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THERMOCHEMISTRY

Terms for you to learn that will make this unit understandable:

Energy (E) – the ability to do work or produce heat ; the sum of all potential and kinetic energy in a

system is known as the internal energy of the system

Potential energy – in chemistry this is usually the energy stored in bonds (i.e., when

gasoline burns there are differences in the attractive forces between the nuclei and the

electrons in the reactants and the products)

Kinetic energy – energy of motion, usually of particles, proportional to Kelvin

temperature; kinetic energy depends on the mass and the velocity of the object: KE = ½

mv2

Law of Conservation of Energy – energy never created nor destroyed

AKA energy of the universe is constant

AKA First Law of Thermodynamics

Heat (q) – transfer of energy in a process (flows from a warmer object to a cooler one – heat transfers

because of temperature difference but, remember, temperature is not a measure of energy—it just

reflects the motion of particles)

Enthalpy (H)– heat content at constant pressure

o Enthalpy of reaction (Hrxn) – heat absorbed or released by a chemical reaction

o Enthalpy of combustion (Hcomb) -- heat absorbed or released by burning (usually with O2)

o Enthalpy of formation (Hf) – heat absorbed or released when ONE mole of compound is

formed from elements in their standard states

o Enthalpy of fusion (Hfus) -- heat absorbed to melt 1 mole of solid to liquid @MP

o Enthalpy of vaporization (Hvap) -- heat absorbed to change 1 mole liquid to gas @BP

System – area of the universe we are focusing on (i.e., the experiment)

Surroundings – everything outside of the system

Endothermic – net absorption of energy (heat) by the system; energy is a reactant; (i.e., baking soda

and vinegar when mixed get very cold to the touch)

Exothermic – net release of energy (heat) by the system; energy is a product; (i.e., burning methane gas

in the lab burner produces heat; light sticks give off light which is also energy)

State Function – A property independent of past or future behavior ; (it does not matter which road

brought you to school today–you started at your house and ended here –there are probably lots of ways

for that to happen)

Entropy (S) – measure of disorder in the system (measure of chaos)

Gibb’s Free Energy (G)– criteria for sponteneity and amount of free energy to do work

Thermodynamics – study of energy and its interconversions

Work – force acting over distance

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ENERGY AND WORK

See definition of energy.

E = q(heat) + w(work) Signs of q

+q if heat absorbed

–q if heat released

Signs of w (commonly related to work done by or to gases)

+ w if work done on the system (i.e., compression)

-w if work done by the system (i.e., expansion)

When related to gases, work is a function of pressure

(pressure is force per unit of area) and volume

w = -PV

NOTE: Energy is a state function. (Work and heat are not.)

Exercise 1 Internal Energy

Calculate ∆E for a system undergoing an endothermic process in which 15.6 kJ of heat flows and where 1.4 kJ

of work is done on the system.

= 17.0 kJ

Exercise 2 PV Work

Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of

15 atm.

= -270 Latm

Exercise 3 Internal Energy, Heat, and Work

A balloon is being inflated to its full extent by heating the air inside it. In the final stages of this process, the

volume of the balloon changes from 4.00 × 106

L to 4.50 × 106 L by the addition of 1.3 × 10

8 J of energy as

heat. Assuming that the balloon expands against a constant pressure of 1.0 atm, calculate ∆E for the process.

(To convert between L atm and J, use 1 L atm = 101.3 J.)

= 8 × 107

J

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ENTHALPY

Measure only the change in enthalpy, H ( the difference between the potential energies of the products

and the reactants)

H is a state function

H = q at constant pressure (i.e. atmospheric

pressure)

(true most of the time for us and a very handy

fact!)

Enthalpy can be calculated from several sources

including:

Stoichiometry

Calorimetry

From tables of standard values

Hess’s Law

Bond energies

Stoichiometrically:

Calorimetry:

The process of measuring heat based on observing the temperature change when a body absorbs

or discharges energy as heat.

Types of calorimetry:

(1) Coffee-cup calorimetry – in the lab this is how we experiment to find energy of a

particular system. We use a Styrofoam cup, reactants that begin at the same

temperature and look for change in temperature. After all data is collected (mass or

volume; initial and final temperatures) we can use the specific formula to find the

energy released or absorbed. We refer to this process as constant pressure

calorimetry. ** q = H @ these conditions.**

(2) Bomb calorimetry – weighed reactants are placed inside a steel container and

ignited. Often referred to as constant volume. This is used by industry to

determine number of food calories that we consume!

