Chemical Systems & Equilibrium

Unit 4

Dynamic Equilibrium dynamic equilibrium = a balance

between forward and reverse processes occurring at the same rate

Party Analogy 30 people at a house party 8pm:

16 people in the kitchen 14 people in the living room

10pm: 16 people in the kitchen 14 people in the living room

Different people but same number in each room

At Equilibrium: Closed system – no matter/energy/pressure

changes No macroscopic changes Reactants and products both present (and

usually in different amounts) [reactant] = constant, [product] = constant Can be approached from both sides

Rate of forward reaction = rate of reverse reactions

Dynamic Equilibrium Example: Closed bottle of pop CO2 gas leaving dissolved state and

entering gas state CO2 gas ALSO, leaving gas state and

entering liquid state No visible change

CO2(g) CO2(aq)

Equilibrium Double Arrow equilibrium is symbolized with an

equation containing a forward (→) and a reverse (←) arrow combined into:

N2O4 (g) 2NO2 (g)

Equilibrium Double Arrow forward reaction = in an equilibrium

equation, the left-to-right reaction

reverse reaction = in an equilibrium equation, the right-to-left reaction

CO2(g) CO2(aq) Forward

Reverse

Drinking Bird Equilibrium https://

www.youtube.com/watch?v=Bzw0kWvfVkA At rest the vapor and the liquid inside the tube are in an equilibrium  Wet head of bird with water – as the water evaporates from around the

head, it takes energy with it, head cools down = vapor inside the head cools and contracts = vacuum = pulls liquid up

3 Types of Equilibrium1. Solubility Equilibrium (dissolving process)

2. Phase Equilibrium (change of state)

3. Chemical Reaction Equilibrium (reactants ⇆ products)

Types of Equilibrium #1 solubility equilibrium = a dynamic

equilibrium between a solute and a solvent in a saturated solution in a closed system

I2(s) I2(aq)

Solubility Equilibrium Saturated solution = a solution

containing the maximum quantity of a solute

Beyond the solubility limit, any added solute will remain solid and not dissolve

Solubility Equilibrium kinetic molecular theory states that

particles are always moving and colliding

even if no changes are observed

Dissolution = the process of dissolving

(a) When the solute is first added, many more ions dissociate from the crystal than crystallize onto it.

(b) As more ions come into solution, more ions also crystallize.

(c) At solubility equilibrium, solute ions dissolve and crystallize at the same rate.

Digesting a Precipitate Allow precipitates to sit for long periods of

time before filtering The longer you wait the more pure the crystal,

also the larger the crystal If precipitate forms quickly, impurities maybe

trapped in the precipitate

Types of Equilibrium #2 phase equilibrium = a dynamic equilibrium

between different physical states of a pure substance in a closed system

closed system = a system that may exchange energy but NOT matter with it’s surroundings

H2O(l) H2O(g)

H2O(s) H2O(l)

Phase Equilibrium

Types of Equilibrium #3 chemical reaction equilibrium a

dynamic equilibrium between reactants and products of a chemical reaction in a closed system

reversible reaction = a reaction that can achieve equilibrium in the forward or reverse direction

Chemical Reaction Equilibrium In a Closed SystemN2O4(g) 2 NO2(g)

Reversible Reactions The same dynamic equilibrium

composition is reached whether we start from pure N2O4(g), pure NO2(g), or a mixture of the two, provided that environment, system and total mass remain the same.

Calculating the Equilibrium Constant The equilbrium constant, Keq, is the ratio of

equilibrium concentrations at a particular temp Kc for solution-phase systems or Kp for gas-

phase systems

Keq = [C]c[D]d for the eqn[A]a[B]b aA+bB cC+dD

Note: The equilibrium constant depends ONLY on the concentration of gases (not liquids/solids)

Questions: Equilibrium Law Expression1. Write the equilibrium law expression for the following:a) 2NO2(g) ↔ N2O4(g)

b) 2HI(g) ↔ H2(g) + I2(g)

2. A reaction vessel contains NH3, N2 and H2 gas at equilibrium at a certain temperature. The equilibrium concentrations are [NH3] = 0.25mol/L, [N2] = 0.11mol/L and [H2] = 1.91 mol/L. Calculate the equilibrium constant for the decomposition of ammonia.

