Enthalpy (H)

• The heat transferred sys ↔ surr during a chemical rxn @ constant P

• Can’t measure H, only ΔH

• At constant P, ΔH = q = mCΔT, etc.

• Literally, ΔH = Hproducts - Hreactants

• ΔH = + (endothermic)• Heat goes from surr into sys

• ΔH = - (exothermic)• Heat leaves sys and goes into surr

Surroundings

System

Surroundings

SystemEn

erg

y

Beforereaction

Afterreaction

In this example, the energy of the system (reactants and products) ↑, while the energy of the surroundings ↓Notice that the total energy does not change

Myers, Oldham, Tocci, Chemistry, 2004, page 41

Reactant + Energy Product Endothermic Reaction

Notice that E must be added, and thus is like a reactant

Surroundings

System

Surroundings

System

En

erg

y

Beforereaction

AfterreactionMyers, Oldham, Tocci, Chemistry, 2004, page 41

In this example, the energy of the system (reactants and products) ↓, while the energy of the surroundings ↑Notice again that the total energy does not change

Reactant Product + Energy Exothermic Reaction

Notice that E is released and thus is like a product

Burning of a Match

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 293

Energy released to the surrounding as heat

SurroundingsSystem

(Reactants)

(PE)

Pot

entia

l ene

rgy

(Products)

Exothermic Reaction

PE

Reaction Coordinate Diagrams: Endothermic Reaction

Progress of the Reaction

Reactants

Products(PE)

Ea

ΔHrxn= +

Activation Energy

PE

Reaction Coordinate Diagrams: Exothermic Reaction

Progress of the Reaction

Reactants

Products

(PE)

Ea

Activation Energy

ΔHrxn= -

Reaction Coordinate DiagramsDraw the reaction coordinate diagram for the following rxn:

C(s) + O2(g) CO2 + 458.1kJ

EXOTHERMIC

PE

Progress of the Reaction

C + O2

CO2

(PE)

Ea

Activation Energy

ΔHrxn= -458.1 kJ

• All reactions have some ΔH associated with itH2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ

• How can we interpret this ΔH?• Amount of energy released or absorbed per specific

reaction species• Use balanced equation to find several definitions

- 483.6 kJ

½ mol O2

Enthalpies of Reaction

- 483.6 kJ

1 mol H2

- 483.6 kJ

1 mol H2O

Able to use like conversion factors in stoichiometry

or or

Enthalpies of Reaction

• Formation of waterH2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ

• ΔH is proportional to amount used and will change as amount changes

2H2(g) + O2(g) → 2H2O(l)

• For reverse reactions, sign of ΔH changes2H2O(l) → 2H2(g) + O2(g)

• Treat ΔH like reactant or product H2(g) + ½ O2(g) → H2O(l) ΔH = - 483.6 kJ

ΔH = - 967.2 kJ

ΔH = + 967.2 kJ

H2(g) + ½ O2(g) → H2O(l) + 483.6 kJ (exo)

2.00 mol C

50.0 g O2

Consider the following rxn:

C(s) + 1/2O2(g) CO + 458.1kJIs the ΔH for this reaction positive or negative?

NEGATIVE (E released as a product)What is the ΔH for 2.00 moles of carbon, if all the carbon is used?

= - 916 kJ

What is the ΔH if 50.0g of oxygen is used?

= -1430 kJ

What is the ΔH if 50.0 g of carbon monoxide decompose, in the reverse reaction?

= 818 kJ

- 458.1 kJ1 mol C

1 mol O2

0.5 mol O2

- 458.1 kJ32.0 g O2

50.0 g CO 1 mol CO

1 mol CO458.1 kJ

28.0 g CO

Enthalpies of Reaction Practice

Hess’s law• Hess’s Law states that the enthalpy of a whole

reaction is equivalent to the sum of it’s steps.• For example: C + O2 CO2

This can occur as 2 stepsC + ½O2 CO H = – 110.5 kJCO + ½O2 CO2 H = – 283.0 kJ

C + CO + O2 CO + CO2 H = – 393.5 kJ

I.e. C + O2 CO2 H = – 393.5 kJ • Hess’s law allows us to add equations.• We add all reactants, products, & H values.

