Equilibrium

EquilibriumThe state where the concentrations of

all reactants and products remain constant with time.

Reactions are reversibleA + B C + D ( forward)C + D A + B (reverse)Forward and Reverse Rxns can be

shown by double arrow

A + B C + D

A + B C + D Initially there is only A and B so only the

forward reaction is possibleAs C and D build up, the reverse reaction

speeds up while the forward reaction slows down.

Eventually the rates are equal So concentrations of the reactants and

products no longer change with time

Rea

ctio

n R

ate

Time

Forward Reaction

Reverse reaction

Equilibrium

Static or Dynamic? At equilibrium, forward and reverse reaction rates

are equal May seem like no changes are occurring but

there are changes No NET changes On the molecular level, there is frantic activity.

Equilibrium is not static, but is a highly dynamic situation.

chemical reactions take place ,but concentrations of reactants and products remain unchanged

Analogies and Metaphors to think about

1. In a football game, the number of players on the field is constant although exchange of players (substitution) changes actual persons.

2. Connected fish bowl analogy . Two fish tanks are connected by a tube large enough to allow passage of fish. A number of fish are placed in one of the tanks. At equilibrium, the number of fish in each tank will eventually become unchanged.

3. Two jugglers analogy.

4. Drinking fountain line: (a) Ten students waiting in line to get a drink of water on a hot day. As each

gets a drink, the same student reenters the line (equilibrium in a closed system).

(b) (b) Same situation as "a," except as each student gets a drink and leaves, a new student enters the line (steady-state in an open system).

5. Picture a number of horses and wranglers in a corral. As each wrangler mounts a horse, the wrangler is bucked off. The equilibrium is:

Horse + Wrangler Mounted wrangler

Molecular Simulation In this simulation two gaseous reactants

collide to produce a more dense solid.

A + B C

gaseous R dense P http://www.absorblearning.com/media/atta

chment.action?quick=w8&att=2310

Homo vs HeteroHomogeneous Equilibria all reacting

species are in the same phasegas phase

• equilibrium constant can be expressed in terms of pressure or concentration, Kp or Kc

Solution (aqueous) phase• concentration term for the pure liquid does not appear in the

expression for the equilibrium constant but aqueous substance concentrations do appear

Heterogeneous Equilibria all reacting species are not in the same phase

• concentration term for solid or liquid does not appear in the expression for the equilibrium constant

Equilibrium SummarizedForward and Reverse rates are equalConcentrations are not.Rates are determined by

concentrations and activation energy.Molecular Motion is frantic and

constantly changingMacroscopically no net change is

occurring (We can’t observe any changes)

Can you identify when the system reached equilbrium?

Distinguishing between Physical and Chemical Equilibrium

As with physical and chemical changes, physical and chemical equilibrium follow the same rules:Physical no changes to the chemical

properties of the substances involved• Ex. equilibrium of water vapor with liquid water in a partly

filled sealed bottle

Chemical involve changes in the chemical composition of substances. Bond breaking and bond formation is involved.

• Ex. dissociation of acetic acid water into acetate and hydronium ion

ActivityModel Dynamic Equilibrium with

CoinsNow lets plot the data using excel

Law of Mass Action For a reaction: aA + bB cC + dD⇄

equilibrium constant: K

Pure liquids and pure solids have concentrations of 1.

c

c

Playing with KIf we write the reaction in reverse.

cC + dD aA + bB⇄Then the new equilibrium constant is

c

c

They are simply the inverse of one another.

Forward Reaction

aA + bB cC + dD⇄

So we call this K1

And K1= 1 = K2-1

K2

Reverse Reaction

cC + dD aA + bB⇄

So we call this K2

And K2= 1 = K1-1

K1

The units for KAre determined by the various powers

and units of concentrations. They depend on the reaction.will always have the same value at a

certain temperature (Why will the T effect it?) no matter what amounts are initially

added ratio at equilibrium will always be same

K Has no units Is constant at any given temperature. Is affected by temperature.