K = [N2O4(g)]

[NO2(g) ]2

K = [H2(g)] [I2(g) ]

[HI(g) ]2

K = [N2(g)] [H2(g) ]3

[NH3(g) ]2K =

[0.11] [1.91 ]3

[0.25 ]2

K = 12.3

2NH3(g) ↔ N2(g) + 3H2(g)

Questions: Equilibrium Law Expression3. Nitryl chloride gas, NO2Cl, is in equilibrium at a certain temperature in a closed container with NO2 and Cl2 gases. At equilibrium, [NO2Cl] = 0.00106mol/L and [NO2] = 0.0108mol/L. If K = 0.558, what is the equilibrium concentration of Cl2?

4. Write a balanced equation for the reaction with the following equilibrium law expression:

K = [NO2(g)]2

[NO (g) ]2 [O2 (g) ]

Heterogeneous Equilibria homogeneous equilibria = equilibria in

which all entities are in the same phase Reactants and products are all gas or all

aqueous

heterogeneous equilibria = equilibria in which reactants and products are in more than one phase Reactants and products are in different

phases

Homogenous equilibrium applies to reactions in which all reacting species are in the same phase.

N2O4 (g) 2NO2 (g)

Kc = [NO2]2

[N2O4]Kp = NO2

P2

N2O4P

Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.CaCO3 (s) CaO (s) + CO2 (g)

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

[CaCO3(s)][CaCO3(s)] = constant[CaO(s) ] = constant

Kc = [CO2(g)]

The concentration of solids and pure liquids are considered to be constant and are not included in the expression for the

equilibrium constant.

Kc [CaO(s)]

[CaCO3(s)] = [CO2(g)]

PCO2 = Kp

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

PCO2does not depend on the amount of CaCO3 or CaO

N2O4 (g) 2NO2 (g)

Start with NO2 Start with N2O4 Start with NO2 & N2O4

equilibrium

equilibrium

equilibrium

Equilibrium favors the reactant side

CHECKPOINTThe reaction at 200C between ethanol and ethanoic acid produces ___________________ and __________________.

1. Write the equation for this reaction

2. Determine the equilibrium constant expression for the reaction

Sample Problem:When ammonia is heated it decomposes:

2NH3(g)↔ N2(g) + 3H2(g)

When 4.0 mol of ammonia is introduced in a 2.0L container and heated. Theequilibrium amount of ammonia is 2 0 mol. Determine the equilibrium concentrations of the other two entities.

STEP 1: Determine the concentration (initial and equilibrium) for known valuesSTEP 2: Setup an ICE TableSTEP 3: Determine the value of XSTEP 4: Use x value to determine the other quantities

Calculating Equilibrium Concentrations (when given one concentration)

Setup ICE Table

Determine the concentrations[NH3]initial = 4.0mol/2.0L = 2.0mol/L

[NH3]equilibrium = 2.0mol/2.0L = 1.0mol/L

Determine the value of X[NH3](g)equil = 2 0mol / L - 2x

[NH3](g)equil = 1.0mol/L (from calculations in Step 1)

2.0mol/L – 2x = 1.0mol/L-2x = - 1.0mol/Lx = 0.5mol/L

Use X to determine other quantities

constant

Reversible Reactions

For a given overall system composition, the same equilibrium concentrations are reached whether equilibrium is approached in the forward or the reverse direction

What about Keq will it be the same in fwd/rev?

Equilibrium TubesThe effects of temperature on equilibrium

Heat + N2O4 (g) 2NO2 (g)Colourless

Brown

Very Cold Cold Hot

ENDOTHERMIC Rxn

NO2 is one of the chemicals in smog! In the summer on hot, windless days an orange

haze is seen over the horizon, this is NO2

In the winter, the smog doesn't go away, it is just less noticeable. The cooler temperatures lead to more N2O4 and less NO 2 which we can't see as well!

