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Chemical Equilibrium (Ch. 14)

• After a period of time, some reactions reach equilibrium, meaning that the concentrations of reactants and products no longer change with time.

This does not mean that the reaction has stopped!

A

Equilibrium is a dynamic process

B

• Equilibrium is reached when the forward rate of reaction of A B is equal to the rate of the reverse reaction of B A.

• Both the forward and reverse reactions are elementary reactions

For the reaction

ratefwd = raterev

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Ratefwd = kfwd[A]1 For A B

For B A Raterev = krev[B]1

- At equilibrium the rates of forward and reverse reactions are equal so,

kfwd[A] = krev[B]

- Rearranging the equation, we get:

• Rate laws for elementary reactions are predictable:

kfwd

krev

=[B]

[A]= K

The EquilibriumConstant

(unitless)

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aA + bB cC + dD

• For any reversible reaction at equilibrium:

“reactants” “products” (as written)

K = [C]c[D]d [A]a[B]b (reactants)

(products)

- If K is a big number (>>1), the forward reaction is favored and a large quantity of products are present at equilibrium

- If K close to 1, equal quantities of reactants and products are present at equilibrium

- If K a small number (<<1), the reverse reaction is favored a large quantity of reactants are present at equilibrium

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• Criteria for chemical equilibrium

1. The reaction must be reversible. A reacts to form B, but B can also react to form A

2. The reaction must occur in a closed system.

flame

CaCO3CaO

CO2

- If the reaction occurred in the open, the carbon dioxide escapes into the environment.

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

The carbon dioxide is nolonger present to undergothe reverse reaction.

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• Because the “concentration” of solids and liquids remains the same throughout a reaction, they do not appear in the equilibrium equation.

CaCO3(s) CaO(s) + CO2(g) Kc = [CO2]

flame

CaCO3CaO

CO2

reacts

- In a closed system, equilibrium will establish

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The Reaction Quotient: Q

aA + bB cC + dD

“reactants” “products” (as written)

Q = [C]c[D]d [A]a[B]b (reactants)

(products)

• It has the exact same formula as the equilibrium constant and is used to determine reaction direction

- If Q = K, then the reaction is at equilibrium

- If Q < K, the reaction will proceed forward to produce more products to reach equilibrium

- If Q > K, the reaction will proceed in reverse to produce more reactants to reach equilibrium

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Changes in the Equilibrium State: Le Chatelier’s Principle

• Le Chatelier’s principle states that when equilibrium is disturbed, the reaction will shift to reduce the effect of the disturbance.

2. Changing the pressure/volume of the system.

3. Changing the temperature of the reaction. It is the only stress that changes the K value.

- 3 ways in which reactions can be stressed:

1. Changing the concentration of one or more of the reactants and/or products

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In General:

1. When the concentration is changed, the equilibrium will shift to consume some of the added substance or produce some of the removed substance.

2. An increase in pressure (decrease in volume) shifts the reaction to the side with less moles of gas, and an decrease in pressure (increase in volume) shifts the reaction to the side with more moles of gas.

3. For exothermic reactions, an increase in temperature will decrease the K value (shift to the left), and a decrease in temperature will increase the K value (shift to the right). For endothermic reactions, an increase in temperature will increase the K value (shift to the right), and a decrease in temperature will decrease the K value (shift to the left).

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• When it comes to equilibrium in solutions, the volume of the solution itself defines the “closed system” allowing for aqueous reversible reactions to reach equilibrium.

500 ml

Fe3+(aq)

SCN-(aq)

FeSCN2+(aq)+

Eg.

(red/orange)

- It is possible to “remove” soluble ions from solution by precipitating them out.

Eg. If a hydroxide source is added, it will ppt with Fe3+(aq)

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Solving Equilbrium Problems: ICE Table

2 HI(g) H2(g) + I2(g) Problem 1:

- At equlibrium, [HI] = 0.078 M. Calculate K.

- 0.200 mol of HI gas is added to a 2.00 L flask and the reaction is allowed to proceed at 453oC

• It’s nearly the same as the IRF table for irreversible reactions except that the C stands for “concentration and E stands for “equilibrium”.

• ICE tables require the use of variables to fill in unknown information.

• Moles can still be entered instead of concentration, but will need to be converted to concentration in the end.

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CO(g) + H2O(g) CO2(g) + H2(g)

- 2.00 M of CO and 2.00 M of H2O are reacted in a sealed flask at 900 oK. What will be the composition of the reaction mixture at equilibrium?

- At this temperature the K is 1.56

• Sometimes the quadratic equation is required to solve equilibrium concentrations

- When the equation is rearranged such that:

a x2 + b x + c = 0

- b ± √b2 - 4ac

2ax = Then

Problem 2:

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• When K is a small number (less than ~1.0 x 10-4), a simplifying assumption can be made

COCl2(g) CO(g) + Cl2(g) K = 8.3 x 10-4

- Calculate the concentration of the reaction mixture at equilibrium when starting with 5.00 mol of COCl2 in a 10.0 L flask

The assumption is considered valid if the percent erroris less than 5%

Problem 3:

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• The properties of acids and bases have been known for a long time, and many where characteristics back in 1661 by English scientist Robert Boyle.

