Forward and reverse reactions taking place at equal rates

It is a dynamic state - reactions are constantly occurring

Chemical Equilibrium

(a) Start: 10 goldfish in the left tank and 10 guppies in the right.(b) Equilibrium state. with 5 of each kind of fish in each tank. The equilibrium is dynamic;

an averaged state and not a static condition . The fish do not stop swimming when they have become evenly mixed.

(c) If we were to observe one single fish (here a guppy among goldfish). we would find that it spends half its time in each tank .

Brightstorm videosChemical Equilibrium Definition 5:01http://www.youtube.com/watch?v=FYc_SoW2M40&list=PL06C3C4E3F84C6A24&index=42

Crash course chemistryhttp://www.youtube.com/watch?v=g5wNg_dKsYY 10:56 Equilibriumhttp://www.youtube.com/watch?v=DP-vWN1yXrY 9:28 Equilibrium equationsYou don’t need to know how to do the RICE table starting at 4:40

Isaacs Teach http://www.youtube.com/watch?v=g4TKRInLdPA 10:09 EquilibriumGood basic explanation!

http://www.youtube.com/watch?v=4z4_rc6nsKU 12:46 What is the equilibrium constant, Keq? Also very good explanation

Equilibrium constant expressions

aA + bB cC + dDKeq = [C]c[D]d

[A]a[B]b

General information about the Keq expression

• Equilibrium [ ] of products are placed in the numerator.

• Equilibrium [ ] of reactants are placed in the denominator.

• Each [ ] term is raised to an exponent equal to its coefficient in the balanced equation.

• If there is more than 1 product or reactant, the terms are multiplied.

• Solids and liquids (pure substances) are not included in the Keq expression. This is because their [ ] are their densities. The density of a substance does not change with changing temperatures.

• Keq is constant for a given reaction at a given temperature. There are no units associated with the value of Keq.

• The value of Keq is independent of the: – individual [ ] of reactants and products– original [ ] of reactants and products– volume of the container.

• The value of Keq is dependent on temperature.

• What does the value of Keq tell you about a reaction?

Keq >1: more products than reactants at equilibrium

Keq < 1: more reactants than products at equilibrium

Using equilibrium constantsCalculating equilibrium concentrations:

Example: At 1405 K, hydrogen sulfide, also called rotten egg gas (because of its bad odor), decomposes to form hydrogen and a diatomic sulfur molecule,S2. Keq = 2.27 x 10-3.

(a) Write the balanced equation for the reaction described above. Write out the Keq

expression.

(b) Calculate the concentration of hydrogen gas if [S2] = 0.0540 M and [H2S] = 0.184 M.

Solving the problem – part (a)

2H2S (g) 2H2 (g) + S2 (g)

Keq = [H2]2[S2]

[H2S]2

Solution – part (b)

[H2]2= Keq [H2S]2 =

[S2]

(2.27 x 10-3)(0.184 M)2 =

0.0540 M

[H2]2= 1.42 x 10-3 M

[H2] = 3.77 x 10-2 M

Le Châtelier’s principle:1884 - Henri Le Châtelier

When a stress is applied to a system at

equilibrium, the system shifts in the direction

that relieves the stress.

Δ in concentration• Adding more of a reactant or product: the

reaction will shift in the direction to consume a portion of what was added. – more reactant shifts right – more product shifts left

• Removing some of a reactant or product: the reaction will shift in a direction to restore part of what was removed. – reactants removed reaction shifts left (i.e. the

reverse reaction)– products removed reaction shifts right (i.e. the

forward reaction).

Δ in volumeRelevant when discussing gaseous equilibria and when

the number of moles of gaseous reactants differ from the

number of moles of gaseous products. The change in

volume is a result of a change in pressure of the gaseous

system.• When P↓, the reaction will shift in a direction to↑ number of

moles of gas.

PCl5 (g) PCl3 (g) + Cl2 (g) 1 mol 2 mol

2NH3(g) N2(g) + 3H2(g) 2 mol 4 mol

• When P↑, the reaction will shift in a direction to ↓ number of moles gas.

PCl5 (g) PCl3 (g) + Cl2 (g) 1 mol 2 mol

2NH3(g) N2(g) + 3H2(g) 2 mol 4 mol

Δ in temperature View changes in temperature as reactants or products.

When the temperature of an equilibrium system is ↑ the reaction that is endothermic (ΔH>0) will take place.* forward rxn is endothermic more product (shifts to the right). * reverse rxn is endothermic less product (shifts to the left)

When the temperature of an equilibrium system is ↓, the rxn which is exothermic (ΔH<0) will take place.* forward rxn is exothermic – more product (shifts to the right). * reverse rxn is exothermic – less product (shifts to the left)

General rule: if the forward rxn is endothermic,↑K.If the forward rxn is exothermic ↓K.