Reactants Products

The change in enthalpy is the same whether the reaction takes place in one step or a series of steps

Why? Because enthalpy is a state function

Hess’s Law

Victor Hess

2. If the coefficients of a reaction are multiplied by an integer, ΔH is multiplied by that same integer

2 CH3OH 2 CH4 + O2 ΔHrxn = +328 kJ

2 CH4 + O2 2 CH3OH ΔHrxn = -328 kJ

To review:

1. If a reaction is reversed, ΔH is also reversed

2(CH4 + 2 O2 CO2 + 2 H2O) ΔHrxn = -1605 kJ

CH4 + 2 O2 CO2 + 2 H2O ΔHrxn = -802.5 kJ

Example: Methanol-Powered Cars

2 CH3OH(l) + 2 CH4(g) + 4 O2(g) 2 CH4(g) + O2(g) + 2 CO2(g) + 4 H2O(g)

2(CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)) ΔHrxn = -1605 kJ

2 CH3OH(l) 2 CH4(g) + O2(g) ΔHrxn = +328 kJ

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g) ΔHrxn = -802.5 kJ

2 CH4(g) + O2(g) 2 CH3OH(l) ΔHrxn = -328 kJ

2 CH3OH(l) + 3 O2(g) 2 CO2(g) + 4 H2O(g) ΔHrxn = ?

2 CH3OH + 3 O2 2 CO2 + 4 H2O ΔHrxn = -1277 kJ

3

Tips for applying Hess’s Law…

Look at the final equation that you are

trying to create first…

• Find a molecule from that eq. that is only in one of the given equations• Make whatever alterations are necessary to those• Once you alter a given equation, you will not alter it again

• Continue to do this until there are no other options

• Next, alter remaining equations to get things to cancel that do not appear in the final equation

1. Given the following data:S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ

2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ

.Calculate ΔH for the following reaction: S(s) + O2(g) → SO2(g)

S(s) + 3/2O2(g) → SO3(g) ΔH = -395.2 kJ 2SO2(g) + O2(g) → 2SO3(g) ΔH = -198.2 kJ 2SO3(g) → O2(g) + 2SO2(g) ΔH = +198.2 kJ SO3(g) → ½ O2(g) + SO2(g) ΔH = +99.1 kJ

S(s) + O2(g) → SO2(g) ΔH = -296.1kJ

2. Given the following data:C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300 kJC(s) + O2(g) → CO2(g) ΔH = -394 kJH2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ Calculate ΔH for the following reaction: 2C(s) + H2(g) → C2H2(g)

C(s) + O2(g) → CO2(g) ΔH = -394 kJ

C2H2(g) + 5/2O2(g) → 2CO2(g) + H2O(l) ΔH = -1300 kJ2CO2(g) + H2O(l) → C2H2(g) + 5/2O2(g) ΔH = +1300 kJ H2(g) + 1/2O2(g) → H2O(l) ΔH = -286 kJ

2C(s) + H2(g) → C2H2(g) ΔH = +226kJ

2C(s) + 2O2(g) → 2CO2(g) ΔH = -788 kJ

NO(g) + O(g) → NO2(g) ΔH = +233kJ

3. Given the following data:  2O3(g) → 3O2(g) ΔH = - 427 kJ O2(g) → 2O(g) ΔH = + 495 kJNO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJ Calculate ΔH for the following reaction:

NO(g) + O(g) → NO2(g)

NO(g) + O3(g) → NO2(g) + O2(g) ΔH = - 199 kJ O2(g) → 2O(g) ΔH = + 495 kJ

2O3(g) → 3O2(g) ΔH = - 427 kJO(g) → ½ O2(g) ΔH = - 247.5 kJ

3/2 O2(g) → O3(g) ΔH = + 213.5 kJ

Heats of Formation, ΔH°f

The enthalpy change when one mole of a compound is formed from the elements in their standard states

° = standard conditions

• Gases at 1 atm pressure

• All solutes at 1 M concentration

• Pure solids and pure liquids

f = a formation reaction

• 1 mole of product formed

• From the elements in their standard states (1 atm, 25°C)

For all elements in their standard states, ΔH°f = 0

C9H12NO3(s)O2(g)N2(g) +H2(g) +Cgr +1/29 6 3/2

What’s the formation reaction for adrenaline, C9H12NO3(s)?

Thermite Reaction

Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

ΔHrxn = ?