Acids Bases

Have sour taste Have bitter taste

Are corrosive to metals Have a soapy feel/disposition

Can have sharp pungent odour

Can have sharp pungent odour

Become less acidic when combined with base

Become less basic when combined with acid

Changes the color of litmus dye to red

Changes the color of litmus dye to blue

Acid-Base Equilibrium (Ch. 15)

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• Classical, or Arrhenius definition of acids and bases in water:

- An acid is a substance that contains H in its formula and dissociates in water to yield H+ ions (or H3O+, hydronium ion).

- A base is a substance that contains OH in its formula and dissociates in water to yield OH- ions.

Eg. HCl, HNO3, and HCN

Eg. NaOH, KOH, and Ba(OH)2

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• The modern and more accepted view of an acid and base is now called the BrØnsted-Lowry acid-base definition:

- An acid is a proton donor, any species that donates an H+ ion. An acid must contain an H in its formula in order for there to be a possibility of H+ donation.

- Arrhenius acids are also Bronsted-Lowry acids

H ClO

H

H

Lone pair electrons bind the H+

Cl- OH

HH +

HCl(g) + H2O(l) H3O+(aq) + Cl-(aq)

Eg.

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• A base is a proton acceptor, any species that accepts an H+ ion. A base must contain a lone pair of electrons to bind the H+ ion, forming a new covalent bond. This is why bases are sometimes also thought of as an electron donor.

- This is a very different concept compared to Arrhenius’. For example, Arrhenius would call NaOH a base, but in the B-L definition, it’s the OH- that acts as the base. The compound NaOH is simply the source of the hydroxide ion.

NH

H

HO

H

HO

H+NH

H

HH

-

NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) Eg.

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Conjugate acid-base pairs

NH3(g) + H2O(l) NH4+(aq) + OH-(aq)

base acid

NH3(g) + H2O(l) NH4+(aq) + OH-(aq)

- However,

acid base

- The conjugate base has one fewer H and one more minus charge than the acid.

- The conjugate acid has one more H and one fewer minus charge than the base.

• Conjugate pairs differ by 1 H atom:

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The strength of conjugate acid-basepairs is inverselyproportional.

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• pH is a measure of how much H+ (H3O+) is in solution

pH = - log [H+]

1. What is the pH of a solution that is 4.2 x 10-3 M HCl(aq)?

2. A solution has a pH of 5.5. What is the concentration of H3O

+ in this solution?

• pOH is a measure of how much OH- is in solution

pOH = - log [OH-]

3. What is the pOH of a solution of 0.125 M sodium hydroxide?

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The Extent of Dissociation of Weak Acids and Bases

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

• For any generic weak acid HA in water, we have

[H3O+] [A-]

[HA] [H2O] K =General

equilibrium

- However, since the concentration of water is a constant, it is factored out of the equation to leave a new constant Ka, the acid-dissociation constant

[H3O+] [A-]

[HA] Ka =

• Strong acids dissociate 100% into ions (irreversible arrow), but weak acids do not, so their reaction in water is reversible.

The bigger the Ka, the stronger the acid

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• Similarly, for any weak base B in water, we have the reaction

B(aq) + H2O(l) BH+(aq) + OH-(aq)

[BH+] [OH-]

[B] Kb =

- Again, since the concentration of water is a constant, it is factored out of the equation to leave a new constant Kb, the base-dissociation constant

The bigger the Kb, the stronger the base

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Autoionization of Water

• Water does not exist purely as the neutral molecule H2O.

It actually reacts with itself in an Bronsted-Lowry acid- base manner to form ions.

To describe this equilibrium, we use the ion-productconstant for water, Kw

Kw = [H3O+][OH-] = 1.0 x 10-14 (at 25oC)

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• Some derivations:

- Taking the –log of this equation [H3O+][OH-] = 1.0 x 10-14

pH + pOH = 14

-log[H3O+] + -log[OH-] = -log(1.0 x 10-14)

Ka x Kb = Kw

- For a conjugate pair

And so pKa + pKb = 14

(-logKa) (-logKb)

The smaller the pKa, the stronger the acid

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pH of Soluble Ionic Salts

• Review: Salts are solid ionic compounds that consist of ions (cations and anions) in a crystal lattice.

Most salt solutions have a pH close to 7.0

Eg. NaCl(s) Na+(aq) + Cl -(aq) H2O

Most metal cationsdo not have any acidor base properties (NoH+ to donate, and cannotaccept an H+ either)

No H so cannot be an acid (proton donor), but it could be a base (proton acceptor using lone pair electrons). However, it is the conjugate base of the strong acid HCl, so Cl- has negligible base properties.

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• Some salts contain the ion of a conjugate base or conjugate acid.

Eg. 1.) Sodium acetate: CH3COONa (sometimes NaCH3COO)

2.) Ammonium chloride: NH4Cl

- When these types of salts dissolve in water, the pH of the solution will be affected.