N2O4 (g) 2NO2 (g)Colourless

Brown

Qualitative Changes in Equilibrium SystemsYou should be familiar with your own body’s attempt at maintaining equilibrium or “homeostasis”:

If body T too high sweat, surface blood vessels dilate

If body T too low shiver, surface blood vessels constrict

If blood CO2 levels ↑ breathe deeper & faster

If blood sugar levels ↑ insulin released to remove excess glucose

Le Châtelier’s Principle When a chemical system at equilibrium

is disturbed by a change in a property, the system adjusts in a way that opposes the change.

In other words: If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

Le Châtelier’s PrincipleLe Chatelier’s Principle: if you disturb

an equilibrium, it will shift to undo the disturbance.

equilibrium shift = movement of a system at equilibrium, resulting in a change in the concentrations of reactants and products

https://www.youtube.com/watch?v=dIDgPFEucFM

1. System starts at equilibrium.2. A change/stress is then made to

system at equilibrium. Change in concentration Change in temperature Change in volume/pressure

3. System responds by shifting to reactant or product side to restore equilibrium.

Le Châtelier’s Principle

Le Châtelier’s PrincipleChange in Reactant or Product

Concentrations• Adding a reactant or product shifts the

equilibrium away from the increase.• Removing a reactant or product shifts the

equilibrium towards the decrease.• To optimize the amount of product at

equilibrium, we need to flood the reaction vessel with reactant and continuously remove product.

Le Châtelier’s PrincipleChange in Reactant or Product

Concentrations

• If H2 is added while the system is at equilibrium, the system must respond to counteract the added H2

• That is, the system must consume the H2 and produce products until a new equilibrium is established.

• Equilibrium shifts to the right.• Therefore, [H2] and [N2] will decrease and [NH3]

increases.

N2 (g) + 3H2 (g) ↔ 2NH3 (g)

Change in Reactant or Product Concentrations

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

AddNH3

Equilibrium shifts left to offset stress

Change in Reactant or Product Concentrations

Change Shifts the EquilibriumIncrease concentration of product(s) leftDecrease concentration of product(s) right

Decrease concentration of reactant(s)Increase concentration of reactant(s) right

left

aA + bB cC + dD

Le Châtelier’s PrincipleEffect of Temperature Changes• The equilibrium constant is

temperature dependent.• For an endothermic reaction, H > 0

and heat can be considered as a reactant.

• For an exothermic reaction, H < 0 and heat can be considered as a product.

Effect of Temperature Changes

Effect of Temperature Changes

Adding heat (i.e. heating the vessel) favors away from the increase:– if H = + (Endothermic), adding heat

favors the forward reaction,– if H = - (Exothermic), adding heat favors

the reverse reaction.Removing heat (i.e. cooling the vessel), favors

towards the decrease:– if H = + (Endothermic), cooling favors

the reverse reaction,– if H = - , (Exoothermic), cooling favors

the forward reaction.

Gas Law – Boyle’s LawRelationship: Pressure & Volume As pressure on a gas increases, the

volume of the gas decreases

Le Châtelier’s PrincipleEffects of Volume and Pressure• As volume is decreased pressure increases.• The system shifts to decrease pressure.• An increase in pressure favors the direction that

has fewer moles of gas. • Decreasing the number of molecules in a

container reduces the pressure.• In a reaction with the same number of product

and reactant moles of gas, pressure has no effect.

Only a factor with gases.

Effects of Volume and Pressure

Change Shifts the EquilibriumIncrease pressure Side with fewest moles of gasDecrease pressure Side with most moles of gas

Decrease volumeIncrease volume Side with most moles of gas

Side with fewest moles of gas

A (g) + B (g) C (g)

Le Châtelier’s PrincipleAdding a Catalyst

• does not shift the position of an equilibrium system

• system will reach equilibrium sooner

Le Châtelier’s PrincipleAdding a Catalyst

lowers the activation energy for both forward and reverse reactions by an equal amount, so the equilibrium establishes much more rapidly but at the same position as it would without the catalyst