Welding railroad tracks

Thermite Reaction

Fe2O3(s) + 2 Al(s) Al2O3(s) + 2 Fe(l)

2 Al(s)

Fe2O3(s) 2 Fe(s) 2 Fe(l)

2 Al(s)3/2 O2(g) Al2O3(s)

Reactants Elements(standard states)

Products

ΔHrxn = 2ΔH°f(Fe(l)) + ΔH°f(Al2O3(s)) - ΔH°f(Fe2O3(s)) - 2ΔH°f(Al(s))

ΔHrxn = [2(15 kJ) + (-1676 kJ)] – [(-822 kJ) – 2(0)]

ΔHrxn = -824 kJ

ΔHrxn = nΔH°f(products) - nΔH°f(reactants)

∆Hrxn = Σ n∆Hof Products - Σn∆Ho

f Reactants

(-74.8) (0) (-106.7) (0)

1. CH4(g) + 2 Cl2(g) CCl4(g) + 2 H2(g)

∆H = [(-106.7) + 0] – [(-74.8)+0]

= -106.7 + 74.8

= -31.9 kJ/mol

2. 2 KCl(s) + 3 O2(g) 2KClO3(s)

(-435.9) (0) (-391.2)

∆H = [(2)(-391.2)] – [(2)(-435.9) + (3)(0)]

= -782.4 + 871.8

= 89.4 kJ/mol

2 3 2

2 2

ΔH°f Example Problems

ΔHrxn = ?

ΔHrxn = ?

(-124.4) (-407.1) (-127.0) (-446.2)

3. AgNO3(s) + NaCl (aq) AgCl(s) + NaNO3(aq)

∆H = [(-127.0) + (-446.2)] – [(-124.4) + (-407.1)]

= -573.2 + 531.5

= - 41.7 kJ/mol

4. C2H5OH(l) + 7/2 O2(g) 2CO2(g) + 3H2O(g)

(-277.7) (0) (-393.5)

∆H = [(2)(-393.5) + (3)(-241.8)] – [(-277.7) + (7/2)(0)]

= -1512.4 + 277.7

= -1234.7 kJ/mol

(3)(7/2) (2) (-241.8)

∆Hrxn = Σ n∆Hof Products - Σn∆Ho

f Reactants

ΔH°f Example Problems

ΔHrxn = ?

ΔHrxn = ?

MATTER IS ENERGY.

ENERGY IS INFORMATION.

EVERYTHING IS INFORMATION.

PHYSICS SAYS THAT

STRUCTURES...

BUILDINGS, SOCIETIES,

IDEOLOGIES... WILL SEEK

THEIR POINT OF LEAST

ENERGY.

THIS MEANS THAT

THINGS FALL.

THEY FALL FROM HEIGHTS

OF ENERGY AND

STRUCTURED INFORMATION

INTO MEANINGLESS,

POWERLESS DISORDER.

THIS IS CALLED

ENTROPY.

Entropy (S)= a measure of randomness or disorder

Entropy: Tendency toward disorder

Entropy: Tendency toward disorder

• In any spontaneous process, the entropy of the universe increases or is +

ΔSuniverse > 0

Second Law of Thermodynamics

occurs without outside intervention

ΔSuniverse = ΔSsystem + ΔSsurroundings

Entropy of the Universe

ΔSuniverse = ΔSsystem + ΔSsurroundings

Positional disorder Energetic disorder

ΔSuniverse > 0 spontaneous process

Both ΔSsys and ΔSsurr positive

Both ΔSsys and ΔSsurr negative

ΔSsys negative, ΔSsurr positive

ΔSsys positive, ΔSsurr negative

spontaneous process

nonspontaneous process

depends

depends

ΔSsys: Positional Disorder and Probability

Probability of 1 particle in left bulb = ½

" 2 particles both in left bulb = (½)(½) = ¼

" 3 particles all in left bulb = (½)(½)(½) = 1/8

" 4 " all " = (½) 4 = 1/16

" 10 " all " = (½)10 = 1/1024

" 20 " all " = (½)20 = 1/1048576

" a mole of " all " = (½)6.02 x1023

The arrangement with the greatest entropy is the one with the highest probability (most “spread out”).