Equilibrium constants are reaction, phase, temperature and pressure dependent

There is a K for each temperature. Equilibrium constant values are thus established

for a specific reaction in a specific system and will be unchanging (constant) in that system, providing the temperatures does not change.

What does the size of my K mean?

Large K > 1 products are "favored“Ex. 1 x 1034

K = 1 neither reactants nor products are

favoredSmall K < 1

reactants are "favored“Ex. 4 x 10-41

Now Let’s Calculate Your KUsing your data from the simulation

calculate K

Different Equilibrium Constants (All of these are known as Keq) Kc is the most used general form with molar

concentrations. Kp can be used with partial pressures when

working with a gas phase reaction. Ka is used for the dissociation of weak acids in

water. Kb is used for the dissociation of weak bases in

water. Kw is the equilibrium expression for the dissociation

of water into its ions. Ksp is used for the dissociation into ions of sparingly

soluble solids in water.

Practice Writing the Equilibrium Expression

4NH3(g) + 7O2(g) 4NO2(g) + 6H2O(g)

First write the equilibrium expression using no concentration values.

What is the value for K if the concentrations are as follows

NH3 1.0 MO2 1.0 MNO2 1.4 MH2O 1.8 M

Equilibrium with GasesEquilibria involving only gases can be

described using pressures or concentrations If using pressures,

use pA not [A]KP not KC

be sure all pressure are in the same units

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

Calculating Kc from Kp

whereΔn is the difference in moles of gas on either side

of the equation (np – nr)R is the gas law constant: 0.08206T is Kelvin temperatureFor: N2(g) + 3H2(g) 2NH3(g) Δ n = (2) – (1+3) = -2

h

PracticeSetup the expression for KP in terms

of KC, R and T

2NO(g) + Cl2(g) 2NOCl(g)

What the equilibrium constant tells us…

if we know the value of K, we can predict: tendency of a reaction to occur if a set of concentrations could be at

equilibriumequilibrium position, given initial

concentrations If you start a reaction with only reactants:

concentration of reactants will decrease by a certain amount

concentration of products will increase by a same amount

Using this we can make ICE charts

The following reaction has a K of 16. You are starting reaction with 9 O3 molecules and 12 CO molecules.

Find the amount of each species at equilibrium.

Consider the following reaction at 600ºC 2SO2(g) + O2(g) 2SO3(g)

In a certain experiment 2.00 mol of SO2,

1.50 mol of O2 and 3.00 mol of SO3 were

placed in a 1.00 L flask. At equilibrium 3.50 mol were found to be present. Calculate

The equilibrium concentrations of O2 and

SO2, K and KP

Practice 1 ICE Charts

Consider the same reaction at 600ºCIn a different experiment .500 mol SO2

and .350 mol SO3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O2 are present.

Calculate the final concentrations of SO2 and SO3 and K

Practice Problem 2 ICE Charts

The Reaction Quotient (Q)Tells you the directing the reaction

will go to reach equilibriumCalculated the same as the

equilibrium constant, but for a system not at equilibrium

Q = [Products]coefficient

[Reactants] coefficient

Compare value to equilibrium constant

What Q tells us IF THEN

Q = K reaction is at equilibrium

Q > K too much products,

left shift

Q < K too much reactants,

right shift

Example 1 Reaction QuotientFor the synthesis of ammonia at 500°C,

the equilibrium constant is 6.0 x 10-2. Predict the direction the system will shift to reach equilibrium in the following case:

Ex 1 Cont.

Example 2In the gas phase, dinitrogen tetroxidedecomposes to gaseous nitrogen dioxide:

Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a T where KP = 0.133. At equilibrium, the pressure of N2O4 was found to be 2.71 atm. Calculate the equilibrium pressure of NO2.

Example 3

At a certain temperature a 1.00 L flask initially contained 0.298 mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in the flask.

PCl5(g) PCl3(g) + Cl2(g)

Calculate the equilibrium concentrations of all the species and the value of K.