Adding a Catalyst

Le Châtelier’s PrincipleAdding Inert Gases

pressure of a gaseous system at equilibrium can be changed by adding a gas while keeping the volume constant

If the gas is inert in the system, for example, if it is a noble gas or if it cannot react with the entities in the system, the equilibrium position of the system will not change

Le Châtelier’s Principle Summary

No Sweat! Chickens cannot perspire. When a chicken gets hot, it pants like a dog. Farmers have known for a long time that

chickens lay eggs with thin shells in hot weather. These fragile eggs are easily damaged. Eggshell is primarily composed of calcium

carbonate, CaCO3(s). The source of the carbonate portion of this

chalky material is carbon dioxide, CO2, produced as a waste product of cellular respiration.

No Sweat! The carbon dioxide dissolves in body

fluids forming the following equilibrium system:

No Sweat! When chickens pant, blood carbon

dioxide concentrations are reduced, causing a shift through all four equilibria to the left and a reduction in the amount of calcium carbonate available for making eggshells.

Solution: Give the chickens carbonated water to drink in the summer.

This shifts the equilibria to the right, compensating for the leftward shift caused by panting.

Le Chatelier’s Principle: Warm-up Page 459 # 4 & 6

AgNO3 – Hint: check solubility table

Graph for Question #4

Le Chatelier’s Principle: Warm-up Page 459

A = ↑ volumeB = ↑ temperatureC = ↑ [C2H6]D = catalyst/inert gasE = ↓ [C2H4]

At constant temperature, regardless of initial concentrations the concentrations of reactants and products always give a constant value K

aA + bB cC + dD

K = [C]c[D]d

[A]a[B]b

ProductsReactants

The Equilibrium Law Expression &The Equilibrium Constant, K

Equilibrium Law Expression The molar concentrations of the products are

always multiplied by one another and written in the numerator, and the molar concentrations of the reactants are always multiplied by one another and written in the denominator.

The coefficients in the balanced chemical equation are equal to the exponents of the equilibrium law expression.

The concentrations in the equilibrium law expression are the molar concentrations of the entities at equilibrium.

Recall: Equilibrium achieved from any combination of reactants and products

constant

N2O4 (g) 2NO2 (g)

= 4.6 x 10-3K = [NO2]2

[N2O4]

Regardless of initial concentrations, at a given temperature, the relationship of the equilibrium concentrations of reactants and products always yields a CONSTANT value, K

Ways Different States of Matter Can Appear in the Equilibrium Constant, K

Molarity Partial Pressure

gas, (g) YES YES

aqueous, (aq) YES ---

liquid, (l) --- ---

solid, (s) --- ---

Questions: Equilibrium Law ExpressionsWrite the equilibrium law expressions for the following reactions:1. NH4Cl(s) ↔ NH3(g) + HCl(g)

2. 2H2O(l) ↔ 2H2(g) + O2 (g)

3. 2NaHCO2(s) ↔ Na2CO3(s) + H2O(g) + CO2(g)

K = [NH3 (g) ] [HCl(g)]

K = [H2 (g) ]2 [O2(g) ]

K = [H2O (g) ] [CO2(g) ]

Equilibrium Law Expression Note: equilibrium constants give no

information about the rate of a reaction; they provide only a measure of the equilibrium position of the reaction

K is independent of the initial concentration of the reactants and products, but on the concentrations at the equilibrium

What Does the Value of K Mean? If K >> 1, the reaction is product-

favoured; product predominates at equilibrium.

If K << 1, the reaction is reactants-favoured; reactants predominates at equilibrium.

Products

Reactants

Question: Magnitude of K Consider the reaction: H2(g) + I2(g) ↔ 2HI(g) + heat

At 448⁰C, K=50.5. Would you predict the value of K to be higher or lower at 300⁰C?

At 448⁰C K >> 1 = PRODUCTS favoured Heat lowered = rxn shifts to PRODUCT side At 300 ⁰C K > 50.5

Equilibrium Reactions in Solution In addition to gas-phase and heterogeneous

reactions, equilibrium reacts can also take place in solution.