Ssolid < Sliquid <<<< Sgas

Entropy of the System: Positional Disorder

Ludwig Boltzmann Orderedstate

Disorderedstate

Low probability(few ways)

High probability(many ways)

Low S

High S

• Ssystem is proportional to positional disorder

• S increases with increasing # of possible positions

Ludwig Boltzmann

Entropy of the Surroundings(Energetic Disorder)

SystemHeat Entropy

Surroundings

SystemHeat Entropy

Surroundings

T

ΔHΔS sys

surr Low T → large entropy change (surroundings)

High T → small entropy change (surroundings)

ΔHsys < 0

ΔHsys > 0

ΔSsurr > 0

ΔSsurr < 0

The Third Law:The entropy of a perfect

crystal at 0 K is zero

• Everything locked into place• No molecular motion whatsoever

The Third Law of Thermodynamics

Crystallization of Water into Ice

Entropy Curve

Solid GasLiquid

S(J/K)

Temperature (K)0

0

Δfus (s ↔ l)

ΔHvap (l ↔ g)

S° (absolute entropy) can be calculated for any substance

Entropy Increases with...• Melting (fusion) Sliquid > Ssolid

ΔHfus/Tfus = ΔSfus

• Vaporization Sgas > Sliquid

ΔHvap/Tvap = ΔSvap

• Increasing ngas in a reaction

• Heating ST2 > ST1 if T2 > T1

• Dissolving (usually) Ssolution > (Ssolvent + Ssolute)

• Molecular complexity more bonds, more entropy

• Atomic complexity more e-, protons, neutrons

Entropy Practice

For the above reaction, predict the sign of ΔSsys

We can predict S values based on phases…

Remember that Ssolid< Sliquid <<<<<< Sgas and that

ΔSsys = Sproducts – Sreactants

ΔSsys. = S (12 mol gas) – S (6 mol gas + 1 mol solid)

Solid is negligible, more gas products, ↑ disorder

Therefore…ΔSsys = +

C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(g)

Entropy Practice

For the above reaction, predict the sign of ΔSsys

We can predict S values based on phases…

Remember that Ssolid< Sliquid <<<<<<< Sgas

ΔSsys = Sproducts – Sreactants

ΔSsys = S (2 mol solid) – S (4 mol solid + 3 mol gas)

Solid is negligible, gas R → solid P , ↓ disorder

Therefore…ΔSsys = -

4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)

Predicting ΔSsys sign summary

• Use relative S values for phases

Ssolid< Saqueous< Sliquid < Sgas

• Gases always have greater entropy

• Consider number of moles, especially with gases

• You cannot predict for some reactions

H2(g) + Cl2(g) → 2HCl(g)

2 mol gas → 2 mol gas

Unable to predict sign ΔSsys for this reaction

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

CompoundC6H12O6(s)

O2(g)

CO2(g)

H2O(g)

S° (J/mol K)212205214189

ΔS°sys = ΣnS°(products) - ΣnS°(reactants)

= [6 S°(CO2(g)) + 6 S°(H2O(g))] – [S°(C6H12O6(s)) + 6 S°(O2(g))]

= [6(214) + 6(189)] – [(212) + 6(205)] J/K

ΔS°sys = 976 J/K

Calculating Entropy Quantitatively

Calculate the value of ΔS°sys:

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(g)

CompoundC6H12O6(s)

O2(g)

CO2(g)

H2O(g)

ΔH°f (kJ/mol)-12750-393.5-242

S° (J/mol K)212205214189

ΔH°rxn = ΣnΔH°f (products) - ΣnΔH°f(reactants)

= [6 ΔH°f(CO2(g)) + 6 ΔH°f(H2O(g))] – [ΔH°f(C6H12O6(s)) + 6 ΔH°f(O2(g))]

= [6(-393.5) + 6(-242)] – [(-1275) + 6(0)] kJ

ΔH°rxn = -2538 kJ

Is this reaction spontaneous at 298K?

ΔSuniverse = ΔSsys + ΔSsurr ΔSsurr = - ΔH/Tand

ΔSsys = 976 J/K from last problem

ΔSuniverse = 0.976 kJ/K + -(-2538 kJ)/298K = 9.49 kJ/K

YES!