Animation demonstration http://www.learnerstv.com/animation/animation.php?ani=120&cat=chemistry

http://www.youtube.com/watch?v=PciV_Wuh9V8 7:00 Le Chatelier’s Principle; good explanations with visuals and excellent discussion on how to increase yield of a reaction

http://www.youtube.com/watch?v=dd5p0VZ-MZg 5:36 Equilibrium disturbancesThis one will help you with the lab we’re doing. He also discusses the effect of disturbances (changes) in an equilibrium system and how they affect the value of K (the equilibrium constant)

Reactions that go to completion

Formation of a gas

H2CO3 (aq) H2O (l) + CO2 (g)

Formation of precipitate (Double displacement reactions)

Formation of a slightly ionized product; often times H2O (i.e. in a neutralization reaction)

H3O+ + OH- 2H2O (l)

Solubility equilibria

http://www.youtube.com/watch?v=YJ-dyEtB66A&feature=topics

The Solubility Product Constant, Ksp

• Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution.

• The solubility product constant, Ksp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.

The Solubility Equilibrium Equation And Ksp

CaF2 (s) Ca2+ (aq) + 2F- (aq)

Ksp = [Ca2+][F-]2 Ksp = 5.3x10-9

As2S3 (s) 2As3+ (aq) + 3S2- (aq)

Ksp = [As3+]2[S2-]3 Ksp = 6 x 10-51

Ksp And Molar Solubility

• The solubility product constant is related to the solubility of an ionic solute, but Ksp and molar solubility - the molarity of a solute in a saturated aqueous solution - are not the same thing.

• Calculating solubility equilibria fall into two categories: – determining a value of Ksp from experimental data

– calculating equilibrium concentrations when Ksp is known.

Calculating Ksp From Molar Solubility

It is found that 1.2x10-3 mol of lead (II) iodide, PbI2, dissolves in 1.0 L of aqueous solution at 25 oC. What is the Ksp at this temperature?

Solution:

PbI2 (s) Pb2+ (aq) + 2I- (aq)

Ksp = [Pb2+] [I-]2

Ksp = (1.2 x 10-3 M) (2 x 1.2 x 10-3 M)2

Ksp = 6.9 x10-9

Calculating Molar Solubility From Ksp

Calculate the molar solubility of silver chromate,

Ag2CrO4, in water from Ksp = 1.1x10-12 for Ag2CrO4.

Solution:

Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq)

Ksp = [Ag+]2 [CrO4 2-]

Ksp = (2x)2(x) = 1.1 x 10-12

4x3 = 1.1 x 10-12

X = 6.5 x 10-5 M

The Common Ion Effect In Solubility Equilibria

• The common ion effect also affects solubility equilibria.

• Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution.

The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.

Solubility Equilibrium Calculation-The Common Ion Effect

What is the solubility of Ag2CrO4 in 0.10 M K2CrO4? Ksp = 1.1x10-12 for Ag2CrO4.

Ag2CrO4 (s) 2Ag+ (aq) + CrO4 2- (aq)

Ksp = [Ag+]2 [CrO4 2-]

Ksp = (2x)2(0.10) = 1.1 x 10-12

x = 1.65 x 10-6 M

Comparison of solubility of Ag2CrO4

In pure water: 6.5 x 10-5 M (prior slide)In 0.10 M K2CrO4: 1.7 x 10-6 MThe common ion effect!!

Determining Whether Precipitation Occurs

• Q is the ion product reaction quotient and is based on initial conditions of the reaction.

• Q can then be compared to Ksp.

• To predict if a precipitation occurs:

- Precipitation should occur if Q > Ksp.

- Precipitation cannot occur if Q < Ksp.

- A solution is just saturated if Q = Ksp.

DR lab: unexpected PPT according to solubility rules!

Ca(OH)2 (s) Ca2+ (aq) + 2OH- (aq)

Ksp = [Ca2+][OH-]2 Ksp = 6.5 x 10-6

Determining Whether Precipitation Occurs – An Example

The concentration of calcium ion in blood plasma is

0.0025 M. If the concentration of oxalate ion is

1.0x10-7 M, do you expect calcium oxalate to

precipitate? Ksp = 2.3x10-9.

Three steps:

(1) Determine the initial concentrations of ions.

(2) Evaluate the reaction quotient Q.

(3) Compare Q with Ksp.

SolutionCaC2O4 (s) Ca2+ (aq) + C2O4

2- (aq)

Ksp = [Ca2+] [C2O42-] = 2.3x10-9

Qsp = (2.5 x 10-3 M) (1.0x10-7 M) = 2.5 x 10-10

2.5 x 10-10 < 2.3x10-9

Q < Ksp therefore no ppt will be formed

Summary

• The solubility product constant, Ksp, represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution.

• The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound.

• The solubilities of some slightly soluble compounds depends strongly on pH.

Equilibrium lab

Fe(OH)3 (s) Fe3+ (aq) + 3OH- (aq)

Ksp = [Fe3+][OH-]3 = 4 x 10-38

Q vs. Ksp

Q = [Fe3+][OH-]3 = (0.2M)(6.0M)3 = 43.2

Q >Ksp so a PPT forms to take the Fe3+ out of solution

Qualitative Inorganic Analysis

• Acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation all come into sharp focus in an area of analytical chemistry called classical qualitative inorganic analysis.

• “Qualitative” signifies that the interest is in determining what is present, not how much is present.

• Although classical qualitative analysis is not as widely used today as instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.