It is important to write the reaction components as they ACTUALLY EXIST IN SOLUTION -- represent ions in solution as individual entities

Get the equilibrium law expression from the net ionic equation

Equilibrium Reactions in SolutionExample: Write the equilibrium reaction and equilibrium law expression for the reaction between zinc metal and copper (II) chloride solution.

Cancel out the spectator ions

The Reaction Quotient, Q If a chemical system begins with reactants only, it

is obvious that the reaction will initially proceed to the right, toward products.

If, however, reactants and products are both present, the direction in which the reaction proceeds is usually less obvious.

In such a case, we can substitute the concentrations into the equilibrium law expression to produce a trial value that is called a reaction quotient, Q

The Reaction Quotient, Q reaction quotient, Q = a test

calculation using measured concentration values of a system in the equilibrium expression

think of Q as being similar to K K is calculated using concentrations

at equilibrium Q may or may not be at equilibrium

The Reaction Quotient, Q

aA + bB cC + dD

Q = [C]c[D]d

[A]a[B]b

The Reaction Quotient, Q Q is equal to K, and the system is at equilibrium.

Q is greater than K, and the system must shift left (toward reactants) to reach equilibrium, because the product-to-reactant ratio is too high.

Q is less than K, and the system must shift right (toward products) to reach equilibrium, because the product-to-reactant ratio is too low.

ICE Table Initial, Change, Equilibrium

I = initial concentration of reactants and products before reaction

C = change in the concentrations of reactants and productsthe start and the point at which equilibrium is achieved

E = concentrations of reactants and products at equilibrium.

Solving Equilibrium Problems with ICEExample1:

For the above reaction [N2]i = 0.32mol/L and [H2]i = 0.66mol/L. At a certain T and P, [N2]eq = 0.20mol/L. What is the value of K under these conditions?

Example 2At 150°C, K for the reaction I2(g) + Br2(g) ↔ 2IBr(g) is found to be 1.20x102 . Starting with 4.00mol of each of iodine and bromine in a 2.00L flask, calculate the equilibrium concentrations of all reaction components.

Example 3 Unlike the previous two examples, it is not

always obvious if a system is already at equilibrium, or which way the reaction will shift to reach equilibrium.

In these situations, it is helpful to determine the Reaction Quotient, Q

When the reaction 2HI(g) ↔ H2(g) + I2(g) takes place at 445°C, the value of K is 0.020. If [HI]=0.20mol/L, [H2]=0.15mol/L and [I2]=0.09mol/L, is the system at equilibrium? If not, in which direction will it shift to reach equilibrium?

Example 4 For the reaction H2(g) + F2(g) ↔ 2HF(g) , K is

1.5x102 at SATP. Calculate all equilibrium concentrations if 4.00mol of H2(g) , 4.00mol of F2(g) and 6.00mol of HF(g) are initially placed in a 2.00L reaction vessel.

Calculations with Imperfect Squares Our ability to square both sides of the

equilibrium law equation greatly simplified the calculation of equilibrium concentrations.

In the absence of perfect squares, a different simplification technique helps us solve the problem.

Assumption“The 100 rule” if the concentration to which x is

added or from which x is subtracted is at least 100 times greater than the value of K

initial conc. divided by K If # is greater 100 then drop the x in

the denominator

When the 100 Rule assumption fails We must use the quadratic equation

Example 5If 0.50 mol of N2O4(g) is placed in a 1.0L closed container at 150C, what will be the concentrations of N2O4(g) and NO2(g) at equilibrium? (K = 4.50)

N2O4(g) ↔ 2NO2(g)

Homework Read section 7.5 Questions

Remember Solubility? Solubility = the concentration of a

saturated solution of a solute in solvent at a specific temperature and pressure Solubility is a specific maximum concentration

Degree of Solubility: Unsaturated Saturated Supersaturated

Solubility

Unsaturated solution = a solution containing less than maximum quantity of a solute

Saturated solution = a solution containing the maximum quantity of a solute

Supersaturated solution = a solution whose solute concentration exceeds the equilibrium concentration

Solubility Curve of Solids

The Solubility Product Constant Solubility product constant (Ksp) = the

value obtained from the equilibrium law applied to a saturated solution

Similar to Keq no units At a specific temp.

