Equilibrium Topics in CHM 1046 1.Intermolecular forces (IMF) 2.Themodynamics 3.Chemical Kinetics...

Equilibrium

Topics in CHM 10461. Intermolecular forces (IMF)

2. Themodynamics

3. Chemical Kinetics

4. Chemical Equilibrium

Combination of above:

Unit 11 Intermolecular Forces, Liquids & Solids (T1)Unit 12 Properties of Solutions (T1)

Unit 13 Thermochemistry (T1)Unit 14 Chemical Thermodynamics (T2)

Unit 15 Chemical Kinetics (T2)

Unit 16 Chem. Equilibrium Gases & Heterogeneous (T3)Unit 17.Acid Base Equilibria (T3)Unit 18 Acid Base Equilibria Buffers & Hydrolysis (T3)Unit 19 Acid Base Equilibria Titrations (T4)Unit 20 Aqueous Equilibria Solubility Product (T4)

Unit 21 Electrochemistry (T4)

Equilibrium

Unit 16Chemical Equilibrium:

Gases & Heterogeneous

Dr. Jorge L. AlonsoMiami-Dade College –

Kendall CampusMiami, FL

Textbook Reference: •Chapter # 17•Module # 5 (& Appendix 2)

CHM 1046: General Chemistry and Qualitative Analysis

Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

This does not mean chemicals are found in same concentration!

The Concept of Equilibrium{Non-Equilibrium Reactions}: proceed in one direction, A B

{Equilibrium Reactions}: proceed in both directions, A ↔ B

2 NO (g) + O2 (g) 2 NO2 (g)

(clear gases) (red-brown gas)

(clear gas) (red-brown gas)

N2O4 (g) 2 NO2 (g)

2X

Equilibrium

The Equilibrium Constant (Keq)

• Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

• Rewriting this, it becomes

kf

kr [NO2]2

[N2O4]=

The Equilibrium Constant is the ratio of the rate constants at a particular temperature.

Keq =

N2O4 (g) 2 NO2 (g)

Equilibrium

The Equilibrium Constant (K)

• To generalize this expression, consider the reaction

• The equilibrium expression for this reaction would be

Kc = [C]c[D]d

[A]a[B]b

aA(aq) + bB(aq) cC(aq) + dD(aq)

where [X] = concentration of each chemical in moles/ liter (M).

Equilibrium

What Does the Value of K Mean?

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

Kc = [C]c[D]d

[A]a[B]b

aA + bB cC + dD

Kc =10 = 10

Kc = 1 = 0.1

1

10

aA + bB cC + dD

Kc = [C]c[D]d

[A]a[B]b

10 x

10 x

Equilibrium

The Equilibrium Constant for Gases

Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written

Kp =(PC)c (PD)d

(PA)a (PB)b

aA(g) + bB(g) cC(g) + dD(g)

where Kp = equilibrium constant for gases, and Px = partial pressure of each gas.

Partial press: (75 torrs) (150 torrs) (300 torrs) (600 torrs) = total (1125 torrs)

Kc = [C]c[D]d

[A]a[B]b

Equilibrium

Relationship between Kc and Kp• ideal gas law:

PV = nRT

P = RTnV

Plugging this into the expression for Kp

Kp = Kc (RT)n

Wheren = (moles of gaseous product) − (moles of gaseous reactant)

Kp =(PC)c (PD)d

(PA)a (PB)b

( )c ( )d

( )a ( )b

RTnV RT

nV

RTnV RT

nV

=

Kc = [C]c[D]d

[A]a[B]b

☺

Equilibrium

Equilibrium Can Be Reached from Either Direction

The ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

[NO2]2

[N2O4]=KC

212.0]0045.0[

]0310.0[

]0045.0[

]0310.0[

]0028.0[

]0243.0[

]0014.0[

]0172.0[ 2222

cK

@ 100°C

Equilibrium212.0]0045.0[

]0310.0[

]0028.0[

]0243.0[

]0014.0[

]0172.0[

]ON[

]NO[K

222

42

2

2c

Equilibrium Can Be Reached from Either Direction

N2O4 (g) 2 NO2 (g)

Equilibrium

Manipulating Equilibrium Constants

The equilibrium constant of a reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

KI = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2(g)

N2O4(g)2 NO2 (g)

RULE #1: Reciprocal Rule

KII =

[N2O4][NO2]2

= 4.72 at 100C

1= 0.212

Equilibrium


If the coefficients of a chemical equation are changed (increased or decreased) by a factor n, then the value of K is raised to that power.

