Investigating Chemical Reactions

• N2O4 (g) ⇄ 2 NO2 (g)

• Colorless brown

Closed Container: Reversibility

Groups of Molecules

A state of Dynamic Equilibrium

Equilibrium Concentrations

• Equilibrium is reached when the concentrations of products and reactants remains unchanged with time

• Condition dependent• Qualitative descriptions

– “Equilibrium lies to the left (or right)”

– “Equilibrium favors products (or reactants)”

Initial: all NO2

2 NO2 N2O4

Rates of Reaction• For simple, one step

reaction, reaction rate is due to inherent reactivity and collision rate

• Forward rate = kf [N2O4]

• Reverse rate = kr [NO2]2

Rates change over Course of Reaction

Initial: all NO2

2 NO2 N2O4

Same Principle for all ReactionsCO + H2O CO2 + H2

Law of Mass Balance

General, Empirical Form:

aA + bB cC + dD⇄

Keq = This is called the Equilibrium expression

Law of Mass Balance

• Empirical law—with justification from kinetics

• Example: N2O4 2 NO2

= = Equilibrium Constant

Forward rate = reverse rate

[N2O4] =

This derivation is a simplification, but the outcome generally applies

N2O4 (g) 2 NO⇄ 2 (g)

Keq = [NO2]2/[N2O4]

If equilibrium [R] and [P] are known, Keq can be calculated.

Example: at equilibrium,

–[NO2] = 1.50 x 10-2 M,

–[N2O4] = 1.03 x 10-2 M @ 317K (from experiment)

Keq = (1.50 x 10-2)2/(1.03 x 10-2) = 0.0218

Keq depends on how the equation is written.

• N2O4 (g) 2 NO⇄ 2 (g)– K1 = [NO2]2/[N2O4] = 0.0218

• 2 NO2 (g) N⇄ 2O4 (g)– K2 = [N2O4] / [NO2]2

– K2 = 1/K1

– K2 = 1 / 0.0218 = 45.9 @ 317K

• ½ N2O4 (g) NO⇄ 2 (g)– K3 = [NO2] /[N2O4]1/2

– K3 = (K1)1/2

– K3 = (0.0218)1/2 = 0.148

Small Keq (less than 1)means less product at equilibrium

Large Keq (more than 1)means more product at equilibrium

Test Your UnderstandingCalculate the equilibrium constant for

at a given temperature if a 1-L container that initially held 1 mol of CO and 1 mol of water reached equilibrium and then had 0.13 mol of CO2 and H2.

CO + H2O CO2 + H2

Answer: Keq = 0.022

Units?

Keq describes ratio of reactants and products

• A B Keq =0.33⇄• Which of the following systems are at

equilibrium?– [A] = 3.0 M, [B] = 1.0 M– [A] = 7.5 x 10-3 M, [B] = 2.5 x 10-3 M– [A] = 12.0 M, [B] = 4.0 M

• Keq describes ratio, not absolute concentrations

Concentrations are dependent on starting point,Ratio of concentrations is not!

Initial: all NO2

All proceed until equilibrium is reached

One Equilibrium Constant, Many Equilibrium Positions

all N2O4 both

2 NO2 N⇄ 2O4 K1 = [N2O4] / [NO2]2

N2O4 + O2 2 NO⇄ 3 K2 = [NO3]2 / [N2O4][O2]

2 NO2 + O2 2 NO⇄ 3 K3 = [NO3]2 / [NO3]2 [O2]

K3 = K1K2

Multiple EquilibriaWhat if the products of one reaction act as reactants in a subsequent reaction?

The overall Keq can be calculated from individual steps.

• N2O4 (g) 2 NO⇄ 2 (g)– Concentrations may be measured in molarity or

pressure• Kc = [NO2]2/[N2O4] --measured in molarity• Kp = PNO2

2 / PN2O4 --measured in pressure

Units of Concentration

Relationship between Kp and Kc

– Describes same phenomenon in different ways, so quantity may be different

– Example: A B– P =(n/V)RT = [C]RT

– Kp = Kc(RT)Δn – where Δn = # mol gas product - # mol gas reactant

𝐾 𝑐=[𝐵 ]𝑏

[ 𝐴 ]𝑎=⌈ 𝑃 𝑏𝑅𝑇

⌉𝑏

[ 𝑃 𝑎𝑅𝑇 ]𝑎

𝐾 𝑃=𝑃𝑏𝑏

𝑃 𝑎𝑎

𝐾 𝑐=𝐾 𝑃

⌈ 1𝑅𝑇

⌉𝑏

[ 1𝑅𝑇 ]

𝑎

Converting Between Kc and Kp

• For N2O4 (g) 2 NO⇄ 2 (g) @ 317 K– Keq = (1.50 x 10-2)2/(1.03 x 10-2) = 0.0218

– Δn = 1– Kp = 0.0218[(0.08206)(317)]1 = 0.567

• What is the relationship between Kc and Kp when there is no change in number of moles of gas? (Ex: NO2 + CO NO + CO2)