Example: AgCl(s) ↔ Ag+(aq) + Cl-

(aq)

Ksp = [Ag+(aq) ] [Cl-

(aq) ] = 1.8x10-10 at 25⁰C

Equilibrium exists between a saturated solution and excess solute.

Dissolving

Precipitation

Saturated SolutionExcess

Solute

Solubility vs. Solubility Product Solubility = the amount of a salt that

dissolves in a given amount of solvent to give a saturated solution mol/L or g/100mL

Solubility Product = the product of the molar concentrations of a the ions in the saturated solution Ksp has no units

Table of Ksp

Appendix C8 (page 802)

Usually only for low solubility ionic compounds

High solubility compounds form solutions that do not tend to be saturated & no equilibrium is established

Calculating Solubility using the Ksp Value Example 1:Calculate the molar solubility of cobalt (II) hydroxide at 25⁰C if Ksp = 1.1x10-15 at this temperature.

Calculating Ksp using Solubility values Example 2:Calculate Ksp for silver chromate (Ag2CrO4) if its solubility is 0.29g/L at 25 ⁰ C.

Predicting Precipitation Instead of using a solubility table… using Q to determine whether, after mixing,

the ions are present in too high a concentration, in which case a precipitate will form

Trial Ion Product = the reaction quotient applied to the ion concentrations of a slightly soluble salt

Using Q to Predict Solubility

Q is greater than Ksp

supersaturated solution

Precipitate will from

Q is equal to Ksp Saturated Precipitate will not form

Q is less than Ksp unsaturated Precipitate will not form

Demo: KI + Pb(NO3)2

Calculations involving the prediction of a precipitate (using Q)

Example 3:If 500mL of a 4.0x10-6 mol/L CaCl2 solution is mixed with 300mL of a 0.0040mol/L AgNO3 solution, will a precipitate form?

Homework: Page 486 #1,2,4 Page 488 #5 Worksheet: Extra Solubility Problems

Quiz on Thursday April 18 ICE problem + solubility

Common Ion Effect Common Ion Effect = A reduction in the

solubility of a salt caused by the presence of another slat having a common ion

Energy & Equilibrium:

The Laws of Thermodynamics

Thermodynamics Thermodynamics = the study of

energy transformation

3 fundamental laws of thermodynamics

Laws used to understand why certain changes occur but others do not

First Law of Thermodynamics“Conservation of Energy”

The total amount of energy in the universe is constant. Energy can be neither created nor destroyed, but can be transferred from one object or place to another, or transformed from one form to another.

First Law of ThermodynamicsRemember:

Total energy of the universe = system + surrounding

Hess’s Law = the value of ∆H for any reaction that can be written in steps equals the sum of the ∆H values for each of the individual steps

Enthalpy Changes & Spontaneity bond energy = the minimum energy

required to break one mole of bonds between two particular atoms; a measure of the stability of a chemical bond

It is also equal to the amount of energy released when a mole of a particular bond is formed.

It is measured in kilojoules (kJ) and is equal to the minimum energy required to break the intramolecular bonds between one mole of molecules of a pure substance.

Bond Energy Bond energy is measured in kilojoules (kJ)

and is equal to the minimum energy required to break the intramolecular bonds between one mole of molecules of a pure substance.

Energy is absorbed when reactant bonds break

Energy is released when product bonds form

Bond Energy

A bond that has a higher bond energy (i.e. Requires more energy to break) is more stable.

Enthalpy & Entropy Changes Together Determine Spontaneity Endothermic = + ∆H Exothermic = - ∆H

Exothermic reactions tend to proceed spontaneously

Spontaneous Reaction spontaneous reaction = one that,

given the necessary activation energy, proceeds without continuous outside assistance

Example: a sparkler Needs light from a flame for activation Once lit, the available fuel combusts

quickly and completely, releasing large amounts of energy as heat and light

Entropy enthalpy is not the only factor that

determines whether a chemical or physical change occurs spontaneously

entropy, S = a measure of the randomness or disorder of a system, or the surroundings

Entropy Increase entropy = increase randomness

= +∆S

When entropy increases in a reaction, the entropy of the products, Sproducts, is greater than the entropy of the reactants, Sreactants, yielding an overall positive change in entropy, S.