KI = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

KIII = = (0.212)2 at 100C[NO2]4

[N2O4]22 N2O4 (g) 4 NO2 (g)

RULE #2: Coefficient Rule

KIV = ½N2O4(g) NO2(g)C100 at 212.0

]O[N

]NO[

42

2

2

Equilibrium


When two or more equations are added to obtain a final resultant equation, the equilibrium constant for the resultant equation is product of the constants of the added equations.

]B[]A[

]C[2

4V K

RULE #3: Multiple Equilibria Rule

2 A(aq) + B(aq) ↔ 4 C(aq)

4 C(aq) + E(aq) ↔ 2 F(aq)

2 A(aq) + B(aq) + E ↔ 2 F(aq)

]E[]C[

]F[4

2VI K

]E][B[]A[

]F[2

2VII K

Equilibrium

Homogeneous

vs.Heterogeneous

Equilibria

Reactions involving only gasses or only solutions

Reactions involving different phases (s, l, g, aq) of matter

N2O4 (g) 2 NO2(g)

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

Kc = = 0.212 at 100C[NO2]2

[N2O4]

][ 2COKC

Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant

The concentrations of solids and liquids do not appear in equilibrium expressions because their concentration does not change.

Kc = [Pb2+] [Cl−]2

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Concentration of both solids and liquids can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

L

mol

LMM

g

MML

g

MM

DMionConcentrat )(

Write the equilibrium expression!

Equilibrium

What are the Equilibrium Expressions for these

Heterogeneous Equilibria?

2

2

2

CO

COp

P

PK

2COp PK

] [

][Zn 2

2

Cu

Kc

EquilibriumAs long as some CaCO3 or CaO remain in the system, the amount of CO2 above the solid will remain the same.

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

How does the concentration of CO2 differ in the two bell jars?

2COp PK

Equilibrium

Equilibrium Calculations

(2) Calculating Equilibrium Concentrations [ ]e when Kc & [ ]i are known

(1) Calculating Kc from experimental data: [ ]i & [ ]e

I [A]i [B]i 0 0

C - change

- change

+ change

+ change

E [A]e [B]e [C]e [D]e

[A] + [B] [C] + [D]

bequi

aequi

dequi

cequi

c BA

DCK

][][

][][

ICE Tables:Initial

Change

Equilibrium

bequii

aequii

dequi

cequi

changeBchangeA

DC

]][[]][[

][][

Equilibrium


A closed system initially containing 1.000 x 10−3 M H2 and 2.000 x 10−3 M I2 at 448C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10−3 M. Calculate Kc at 448C for the reaction taking place, which is

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

[HI] Increases by 1.87 x 10-3 M

[H2], M + [I2], M 2 [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At Equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Stoichiometry tells us [H2] and [I2] decrease by half as much

Kc =[HI]2

[H2] [I2]

Equilibrium

…and, therefore, the equilibrium constant

= 51

=(1.87 x 10-3)2

(6.5 x 10-5)(1.065 x 10-3)

[H2], M + [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At Equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

Kc =[HI]2

[H2] [I2]

EquilibriumThe ratio of [NO2]2 to [N2O4] remains constant at this temperature no matter what the initial concentrations of NO2 and N2O4 are.

212.0]0045.0[

)2]00.([

]0045.0[

)2]04.0([

]0028.0[

)2]03.0([

]0014.0[

)2]02.0([ 2222

xxxx

K iiiic

@ 100°C

Calculating Kc from experimental data: [ ]i & [ ]e

N2O4 (g) 2 NO2 (g)

- 2x- 2x- 2x+

2x

+ x+ x+ x - x

x

[NO2]2

[N2O4]=Kc

I 0.0 0.0200

C + x - 2x

E 0.0014 0.0200-2x

Equilibrium

Practice Problem

2006B Q5

{ }

Equilibrium

Equilibrium

(2) Calculating Equilibrium Concentrations when K & [ ]i are known



[H2], M + [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At Equilibrium 1.87 x 10-3

I 0.00623 0.00414 0.0224

C - x - x + 2xE 0.00632-

x0.00414-

x0.0224 + 2x

H2(g) + I2(g) ↔ 2HI(g)

ee

e

K]I[]H[

]HI[

22

2

51 =

Equilibrium

(2) Calculating Equilibrium Concentrations [ ]e when K & [ ]i are known

I [A]i [B]i 0 0

C - ax - bx + cx + dx

E [A]i- ax [B]i- bx cx dx

bi

ai

dc

bequi

aequi

eequi

cequi

bxax

xx

BA

DCK

)]B([)]A([

)d()c(

][][

][][

aA + bB cC + dD

0 2 cbxax

Solve for x by using the quadratic equation:

a

acbbx

2

4

2

Equilibrium

Calculating Equilibrium Concentrations when K & [ ]i are known

Problem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54.3 @ 430ºC. Suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium.