Sample Problem A: Upon adding solid potassium hydroxide pellets to water the following reaction takes

place:

KOH (s) KOH(aq) + 43 kJ

Answer the following questions regarding the addition of 14.0 g of KOH to water:

a) Does the beaker get warmer or colder?

b) Is the reaction endothermic or exothermic?

c) What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

Answers: (a) warmer (b) exothermic (c) -10.7 kJ

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Terms to know:

Heat capacity – energy required to raise temp. by 1 degree (Joules/ C)

Specific heat capacity (Cp) – same as above but specific to 1 gram of substance

change) re temperatuof (degrees material) of (

ferredheat trans ofquantity heat specific

g

Molar heat capacity -- same as above but specific to one mole of substance

(J/mol K or J/mol C )

Energy (q) released or gained -- q = mCpT

q = quantity of heat ( Joules or calories)

m = mass in grams

ΔT = Tf - Ti (final – initial)

Cp = specific heat capacity ( J/gC)

Specific heat of water (liquid state) = 4.184 J/gC ( or 1.00 cal/g C)

Water has one of the highest specific heats known! That is why the earth stays at such an

even temperature all year round! Cool huh?

Heat lost by substance = heat gained by water

(if this does not happen, calculate the heat capacity of the substance)

Units of Energy:

calorie--amount of heat needed to raise the temp. of 1.00 gram of water 1.00 C

kilocalorie--duh!; the food calorie with a capital C.

Js

mkgmvKE 0.12/1

2

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joule--SI unit of energy; 1 cal = 4.184 J

Sample Problem B:

In a coffee cup calorimeter, 100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed. Both

solutions were originally at 24.6C. After the reaction, the final temperature is 31.3C. Assuming that all

solutions have a density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/gC, calculate the enthalpy

change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or the

calorimeter.

Answer: -5.6 kJ/mol

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Exercise 4 Enthalpy When 1 mole of methane (CH4) is burned at constant pressure, 890 kJ of energy is released as heat. Calculate

∆H for a process in which a 5.8-g sample of methane is burned at constant pressure.

∆H = heat flow = -320 kJ

Exercise 5 Constant-Pressure Calorimetry

When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25C in

a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1C. Assuming

that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is

4.18 J/C g, and that the density of the final solution is 1.0 g/mL, calculate the enthalpy change per mole of

BaSO4 formed.

= -26 kJ/mol

Exercise 6 Constant-Volume Calorimetry

It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural

gas (principally methane). To compare the energies of combustion of these fuels, the following experiment was

carried out using a bomb calorimeter with a heat capacity of 11.3 kJ/C. When a 1.50-g sample of methane gas

was burned with excess oxygen in the calorimeter, the temperature increased by 7.3C. When a 1.15-g sample

of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3C. Calculate the energy of

combustion (per gram) for hydrogen and methane.

Methane = 55kJ/g

Hydrogen = 141 kJ/g

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Tables:

Hf = enthalpy of formation.

= Production of ONE mole of compound FROM

ITS ELEMENTS in their standard states ()

= 0 for ELEMENTS in standard states

Standard States: 25C (298 K), 1 atm, 1M

Big Mamma Equation: Hrxn = Hf (products) - Hf (reactants)

Sample Problem D: Calculate the Hrxn for the following:

3 Al(s) + 3 NH4ClO4(s) Al2O3(s) + AlCl3(s) + 3 NO(g) + 6 H2O(g)

Given the following values:

Substance Hf (kJ/mol)

NH4ClO4(s) -295

Al2O3(s) -1676

AlCl3(s) -704

NO(g) 90.0

H2O(g) -242

Answer: -2680 kJ (exo)

Sample ProblemC:

Camphor (C10H16O) has a heat of combustion of 5903.6 kJ/mol. When a sample of camphor with mass of

0.1204 g is burned in a bomb calorimeter, the temperature increases by 2.28C. Calculate the heat capacity

of the calorimeter.