Ex 3 Cont

Approximations If K is very small, we can assume that the

change (x) is going to be negligible compared to the initial concentration of the substances

can be used to cancel out when adding or subtracting from a “normal” sized number to simplify algebra

Example 4

At 35°C, K=1.6x10-5 for the reaction 2NOCl(g) 2NO(g) + Cl⇄ 2(g)

Calculate the concentration of all species at equilibrium for the following mixtures

2.0 mol NOCl in 2.0 L flask

1.0 mol NOCl and 1.0 mol NO in 1.0 L flask

2.0 mol NO and 1.0 mol Cl2 in 1.0 L flask

Ex 4 Cont

Ex 4 Cont

Ex 4 Cont

Le Chatelier’s Principle can predict how certain changes or

stresses put on a reaction will affect the position of equilibrium

helps us determine which direction the reaction will progress in to achieve equilibrium again

system will shift away from the added component or towards a removed component

Change Concentration equilibrium position can change but not K system will shift away from the added

component or towards a removed component

Ex: N2 + 3H2 2NH3

if more N2 is added, then equilibrium position shifts to right (creates more products)

if some NH3 is removed, then equilibrium position shifts to right (creates more products)

Adding Gasadding or removing gaseous reactant

or product is same as changing concentration

adding inert or uninvolved gas increase the total pressure doesn’t effect the equilibrium position

Change the Pressure by changing the Volume

only important in gaseous reactions decrease V

requires a decrease in # gas moleculesshifts towards the side of the reaction with

less gas molecules increase V

requires an increase in # of gas moleculesshifts towards the side of the reaction with

more gas molecules

Change in Temperature all other changes alter the concentrations

at equilibrium but don’t actually change value of K

value of K does change with temperature if energy is added, the reaction will shift in

direction that consumes energy treat energy as a

reactant: for endothermic reactionsproduct: for exothermic reactions

CatalystThe use of a catalyst may speed up

a reaction, but it speeds it up in both the forward and reverse direction, therefore a catalyst has no effect on the equilibrium state of the system.

energy + N2(g) + O2(g) 2NO(g)⇄ endo or exo?

endothermic increase temp

to right remove O2

to left increase volume

no shift add N2

to right

Le Chatlier’s Simulationhttp://www.learnerstv.com/

animation/animation.php?ani=120&cat=chemistry

Solution Equilibria

Another Dynamic EquilibriumEquilibrium occurs when the solution is

saturated

Ksp

The solubility product constant (Ksp) is similar to the Keq.

When a solid is added to water, some dissolves (and splits into ions) while some remains a solid.

Ex: NaCl(s) Na+(aq) + Cl-(aq)

The mass action expression for this is:

Ksp = [Na+][Cl-]

(remember… solids are 1)

Common Ion EffectThe solubility of a solid is lowered if

the solution already contains ions common to the solidDissolving silver chloride in a

solution containing silver ionsDissolving silver chloride in a

solution containing chloride ions

PrecipitationOpposite of dissolution Can predict whether precipitation or

dissolution will occurUse Q: ion product

Equals Ksp expression but doesn’t have to be at equilibrium

Q > K: more reactant will form, precipitation until equilibrium reached

Q < K: more product will form, dissolution until equilibrium reached

Example 1

Example 1 Cont.

Example 2

Example 2 Cont

Example 2 Cont

Example 2 Cont

Qualitative AnalysisProcess used to separate a solution

containing different ions using solubilities

Example Problem

A solution of 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If I- is gradually added, which will precipitate out first, CuI or PbI2?

Solution to Example Problem

ExampleThe molar solubility for MgCl2 is

0.0056M. Calculate Ksp

Salt SimulatorIn groups of two use the simulator to

discover the Ksp and Le Chatlier’s Principle

Acid-base Equilibria

Solutions of Acids or Bases Containing a Common Ion

Common Ion Ion provided in solution by an aqueous acid (or base) as

well as a salta. HF(aq) and NaF (F- in common)

b. HF(aq) H+(aq) + F-(aq) Excess F- added by NaF

Equilibrium shifts away from added component. Fewer H+ ions present.

pH is higher than expected.

a. NH4OH and NH4Cl (NH4+ in common)

b. NH3(aq) + H2O(l) NH4 +(aq) + OH-(aq) Equilibrium shifts to the left. pH of the solution decreases due to a decrease in OH- concentration

Equilibrium CalculationsConsider initial concentration of ion

from salt when calculating values for H+ and OH-

Acid + Base Salt + Water

Acid + Base Conjugate Base + Conjugate Acid

Some solvents are amphiproticWater can act as an acid and a base!Methanol can act as an acid and a base!

