = 0.12 mol

n = 3 0 x = 2 88 mol n = 2x = 0 24 molnN2O4 = 3.0 – x = 2.88 mol nNO2 = 2x = 0.24 mol

α N2O4 = 0.12 mol / 3.0 mol = 0.04

28

PracticeQ2: For a reaction: SO2Cl2 (g) SO2 (g) + Cl2 (g)

At 375 K, K = 2.4

If we put 6.7 gram of SO2Cl2 (g) inside a 1-liter

container. (MSO2Cl2 = 135 g/mol)container. (MSO2Cl2 135 g/mol)

(a) When the reaction reaches equilibrium, what is the

partial pressure of each gas species?

(b) If dd th 1 t f Cl ( ) i t th ilib i(b) If we add another 1 atm of Cl2 (g) into the equilibrium

system above, what is the partial pressure of each gas

29at the new equilibrium?

Solution:(a) Po = nRT / V = (6.7 / 135) x 0.082 x 375 /1 = 1.526 atm( ) o ( )

SO2Cl2 (g) SO2 (g) + Cl2 (g)

Initial (atm): 1.526 0 0Initial (atm): 1.526 0 0

Equilibrium: 1.526 – x x x

K = 2 4 = x2 / 1 526 x so: x = 1 058 atmK = 2.4 = x2 / 1.526 – x so: x = 1.058 atm

PSO2Cl2 = 1.526 – 1.058 = 0.467 atm

P P 1 058 tPSO2 = PCl2 = 1.058 atm

(b) SO2Cl2 (g) SO2 (g) + Cl2 (g)

Initial (atm): 1.526 0 1

Equilibrium: 1.526 – y y 1+y q y y y

K = 2.4 = y (1+ y) / (1.526 – y) so: y = 0.859 atm

30

PSO2Cl2 = 1.526 –0.859 = 0.667 atm

PSO2= 0.859 atm PCl2= 1 + y = 1. 859 atm

(b) Alternatively:( ) y

SO Cl (g) SO (g) + Cl (g)SO2Cl2(g) SO2 (g) + Cl2 (g)

Initial (atm): 0.467 1.058 1.058 + 1

E ilib i 0 467 1 058 2 058Equilibrium: 0.467 + x 1.058 – x 2.058 – x

K = 2.4 = (1.058 – x) · ( 2.058 – x) / (0.467 + x)

so: x = 0.199 atm

PSO2Cl2 = 0.467 + x = 0.667 atm

PSO2 = 1.058 – x = 0.859 atm

PCl2 = 2.058 – x = 1.859 atm

31

Practice

Q3: For a reaction, N2O4 (g) 2 NO2 (g) , at 27ºC, 2 4 2

Ptotal = 1 atm, 20% of N2O4 dissociates into NO2 .

(a) What is K of this reaction at 27ºC ?

(b) In an equilibrium at 27ºC, Ptotal = 0.1 atm, what is the

dissociation percentage of N O (g)?dissociation percentage of N2O4 (g)?

(c) If at 27ºC, put 69 grams of N2O4 (g) in a 20-liter ( ) , p g 2 4 (g)

container. What is the dissociation percentage of N2O4

? ( )32

when the system reaches the equilibrium? (MWN2O4 = 92)

Solution:

(1) N2O4 2 NO2

1 – α 2α

PNO2 = Ptotal · XNO2 , XNO2 = 2α / (1 α + 2α) = 2α / (1 + α) ,

PNO2 = Ptotal · 2α / (1+ α)

PN2O4 = Ptotal · XN2O4, XN2O4= (1 – α) / (1+ α )

PN2O4 = Ptotal · (1 – α) / (1+ α )

So: K = (PNO2)2/ PN2O4 = Ptotal2 [ 2α / (1+ α )]2 / [ Ptotal (1 – α ) / (1+ α ) ]

= Ptotal (2α)2 / (1 – α )(1+ α ) = 1 ( 2 x 0.2 )2 / [ (1 – 0.2 ) (1+ 0.2 ) ]

= 0 167

33

= 0.167

(2) K = 0.167 = Ptotal (2α)2 / (1 – α )(1+ α )

0.167 = 0.1 (2α)2 / (1 – α2)

α = 0.542

(3) Pintial = nRT / V = ( 69 / 92 ) x 0.082 x 300 / 20.0 = 0.9225 atm

K = (P )2 / PK = (PNO2)2 / PN2O4

0.167 = (2x)2 / (0.9225 – x)

so: x = 0.176 atm

α = 0 176 atm / 0 9225 atm = 0 191

34

α 0.176 atm / 0.9225 atm 0.191

PracticeQ4: For hemoglobin protein’s reaction with O2 (g):

Hb (aq) + O2 (g) HbO2 (aq), K = 85.5

When PO2 in the air is 20.2 kPa, the solubility of O2 in water O2 y 2is 2.3 x 10-4 mol/L. Calculate the equilibrium constant K for:

Hb (aq) + O2 (aq) HbO2 (aq)Hb (aq) O2 (aq) HbO2 (aq)

Solution:(1) Hb ( ) O ( ) HbO ( ) K 85 5(1) Hb (aq) + O2 (g) HbO2 (aq) K1 = 85.5

(2) O2 (g) O2 (aq)

K2 = [O2] / PO2 = 2.3 x 10-4 / (20.2/101.3) = 1.15 x 10-3

(3) Hb (aq) + O2 (aq) HbO2 (aq) K3 = ?

35As (3) = (1) – (2), so: K3 = K1 / K2 = 85.5 / 1.15x10-3 = 7.4 x 104