I 0.00623 0.00414 0.0224

C - x - x + 2x

E 0.00632- x

0.00414- x

0.0224 + 2x

) -0.00414]([) -0.00632]([

)2 0.0224(3.54

2

xx

x

)0104.010 x 58.2(3.54 215 xx

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

010 x 98.8654.03.50 -42 xx

ee

e

K]I[]H[

]HI[

22

2

0 2 cbxax

2-4 40896.010 x 02.5 xx

a

acbbx

2

4

2

Deciding if x is negligible:

Equilibrium

Problem: for the reaction H2(g) + I2(g) 2HI(g) , K= 54.3 @ 430ºC. Suppose that the initial concentrations of H2, I2, and HI are 0.00623 M, 0.00414 M, and 0.0224 M, respectively. Calculate the concentrations of these species at equilibrium.

) -0.00414]([) -0.00632]([

)2 0.0224(3.54

2

xx

x

010 x 98.8654.03.50 -42 xx

2.24 54.3 002240.0

100x [A] 100

[A] i

i K


0 2 cbxax 1. Take the largest [A]i and multiply it by 100 and also divide it by 100.

2. Compare the answers of above to the value of K

3. If comparison are not significant as compared to K, then x is negligible and you can delete the x that is subtracted from [A]I

4. If these calc values are significant compared to K, then x is not negligible and you must use the quadratic to solve for x.

50.3

10 x 98.8

3.50

3.50 -42

x

00423.0x

-52 10 x 79.1x

YOU MUST USE QUADRATIC!

Equilibrium

Calculating Equilibrium Concentrations when K & [ ]i are known

010 x 98.8654.03.50 -42 xx

)3.50(2

)10 x 98.8)(3.50(4)654.0( 654.0

-42 x

0 2 cbxaxa

acbbx

2

4

2

M 00156.0or M 0114.0 xx

[H2]e = 0.00623 – 0.00156) = 0.00467 M

[I2]e = (0.00414 – 0.00156) = 0.00258 M[HI]e = (0.0242 + 2 x 0.00156) = 0.0255 M

First answer is physically impossible, since conc. of H2 and I2 would be more than original conc.

[H2]i = 0.00623

[I2]i = 0.00414 E 0.00632- x

0.00414- x

0.0224 + 2x

x calc without quadratic:

M00423.0x

Equilibrium

(2) Calculating Equilibrium Concentrations when K & [ ]i are known



[H2], M + [I2], M 2 [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At Equilibrium 1.87 x 10-3

I 0.00623 0.00414 0.0224

C - x - x + 2xE 0.00632-

x0.00414-

x0.0224 + 2x

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

ee

e

K]I[]H[

]HI[

22

2

51 =

100x [A] 100

[A] i

i K


0 2 cbxaxa

acbbx

2

4

2

e2e2

2

]I[]H[

]HI[ K

e

Equilibrium

Le Châtelier’s Principle

“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

Henri Louis Le Chatelier1850-1936

2 NO2 (g) N2O4 (g)

(clear gas)(red-brown gas)

{L.C.&PressVol}

{Equi.Intro1}

{Equi.Intro2}

The effect of (2) Pressure & Volume

The effect of (1) Concentration

(3) Temperature

[N2O4][NO2]2=KC

Equilibrium

H = -58.0 kJ

{Le Châtelier’s&Temp}

Le Châtelier’s Principle

(clear gas)(red-brown gas)

2 NO2 (g) N2O4 (g)

The effect of (3) Temperature

Why does heat favor the formation of NO2 over N2O4?

{Molecular Explanation}

Keq at different Temp:

+ HEAT

Equilibrium

The Effect of Temperature Changes

Co(H2O)62+

(aq) + 4 Cl(aq) CoCl4 (aq) + 6 H2O (l)

{CoCl}

(pink) (blue)

H= +

Add heat (), which way will the equlibrium be shifted?

+ HEAT

Equilibrium

Briggs-Rauscher reaction

{ClockReaction}

IO3− + 2H2O2 + CH2(COOH)2 + H+ → ICH(COOH)2 + 2O2 + 3H2O

Equilibrium

It was not until the early 20th century that this method was developed to harness the atmospheric abundance of nitrogen to create ammonia, which can then be oxidized to make the nitrates and nitrites essential for the production of fertilizers and ammunitions.