AutoprotolysisSome solvents can react with themselves to

produce an acid and a base• Water is a classical example

Weak acids dissociate partially, weak bases undergo partial hydrolysis. Strong acids and bases are strong electrolytes.

Kw (Dissociation of Water) Water is amphiprotic it also undergoes autoprotolysis

Kw = 1.0E-14 at about 25 ˚C This is where the pH scale we commonly use originates

from!

What is the concentration of hydronium and hydroxide ions in neutral solution? What is the pH? What is the pOH?

H2OBase

+ H2OAcid

Kw

H3OConjugate Acid

+ + OH-

Conjugate Base

Kw [H3O] x [OH- ]

Weak Acid & Weak Base Equilibria

Weak acids produce weak conjugate bases, and weak bases produce weak conjugate acids

Ka is a “special” equilibrium constant for the dissociation of a weak acid (found in standard tables)

Kb is a “special” equilibrium constant for the hydrolysis (or dissociation” of a weak base.

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

-

Ka = [H3O(aq)

+ ] x [A(aq)- ]

[HA(aq) ]

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

-

Kb = [NH4(aq)

+ ] x [OH(aq)- ]

[NH3(aq) ]

FOR ANY Conjugate Acid -Base Pair in Aqueous Solution

Ka x Kb = Kw

Only Ka values appear in most standardized tables.

Calculations…….. What is the pH of a 1.0 M solution of acetic acid

(HAc)?

What assumption can you make? If [acid] is about 1000 times the Ka value, it’s

concentration in solution won’t change much!Use an “I-C-E” table to look at this.The text goes into a more elaborate discussion

of approximations. I will allow approximations if the concentrations or pH values do not change in the hundred’s decimal place.

What is the pH of a 4.0 M solution of phosphate ion?Write reactionCalculate KbSetup “I-C-E” tableMake assumptionsSolve algebraically.

BuffersBuffers resist the change in pH

because they have acid to neutralize bases and bases to neutralize acids.

Made from a weak acid (HA) and the salt of its conjugate base (A-, where the counter ion is gone for example), or a weak base and the salt of its conjugate acid.

Features of BuffersBuffers work best at maintaining pH

near the Ka of the acid component, usually about +/- 1 pH unit. This is their buffer capacity .

Buffers resist pH changes due to dilution.

All seen when we use the “Buffer Equation”

Henderson-Hasselbalch (Buffer) Equation A modification of the equation for the dissociation of a

weak acid.

The pH is the pH of the buffered solution, pKa is the pKa of the weak acid.

What is the pH of a buffer solution made from 1.0 M acetic acid and 0.9 M sodium acetate?

You add .10 moles of sodium hydroxide to the above solution? What is the new pH?

pH = pKa + log A - HA

H-H Equation & Buffers….

If [A-] = [HA] pH = pKa!This is what we see at half-way to

the equivalence point in the titration of a weak acid with a strong base!

Dilution does not change the ratio of A- to HA, and thus the pH does not change significantly in most cases

How do you prepare buffers? Select a buffer ‘system’ based on the pH you want to maintain. Use the H-H equation to calculate how much acid and

conjugate base you need Take one of three approaches:

Mix acid and base forms, measure pH and adjust with strong acid or strong base

Start with a solution of weak acid, and add strong base until you reach the desired pH

Start with a solution weak base and add strong acid until you reach the desired pH

REMEMBER, STRONG Acids or Bases react completely! Dilute as necessary, adjust pH further if needed with strong

acid or base.

You want 1L of a buffer system that has a pH of 3.90?What acid/conjugate base pair

would you use?How would you go about figuring

out how much of each reagent you might need?

How would you prepare and adjust the pH of this solution?