Entropy (S) Review

• ΔSuniverse > 0 for spontaneous processes

• ΔSuniverse = ΔSsystem + ΔSsurroundings

positional energetic

• We can find the absolute entropy value for a substance

• S° values for elements & compounds in their standard states are tabulated (Thermodynamic Appendix)

• For any chemical reaction, we can calculate ΔS°rxn:

ΔS°rxn = ΣnS°(products) - ΣnS°(reactants)

Recap: Characteristics of Entropy

• S is a state function (we can use final – initial)

• S is extensive (more stuff, more entropy)

• At 0 K, S = 0 (we can determine absolute entropy)

• S > 0 for elements and compounds in their standard states

• Raise T increase S

• Increase ngas increase S

• More complex systems larger S

–ΔG means +ΔSuniv

A process (at constant T, P) is spontaneous if free energy decreases

Gibbs Free Energy (G)

Josiah Gibbs

G = H – TSAt constant temperature,

ΔG = ΔH – TΔS

(system’s point of view)

ΔG = ΔH – TΔS

Divide both sides by –T

-ΔG/T = -ΔH/T + ΔS

ΔSuniverse = ΔS – ΔH/T

ΔG and Chemical Reactions

ΔG = ΔH – TΔS

• If ΔG < 0, the reaction is spontaneous

• If ΔG > 0, the reaction is not spontaneous

(The reverse reaction is spontaneous)

• If ΔG = 0, the reaction is at equilibrium

• Neither the forward nor the reverse reaction is favored

• Both reactions are occurring simultaneously and at equal rates

• ΔG is an extensive, state function

Depends on how much stuff Depends on final and initial states only

Ba(OH)2(s) + 2NH4Cl(s) BaCl2(s) + 2NH3(g) + 2 H2O(l)

ΔH°rxn = 50.0 kJ (per mole Ba(OH)2)

ΔS°rxn = 328 J/K (per mole Ba(OH)2)

ΔG = ΔH - TΔS

ΔG° = 50.0 kJ – 298 K(0.328 kJ/K)

ΔG° = – 47.7 kJ

At what T does the reaction stop being spontaneous?

The T where ΔG = 0

ΔG = 0 = 50.0 kJ – T(0.328 kJ/K)

50.0 kJ = T(0.328 kJ/K)

T = 152 K not spontaneous below 152 K

Spontaneous

Is the following reaction spontaneous at 298 K?

Effect of ΔH and ΔS on Spontaneity

ΔH

–

+

–

+

ΔS

+

+

–

–

Spontaneous?

Spontaneous at all temps

Spontaneous at high temps• Reverse reaction spontaneous at low temps

Spontaneous at low temps• Reverse reaction spontaneous at high temps

Not spontaneous at any temp

ΔG = ΔH – TΔSΔG (-) → spontaneous reaction

1. ΔG° = nΔG°f(products) - nΔG°f(reactants)

• ΔG°f = free energy change when forming 1 mole of

compound from elements in their standard states (see Thermodynamics Appendix for values)

2. ΔG° = ΔH° - TΔS°

3. ΔG° can be calculated by combining ΔG° values for several reactions

Using Hess’s Law! Your favorite!

Ways to Calculate ΔG°rxn

2 H2(g) + O2(g) 2 H2O(g)

1. ΔG° = ΔG°f(products) - ΔG°f(reactants)

ΔG°f(O2(g)) = 0

ΔG°f(H2(g)) = 0

ΔG°f(H2O(g)) = -229 kJ/mol

ΔG° = (2(-229 kJ) – 2(0) – 0) kJ = -458 kJ

2. ΔG° = ΔH° - TΔS°

ΔH° = -484 kJ

ΔS° = -89 J/K

ΔG° = -484 kJ – 298 K(-0.089 kJ/K) = -457 kJ

Calculate ΔG° for the following reaction:

2H2(g) + O2(g) 2 H2O(g)

3. ΔG° = combination of ΔG° from other reactions (using Hess’s Law)

2H2O(l) 2H2(g) + O2(g) ΔG°1 = 475 kJ

H2O(l) H2O(g) ΔG°2 = 8 kJ

ΔG° = - ΔG°1 + 2(ΔG°2)

ΔG° = -475 kJ + 16 kJ = -459 kJ

Method 1: -458 kJ

Method 2: -457 kJ

Method 3: -459 kJ

Method 1:

Method 2:

Method 3:

What is Free Energy, Really?

• NOT just “another form of energy”

• Free Energy is the energy available to do useful work

• If ΔG is negative, the system can do work (wmax = ΔG)

• If ΔG is positive, then ΔG is the work required to make the process happen

– Example: Photosynthesis

6 CO2 + 6 H2O C6H12O6 + 6 O2

ΔG = 2870 kJ/mol of glucose at 25°C

Thus, 2870 kJ of work is required to photosynthesize 1 mole of glucose