Answer 2.05 kJ/C

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Sample Problem E: Sometimes all values are not found in the table of thermodynamic data. For most

substances it is impossible to go into a lab and directly synthesize a compound from its free

elements. The heat of formation for the substance must be found by working backwards from its

heat of combustion.

Find the Hf of C6H12O6(s) from the following information:

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2800 kJ

Substance Hf (kJ/mol)

CO2(g) -393.5

H2O(l) -285.8

Answer: Hf =-1276 kJ/mol for glucose

Exercise 10

Using enthalpies of formation, calculate the standard change in enthalpy for the thermite reaction:

2A1(s)+Fe2O3(s) A12O3(s)+2Fe(s)

This reaction occurs when a mixture of powdered aluminum and iron (III) oxide is ignited with a magnesium

fuse.

= -850. kJ

Hess’s Law

Enthalpy is not dependent on the reaction pathway. If you can find a combination of chemical

equations that add up to give you the desired overall equation, you can also sum up the H’s for the

individual reactions to get the overall Hrxn.

Remember this:

First decide how to rearrange equations so reactants and products are on appropriate sides of

the arrows.

If equations had to be reversed, reverse the sign of H

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Sample Problem G:

Using bond energies, calculate the change in energy that accompanies the following reaction:

H2(g) + F2(g) 2 HF(g)

Bond Type Bond Energy

H-H 432 kJ/mol

F-F 154 kJ/mol

H-F 565 kJ/mol

Answer: -544 kJ

If equations had be multiplied to get a correct coefficient, multiply the H by this coefficient

since H’s are in kJ/MOLE (division applies similarly)

Check to ensure that everything cancels out to give you the exact equation you want.

Hint** It is often helpful to begin your work backwards from the answer that you want!

Sample Problem F: Given the following equations

H3BO3(aq) HBO2(aq) + H2O(l) Hrxn = -0.02 kJ

H2B4O7(aq) + H2O(l) 4 HBO2(aq) Hrxn = -11.3 kJ

H2B4O7(aq) 2 B2O3(s) + H2O(l) Hrxn = 17.5 kJ

find the H for this overall reaction

2 H3BO3(aq) B2O3(s) + 3 H2O(l)

Answer: 14.4 kJ endothermic

Bond Energies

Energy must be added/absorbed to BREAK bonds (endothermic). Energy is released when

bonds are FORMED (exothermic).

H = sum of the energies required to break old bonds (positive signs) plus the sum of the

energies released in the formation of new bonds (negative signs).

H = bonds broken – bonds formed

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SUMMARY FOR ENTHALPY: What does it really tell you about an equation?

H = + reaction is endothermic

H = - reaction is exothermic (favored – nature tends towards lower energy)

Spontaneity

Entropy and Free Energy

WHAT DRIVES A REACTION TO BE SPONTANEOUS?

(1) ENTHALPY (H) – heat content (exothermic reactions are generally favored)

(2) ENTROPY (S) – disorder of a system (more disorder is favored) Nature tends toward

chaos! Think about your room at the end of the week! Your mom will love this law.

Spontaneous reactions are those that occur without outside intervention. They may occur fast OR slow (that is

kinetics). Some reactions are very fast (like combustion of hydrogen) other reactions are very slow (like

graphite turning to diamond)

ENTROPY:

The second law of thermodynamics: the universe is constantly increasing disorder. Rudolph Clausius

(you=ll hear lots about him later when we study vapor pressures) Adiscovered@ it and gave it its symbol.)

The third law of thermodynamics: the entropy of a perfect crystal at 0K is zero.

[not a lot of perfect crystals out there so, entropy values are RARELY ever zero—even

elements]

So what? This means the absolute entropy of a substance can then be determined at any temp. higher

than 0 K. (Handy to know if you ever need to defend why G & H for elements = 0. . . . BUT S does

not!)

Predicting the entropy of a system based on physical evidence:

1) The greater the disorder or randomness in a system, the larger the entropy.

2) The entropy of a substance always increases as it changes from solid to liquid to

gas.

3) When a pure solid or liquid dissolves in a solvent, the entropy of the substance

increases (carbonates are an exception!--they interact with water and actually

bring MORE order to the system)

4) When a gas molecule escapes from a solvent, the entropy increases

5) Entropy generally increases with increasing molecular complexity (crystal

structure: KCl vs CaCl2) since there are more MOVING electrons!