1. a) The Ka of acetic acid is 1.8 x 10-5. What is the Kb

for its conjugate base acetate?

b) 0.15 moles of sodium acetate is dissolved in a total of 1.0 L of water. What is the pH of the solution?

2. The Kb for ammonia is 1.8 x 10-5. What would the pH be of a solution of 0.23 M ammonium chloride?

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Acid-Base Buffers

• An acid-base buffer is a solution that resists changes in pH when a small amount of strong acid or base is added.

- a solution can act as a buffer it in contains an acid- base conjugate pair, in similar concentrations

• Buffer range: Buffers work best when there are equal concentrations of conjugate pairs. When this occurs, the pH = pKa

A buffer is effective at a pH within approximately ± 1 of the pKa

• Buffer capacity: The more highly concentrated the buffer solution, the more acid or base it can “consume”.

Buffer pH is NOT affected by addition of water

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H2O + CO2 H2CO3 HCO3- + H+

Carbonic acid Bicarbonate

• A Biological Buffer: Controlling pH in blood plasma

Controlled throughrespiration

Controlled by the liver

Basic substances are consumed by the carbonic acid, andacidic substances are consumed by the bicarbonate.

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• Ways of Making Buffers

1. Adding both conjugates into solution.

2. Generating the conjugate pair through neutralization (more common method)

- The ionic species typically comes from a salt.

Eg. Acetic acid (solution) with sodium acetate. Ammonia (solution) with ammonium chloride.

- When a weak acid is partially neutralized by a strong base, a certain quantity of conjugate base will form. (Weak bases require partial neutralization by strong acids).

Eg. CH3COOH(aq) + NaOH(aq) CH3COO-(aq) + Na+(aq) + H2O

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A buffer is made by dissolving 7.38 grams of solid sodium acetate to a total of 1.0 L solution with 0.100 mol of acetic acid. What is the pH of the buffer solution?

Formula weight of sodium acetate = 82.03Ka of acetic acid is 1.8 x 10-5

1. a)

b) Calculate the pH if 500.0 ml of water is added to the buffer.

c) Calculate the pH if 0.400 g of solid sodium hydroxide is added to the buffer in a) Formula weight of sodium hydroxide = 40.00

d) Calculate the pH if 0.400 g of solid sodium hydroxide is added to 1.0 L of pure water.

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2. You want to make 1.00 L of a 0.100 M solution buffered at pH 9.0 using an ammonia/ammonium buffer system. Kb of ammonia = 1.74 x 10-5

a) How many moles of each conjugate species would you need?

b) The source of ammonia is a 1.00 M solution, and the source of ammonium is from solid ammonium nitrate. (FW = 80.04). How much would you add of each to achieve your buffer?

3. You want to make 1.00 L of a 0.100 M solution buffered at pH 9.0 using using an ammonia/ammonium buffer system. You have only solid ammonium nitrate and 6.00 M NaOH solution. How much of each must be used?

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Titration Curves

• During an acid-base titration, the pH of the solution changes as acid or base is continuously added.

• The plot of the pH versus the quantity of acid or base added, generates a titration curve.

Important points on the titration curve:

1.) Equivalence point: The mid point of large inflections in the curve. (“S” portion of the curve)

2.) Half equivalence point for weak acids and bases: This is when pH = pKa

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HCl(aq)

Magnetic stirrer

Burette containingNaOH(aq)

pHprobeinterfacedwith Labworks

Automated pH Titration Data Acquisition

Example: Strongacid HCl titrated with NaOH

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HAc = acetic acid

Ac- = acetate ion

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Polyprotic Weak Acids

• These acids have multiple ionizable protons (acidic hydrogens)

Eg. H3PO4 – Phosphoric acid has 3 acidic hydrogens H2CO3 – Carbonic acid has 2 acidic hydrogens H2SO4 – Sulfuric acid has 2 acidic hydrogens

• Interestingly, for weak polyprotic acids, the protons ionize one at a time. The first proton has the greatest ionizability, followed by the next proton, etc.

H2SO3 HSO3- SO3

2-

Ka1 Ka2

Where Ka1 >> Ka2

Nearly all polyprotic acids are weak acids except for sulfuric

H+ H+

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Molecular Bases are N Containing Compounds

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Equilibria of Slightly Soluble Ionic Compounds

• On solubility table, these are compounds marked as “ss”

Eg. CaSO4(s) Ca2+(aq) + SO42-(aq)

• In solution, small quantities of solid are dissolving into ions. Once a certain ion concentration is reached, the ions react to reform the solid in an equilibrium

General K [Ca2+] [SO4

2-]

[CaSO4 ]=

Ksp = [Ca2+] [SO42-]

Where Ksp stands for the solubility product constant

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• Shifting solubility is a property of Le Chatelier’s Principle

- This will increase the concentration of “products” in solution and shift the equilibrium to the left.

- The result is that the salt is even less soluble in a solution containing a common ion.

2.) pH differences

- excess H3O+ (low pH) can often react with one of the ions of the salt, removing it from solution.

- This will reduce the concentration of “products” in solution, shifting equilibrium to the right which increases the solubility of the salt.

- Excess OH- (high pH) often acts as a common ion

1.) A common ion in solution.