The Concept of Activity• More accurately, the equilibrium expression

does not depend on the concentration, but the activity

• Activity = Concentration/Reference– Activity of 0.5 M SO2 = 0.5M/1 M = 0.5– Activity of 2.7 atm CO2 = 2.7atm/1 atm = 2.7– Activity of pure liquids and solids: reference is pure

compound• Activity of liquid water = 55M water/55M water = 1• Activities of all pure liquids and solids are unity

Heterogeneous Equilibria• We have been talking about homogeneous

equilibria – where all reactants and products are in the same state. (gases or solutions)

• What about mixed phases?• Ni (s) + 4 CO (g) Ni(CO)⇄ 4 (g)• Kc = [Ni(CO)4] / [CO]4

Summary of Equilibrium Expressions

• Keq tells us about the ratio of reactants and products at equilibrium (not absolute values)

• Must specify equation and temperature with Keq

• If equilibrium concentrations of all reactants and products are known, Keq may be determined.

• The Keq of multiple equilibria may be determined from the Keqs of individual reactions

• Kc and Kp are interconvertable• Solids and pure liquids don’t affect Keq

Quantitative Equilibrium Problems

• Determine whether or not a reaction is at equilibrium

• Calculate an equilibrium constant from equilibrium concentrations

• Given starting concentrations and K, predict the equilibrium positions

Reaction Quotient• Reaction quotient – Q – the same expression as

Keq but with the current concentrations (not equilibrium concentrations)

• aA + bB cC + dD⇄• Q = [C]c[D]d

[A]a[B]b

• At equilibrium, Q = Keq

• Q helps to determine the direction of the reaction• Reactions move toward equilibrium• Q Keq

2 NO2 (g) N⇄ 2O4 (g) Keq = 45.9 @ 317K

[NO2] = 0.50 M [N2O4] = 0.50 M

Q = [N2O4] / [NO2]2 = (0.50)/ (0.50)2 = 2.0

Q = Keq , at equilibrium

Q < Keq , moves to product (as written)

Q > Keq , moves to reactant (as written)

Keq

45.9

Q

2.0

Equilibrium Calculations• To find Keq, equilibrium concentrations

measured.– Practically, only one [reactant] or [product] measured

• To calculate Keq, we will need the equilibrium expression and 3 other pieces of information.– Initial Concentrations – Changes due to reaction – depends on stoichiometry– Equilibrium concentrations

• To do any equilibrium calculation, we will need to set up an ICE table.

1.00 mol of CO and 1.00 mol of H2O are placed in a 50.0 L vessel. At equilibrium, [CO2] is found to be 0.0086 M @ 1273 K. Calc. Keq.

CO (g) + H2O (g) ⇄ CO2 (g) + H2 (g) I

C

E

1.00/50.0 = 0.0200

0.0200 0 0

0.0086

+ 0.0086 + 0.0086

0.0086

- 0.0086 - 0.0086

0.0200 - 0.0086 = 0.0114

0.0200 - 0.0086 = 0.0114

Keq = [CO2][H2] = (0.0086)(0.0086) = 0.569

[CO][H2O] (0.0114)(0.0114)

What are the equilibrium concentrations of each species if 0.500 mol of H2 and I2 are placed into a 1.00 L vessel at 458 oC?

H2 (g) + I2 (g) ⇄ 2 HI (g) Keq = 49.7 @ 458 oC

PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)

If the initial concentration of PCl5 is 1.00 M, calculate the equilibrium concentration of each species at 160 oC if K = 0.0211 at that temperature.

0.030 mol of SO2Cl2, 2.0 mol of SO2 and 1.0 mol of

Cl2 are placed into a 100 L reaction vessel at 173 oC. The K for that temperature is 0.082. Find the equilibrium concentrations of all species. 

SO2Cl2 (g) ⇄ SO2 (g) + Cl2 (g)

 

A 2.00 L container at 463 K contains 0.500 mol of phosgene, COCl2. K is 4.93 x 10-3 for

COCl2 (g) ⇄ CO (g) + Cl2 (g)

Calculate the equilibrium concentrations of all species.

Calculate the partial pressure of SO3, given Kp

is 0.74 (at 2100 K) for

CaSO4 (s) ⇄ CaO (s) + SO3 (g)

LeChatelier’s Principle• “If a stress is applied to a system at

equilibrium, the system reacts to relieve the stress.”