6) Reactions increasing the number of

moles of particles often increase

entropy.

In general, the greater the number of

arrangements, the higher the entropy of the system!

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Exercise 2 Predicting Entropy Changes

Predict the sign of the entropy change for each of the following processes.

A: Solid sugar is added to water to form a solution.

B: Iodine vapor condenses on a cold surface to form crystals.

A: +∆S

B: -∆S

Calculating Entropy from tables of standard values: (Just the same as calculating the enthalpy earlier)

S is + when disorder increases (favored)

S is – when disorder decreases

ENTROPY CHANGES FOR REVERSIBLE PHASE CHANGES that=s a phase change at constant

temperature

ΔS = heat transferred = q

temperature at which change occurs T

**where the heat supplied (endothermic) (q > 0) or evolved (exothermic) (q < 0) is divided by the

temperature in Kelvins

** It is important here to note if the reaction is endothermic or exothermic. The actual significance of

this is really dependent on the temperature at which the process occurs.

(i.e., If you gave a millionaire $100 it would not make much difference in his happiness; if you gave a

poor college student $100 it would create a totally different expression of happiness!)

EX: water (l @ 100) water (g @ 100)

the entropy will increase

Taking favored conditions into consideration, the equation above rearranges into:

S = - H

T

Sample Problem H:

Which of the following has the largest increase in entropy?

a) CO2(s) CO2(g)

b) H2(g) + Cl2(g) 2 HCl(g)

c) KNO3(s) KNO3(l)

d) C(diamond) C(graphite)

Answer: a) the substance changes from a highly organized state to a more disorganized state.

BIG MAMMA, verse 2: Srxn = S (products) - S (reactants)

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Give signs to ΔH following exo/endo guidelines! (If reaction is exo.; entropy of system increases—

makes sense!)

Exercise 4 Determining ∆Ssurr

In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition

of the ore. For example, iron is used to reduce antimony in sulfide ores:

Sb2S3(s)+3Fe(s)2Sb(s)+3FeS(s) ∆H = -125kJ

Carbon is used as the reducing agent for oxide ores:

Sb4O6(s)+6C(s)4Sb(s)+6CO(g) ∆H = 778kJ

Calculate ∆Ssurr for each of these reactions at 25C and 1 atm.

= -2.61 × 103 J/K

SUMMARY ENTROPY:

S = + MORE DISORDER (FAVORED CONDITION)

S = - MORE ORDER

Whether a reaction will occur spontaneously may be determined by looking at the S of the universe.

ΔS system + ΔS surroundings = ΔS universe

IF ΔS universe is +, then reaction is spontaneous

IF ΔS universe is -, then reaction is NONspontaneous

Consider

2 H2 (g) + O2 (g) H2O (g) ignite & rxn is fast!

ΔSsystem = -88.9J/K

entropy declines (due mainly to 32 moles of gas!)

. . . to confirm we need to know entropy of surroundings

ΔSsurroundings = q surroundings this comes from ΔH calc.

T

Hsystem = - 483.6 kJ

First law of thermodynamics demands that this energy is transferred from the system to the surroundings so...

-ΔHsystem = ΔHsurroundings OR

- (-483.6 kJ) = +483.6 kJ

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ΔSsurroundings = ΔHsurroundings = + 483.6 kJ = 1620 J/K

T 298 K

Now we can find ΔSuniverse

ΔS system + ΔS surroundings = ΔS universe

(-88.9 J/K) + (1620 J/K) = 1530 J/K

Even though the entropy of the system declines, the entropy change for the surroundings is SOOO large that the

overall change for the universe is positive.

Bottom line: A process is spontaneous in spite of a negative entropy change as long as it is extremely

exothermic. Sufficient exothermicity offsets system ordering.

FREE ENERGY Calculation of Gibb’s free energy is what ultimately decides whether a reaction is spontaneous or not.

NEGATIVE G’s are spontaneous. G can be calculated one of several ways:

(a) Big Mamma,verse 3: Grxn = G (products) - G (reactants) This works the same way as enthalpy and entropy from tables of standard values! Standard molar free

energy of formation--same song, 3rd verse. ΔGf = 0 for elements in standard state

(b) GRAND Daddy: G = H - TS

This puts together all information thus far! By far, one of the most beneficial equations to

learn for AP exam!

