AP NOTES - Georgetown High School



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

• 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 = -P(V

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 L(atm

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 × 108 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

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!

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

[pic]

• 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 = m(Cp((T

q = quantity of heat ( Joules or calories) m = mass in grams ΔT = Tf - Ti (final – initial)

Cp = specific heat capacity ( J/g(C)

• Specific heat of water (liquid state) = 4.184 J/g(C ( 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.

[pic]

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

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 25.0(C is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25(C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1(C. 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 7.3(C. When a 1.15-g sample of hydrogen gas was burned with excess oxygen, the temperature increase was 14.3(C. Calculate the energy of combustion (per gram) for hydrogen and methane.

Methane = 55kJ/g

Hydrogen = 141 kJ/g

❖ 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: 25(C (298 K), 1 atm, 1M

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

Sample Problem D: Calculate the (H(rxn 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 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

□ 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

❖ 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!

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 Λ 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

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 25(C 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) 6 H2O (g) ignite & rxn is fast!

ΔSsystem = -88.9J/K

entropy declines (due mainly to 362 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

ΔSΕsurroundings = ΔHΕsurroundings = + 483.6 kJ = 1620 J/K

T 298 K

Now we can find ΔSΕuniverse

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: (G(rxn = ( (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. ΔGΕf = 0 for elements in standard state

b) GRAND Daddy: (G = (H - T(S

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/mol(K

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**

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

GRAND Daddy: (G = (H - T(S

Exercise 9 Calculating ∆H(, ∆S(, and ∆G(

Consider the reaction

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

carried out at 25(C 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 25(C)

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

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 25(C 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 25(C.

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!

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 ................
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