Entropy decrease entropy = decrease

randomness = -∆S

When entropy decreases in a reaction, the entropy of the products, Sproducts, is less than the entropy of the reactants, Sreactants, yielding an overall negative change in entropy, S.

Increase in Entropy

Change in Volume of Gaseous Systems

Change in Temperature

Change in State

In Chemical Reactions…

Enthalpy, Entropy, and Spontaneous Change Changes in the enthalpy, ∆H, and entropy, ∆S,

of a system help us to predict whether a change will occur spontaneously

Exothermic reactions (-∆H) involving an increase in entropy (+∆S) occur spontaneously, because both changes are favoured

Endothermic reactions (+∆H) involving a decrease in entropy (-∆S) are not spontaneous because neither change is favoured

Enthalpy, Entropy, and Spontaneous Change But what happens in cases where the

energy change is exothermic (favoured) and the entropy decreases (not favoured)?

Or when the energy change is endothermic (not favoured) but entropy increases (favoured)?

In these situations, the temperature at which the change occurs becomes an important consideration as well as free energy

Free Energy free energy (or Gibbs free energy), G

= energy that is available to do useful work

In general, a change at constant temperature and pressure will occur spontaneously if it is accompanied by a decrease in Gibbs free energy, G

-∆G = spontaneous+∆G = nonspontaneous

Second Law of Thermodynamics“Law of Entropy”

all changes that occur in the universe. All changes, whether spontaneous or not, are accompanied by an increase in the entropy (overall disorder) of the universe

Mathematically, Suniverse > 0

Second Law of Thermodynamics

a system’s entropy, Ssystem, can decrease (the system becomes more ordered), so long as there is a larger increase in the entropy of the surroundings, Ssurroundings, so that the overall entropy change, Suniverse, is positive.

Problem? Living organisms seem to violate the

second law of thermodynamics. Build highly ordered molecules such as

proteins and DNA from a random assortment of amino acids and nucleotides dissolved in cell fluids

building highly ordered structures such as nests, webs, and space huttles.

Not really a problem… Living organisms obey the second law

of thermodynamics because they create order out of chaos in a local area of the universe while creating a greater amount of disorder in the universe as a whole

Oh no! Thermal Death! The second law of thermodynamics predicts that the

universe will eventually experience a final “thermal death” in which all particles and energy move randomly about.

Life will come to an end because there won’t be any sources of free energy to exploit; stars will stop shining. Waterfalls will stop falling.

All energy will have become randomized. All of the energy that there ever was will still be there, except that it will be uniformly distributed throughout the universe, unable to apply an effective push or a pull on anything. According to the second law, a state of perfect equilibrium is the ultimate fate of the universe.

Predicting Spontaneity The spontaneity of any reaction carried

out at constant temperature and pressure can be predicted by calculating the value of G using the following equation, called the Gibbs-Helmholtz equation:

∆G, Spontaneity & Free Energy

∆G = ∆H - T∆S

∆G = - = spontaneous∆G = + = nonspontaneous

Remember: K = ºC + 273

Predicting Spontaneity

+∆H -∆H+∆S Spontaneity

depends on TSpontaneo

us

-∆S nonspontaneous

Spontaneity depends

on T

∆G = ∆H - T∆S-∆G = spontaneous

+∆G= nonspontaneous

Third Law of Thermodynamics“Law of Entropy”

The entropy of a perfectly ordered pure crystalline substance is zero at absolute zero.

Mathematically, S = 0 at T = 0 K

Calculating Standard Entropy Change standard entropy = the entropy of

one mole of a substance at STAP; units (J/molK)

Remember ∆H° Standard enthalpy change of reaction

∆H° vs ∆S°