(c) Hess’s law summation Works same as Hess’s in the enthalpy section—sum up equations using the guidelines as

before.

(d) G = G + RT ln (Q) Define terms: G = free energy not at standard conditions

G = free energy at standard conditions

R = universal gas constant 8.3145 J/molK

T = temp. in Kelvin

ln = natural log

Q = reaction quotient: (for gases this is the partial pressures of the

products divided by the partial pressures of the reactants—all raised to the power of their

coefficients) Q = [products]

[reactants]

(e) “RatLink”: G = -RTlnK

Terms: basically the same as above --- however, here the system is at equilibrium, so G

= 0 and K represents the equilibrium constant under standard conditions. K = [products]

still raised to power of coefficients [reactants]

(f) “nFe”: G = - nFE remember this!! Terms: G = just like above—standard free energy

n = number of moles of electrons transferred (look at ½ reactions)

F = Faraday’s constant 96,485 Coulombs/mole electrons

E = standard voltage ** one volt = joule/coulomb**

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BIG MAMMA, verse 3: Grxn = G (products) - G (reactants)

GRAND Daddy: G = H - TS

Exercise 9 Calculating ∆H, ∆S, and ∆G

Consider the reaction

2SO2(g)+O2(g)2SO3(g)

carried out at 25C and 1 atm. Calculate ∆H, ∆S, and ∆G using the following data:

∆H = -198 kJ

∆S = -187 J/K

∆G = -142 kJ

Hess’s law of summation

Exercise 10 Calculating ∆G

Using the following data (at 25C)

Cdiamond(s)+O2(g)CO2(g) ∆G= -397 kJ (16.5)

Cgraphite(s)+O2(g) CO2(g) ∆G= -394 kJ (16.6)

Calculate ∆G for the reaction

Cdiamond(s)Cgraphite(s)

= -3 kJ

Sample Problem C:

Find the free energy of formation for the oxidation of water to produce hydrogen peroxide.

2 H2O(l) + O2(g) 2 H2O2(l)

Given the following information:

Gof

H2O(l) -56.7 kcal/mol

O2(g) 0 kcal/mol

H2O2(l) -27.2 kcal/mol

Answer: 59.0 kcal/mol

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Exercise 13 Calculating ∆G

One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases:

CO(g)+2H2(g)CH3OH(l)

Calculate ∆G at 25C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are

converted to liquid methanol.

= -38 kJ/mol rxn

“RatLink”: G = -RTlnK

Exercise 15 Free Energy and Equilibrium II

The overall reaction for the corrosion (rusting) of iron by oxygen is

4 Fe (s) + 3 O2 (g) 2 Fe2O3 (s)

Using the following data, calculate the equilibrium constant for this reaction at 25C.

K = 10261

Gibb’s equation can also be used to calculate the phase change temperature of a substance. During the phase

change, there is an equilibrium between phases so the value of G is zero. Really just like what we did earlier

in this unit with enthalpy and entropy!

G = G + RT ln (Q)

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SUMMARY OF FREE ENERGY:

G = + NOT SPONTANEOUS

G = - SPONTANEOUS

Conditions of G: H S Result

negative positive spontaneous at all temperatures

positive positive spontaneous at high temperatures

negative negative spontaneous at low temperatures

positive negative not spontaneous, ever

Relationship to K and E : G K E

0 at equilibrium; K = 1 0

negative >1, products favored positive

positive <1, reactants favored negative

Consider the reaction: A (g) B (g)

a) The change in free energy to reach equilibrium, beginning with 1.0 mol of A(g) at PA = 2.0 atm

b) The change in free energy to reach equilibrium, beginning with 1.0 mol of B (g) at PB = 2.0 atm.

c) The free energy profile for A (g) B (g) in a system containing 1.0 mol (A plus B) at PTOTAL = 2.0 atm.

Each point on the curve corresponds to the total free energy of the system for a given combination of A and

B.

Sample Problem:

Find the thermodynamic boiling point of H2O(l) H2O(g)

Given the following information:

Hvap = +44 kJ Svap = 118.8 J/K

Answer: 370K