• Sets up new equilibrium position• Change in

– Concentration of reactant or product– Pressure of system– Temperature of system

• In this case, changes equilibrium position AND equilibrium constant

Change in Concentration

• 2 NO2 (g) N⇄ 2O4 (g)• Add NO2 (either conc. or pressure)

– Equilibrium shifts to right (more products)• Remove NO2

– Equilibrium shifts to left• Add N2O4

– Equilibrium shifts to left Logic: Think about changing rates

Fig. 14.7Example:Haber Process

Applications of Le Chatelier

OH

Cat. H+

+ H2O

“Forcing” an unfavorable reaction

Applications of Le Chatelier

• Strategies for driving unfavorable metabolic reactions

O

HO

HO

OPO3-2

OH

OH

NH

O

ONO

OHOH

HH

OPO

O-

O

P

O

O-

OP

O

-O

O-

+

OHP

O

O-

OP

O

-O

O-

NH

O

ONO

OHOH

HH

OPO

O-

O

P

O

O-

OHO

HO

O

OH

OH

glycogen

2 OHP

O

O-

O

glucose-1-PUTP

UDP-glucose

LeChatelier’s Principle• Adding/removing gases not involved in the

equilibrium – has no effect on system—no change in partial pressures of reactants

• 2 NO2 (g) N⇄ 2O4 (g)• Decrease the volume?

– A decrease in volume will favor the side with the least # moles of gas

– An increase in volume will favor the side with the greater # moles of gas

– If moles of gas equal, no effect

Changes in Temperature• Changing temperature changes equilibrium

position and the equilibrium constant (Keq)

• CO2 (g) + C (s) 2 CO (g) ⇄ ΔH = +173 kJ– Endothermic--re-write equation as:

• HEAT + CO2 (g) + C (s) 2 CO (g) ⇄• Follow Le Chatelier

– Add heat (inc temp), increases K, shifts to right (products)

– Remove heat (dec temp), decreases K, shift to left

N2O4 (colorless) 2 NO2 (brown)

Test Your Understanding

Is this reaction exothermic or endothermic as written?

Haber Process• Production of ammonia

is big business!• Given the equation and

data, how would you run the process to maximize ammonia output?

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

Equilibria of Processes• Equilibrium can describe process as well as

reactionCH3

CH3

CH3 H3C

OH OH

(ether) (aq)

Conformation

Solubility

Acid-Base Chemistry

• Major application of equilibrium (ch 7-8)• Acid/base reactions reach equilibrium

quickly• Relatively simple reaction with MAJOR

applications

Acid/Base Reactions

• Bronsted-Lowry Definition– Acids are proton (H+) donors– Bases are proton (H+) acceptors (lone pair)

• Limit discussion to aqueous solutions

Acid-Base Reactions

H Cl

H

O

HH O H

H

+ Cl

H O

H

O

HH O H

H

C

O

CH3

O C

O

CH3

H2O + HC2H3O2 H3O+ + C2H3O2-

Base AcidConjugate acid

Conj. Base

Test Your Understanding

• Write an aqueous acid/base reaction for CH4. Then write the equilibrium expression. (The expression for Ka.)

Defining Strong and Weak Acids

• Ka of CH4 = 10-50

• Ka of HC2H3O2 = 10-5

• Ka of HCl = 107

In which direction does each equilibrium lie?What is a strong acid? What is a weak acid?What is a “stronger” and “weaker” acid?

Defining Strong and Weak Bases

H Cl

H

O

HH O H

H

+ Cl

H O

H

O

HH O H

H

C

O

CH3

O C

O

CH3

H2O + HC2H3O2 H3O+ + C2H3O2-

Strong Acids

• H2SO4

• HNO3

• HClO4

• HCl, HBr, HI

At equilibrium, [H+] = [HA]o

Test Your Concept

Working with pKa

• pKa is convenient way to express Ka• pKa = -log Ka

Ka pKa

1 x 10-12

1 x 108

35

-9

4.76 x 10-5

How does pKa value relate to acid strength?

Weak Acid-Base Equilibrium:Nothing New

• What is the % dissociation of the side chain of aspartate in a 0.600M aq. solution (pKa 3.9)?

Answer: 1.4% ionized

pH of Aqueous Solutions

• Neutral water: autoionization• Kw = 1 x 10-14

• [HO-] = [H+] = 10-7

• pH = 7• In concentrated acid solutions, we assume

autoionization is a minor contributor: “all” hydronium from acid dissociation

What is the pH of a 0.45M aq solution of acetic acid (pKa = 5.0)?• 1. Write equilibrium expression• 2. Solve ICE to determine [H+]• 3. Determine pH

Answer: 2.7

• What happens to the pH of a 0.45M aq solution of acetic acid (pKa = 5.0) if sodium acetate is added to give an acetate concentration of 0.05M?

• What happens to the pH of a 0.45M aq solution of acetic acid (pKa = 5.0) if sodium acetate is added to give an acetate concentration of 0.45M?

• What is the pH of a 0.50 M aq solution of acetic acid (pKa = 5.0) to which has been added 0.05M NaOH?

Answers: 4.0, 5.0, 4.0