∆H = -92.4 kJ/n @ 250 C

NH3 + HNO3 NH4NO3 (explosives & fertilizers)

Le Châtelier’s Principle & the Haber Process

Equilibrium

Le Châtelier’s Principle & the Haber Process

This apparatus helps push the equilibrium to the right by removing the ammonia (NH3) from the system as a liquid

Subs b. pt.

NH3 -33°C

N2 -196°C

H2 -253°C

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

Fe

Fe

Shift equilibrium to the right by increasing press & conc. of N2 + H2 and by removing NH3 from system

Removed by liquefaction

Equilibrium

• Increase the rate of both the forward and reverse reactions.• Equilibrium is achieved faster, but the equilibrium

composition remains unaltered.

The Effect of Catalysts on Equilibrium

How does the addition of a catalyst affect the equilibrium between subs. A and B?

catalyst

Equilibrium

Equilibrium Constant (K) vs. Reaction Quotient (Q)

• To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression.

• Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

K = [C]c[D]d

[A]a[B]b Q =

[C]ic[D]id

[A]ia[B]ib


At Equilibrium: At Non-Equilibrium:

Equilibrium

If Q = K, the system is at equilibrium.

K = [C]c[D]d

[A]a[B]b

If Q > K, there is too much product and the equilibrium shifts to the left.

If Q < K, there is too much reactant, and the equilibrium shifts to the right.


Q = [C]i

c[D]id

[A]ia[B]i

b

If Q < K?

If Q > K?

Equilibrium

Kc, Q and Spontaneity (ΔG)

Q = [NH3]i2

[N2]i[H2]i

@ Equilibrium K = Q, ΔG = 0

ΔG= -

ΔG= -

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

Reverse Process

ΔG = +

Reverse Process

ΔG = +

KC = [NH3]2eq

[N2]eq[H2]eq

Equilibrium

Free Energy (G) & Reaction Quotient (Q) under non-Standard Conditions

G implies DIFFERENT CONCENTRATIONS of both reactants ↔ products and under non-standard conditions of concentration, temperature and pressure.

G = - (spontaneity), implies COMPLETE conversion of all reactants to →

products @ standard conditions (25ºC, 1 atm, 1M).

• If [Reactants]↑, then G = -

• If [Products]↑, then G = +

• If at Equilibrium , then G = 0

G = G + RT ln Q = G + RT 2.303 log Q

r

p

][

][Q

Reactants

Products

Non-standard conditions

☺Non-standard

conditions

Equilibrium

G = - RT ln K

= G + RT ln K

= - 2.303 RT log K

At equilibrium, G = 0 and Q = K so……

☺Rearranging ……

G K Product formation

– > 1 Product-favored @ Equilibrium

0 = 1 [Products] = [Reactants] @ Equilibrium (RARE)

+ < 1 Reactant-favored @ Equilibrium

Free Energy (G) & Equilibrium Constant (K) at Equilibrium

G = G + RT ln Q

K = eG/RT

Equilibrium

{G, Q & T↓ NO2 → N2O4}LowT{G, Q & T↑NO2 ← N2O4}HighT

Free Energy (ΔG), the Equilibrium Constant (ΔK),and Temperature

2 NO2 (g) N2O4 (g)(clear gas)(red-brown gas)

H = -58.0 kJ

Low Temperature: High Temperature:

Products only

Reactants only

Q = K

Equilibrium mixture

Low Temp.

High Temp.

G = - RT ln K

G°= - G°= +

r

p

][

][

Reactants

ProductsK @ T ↑, K ↓

How would T ↓or ↑ affect K?

@ T ↓, K ↑ G°= -

G°= +

Equilibrium

2003B Q12 HIH2 + I2

Equilibrium

2003B Q1 Kp = Kc(RT)Δn

Equilibrium

2000A Q1

Equilibrium

Equilibrium

2004B Q1

Equilibrium

Equilibrium

2007B Q1

Equilibrium

2000 1

Equilibrium

Equilibrium

2003 B

Equilibrium

Equilibrium

2004 B

Equilibrium

Equilibrium

2006 A

Equilibrium

Equilibrium

2006 (B)

Equilibrium

Equilibrium

2007 (B)

Equilibrium

Equilibrium Topics in CHM 1046 1.Intermolecular forces (IMF) 2.Themodynamics 3.Chemical Kinetics...

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Transcript of Equilibrium Topics in CHM 1046 1.Intermolecular forces (IMF) 2.Themodynamics 3.Chemical Kinetics...