THERMOCHEMISTRY

CHAPTER 6

6.1- The Nature of Energy Ch. 6 Unit Essential Question:

What does the thermodynamics branch of chemistry study and how is it involved in chemical reactions?

Lesson Essential Question: What aspects must be considered when studying the

area of thermodynamics?

Temperature vs. heat Temperature: measure of average KE of particles. Heat: E transfer between two objects at different

temperatures. Flows from higher energy (hotter object) to lower

energy (cooler object).

State function: depends on the current state of the substance, not how the current state of the substance was reached. Ex. 1: The net distance from A to B is shown, and

does not change (does not depend on the path taken)- state function. But, the total distance travelled from A to B may change depending upon the path taken- path function.

A B

6.1- The Nature of Energy

State FunctionsAccording to this diagram, does it matter how sodium chloride is formed with respect to the net amount of heat energy involved (ΔH)?

No matter the path that’s taken to form a compound, the net amount of heat energy involved will always be the same for that compound- state function!

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions

Ex. 2

Thermodynamics Thermodynamics: study of heat E and its

transformations. Examines internal energy changes of systems- sum of

PE and KE. Use simplest system to examine internal energy: ideal

gas. No attraction between particles = no PE. Thus internal E is all KE, which we said in the last

chapter is directly related to T. We can say that the internal E of a gas increases

with T.Recall: E = 3/2RT

Bonds in reactants and products have PE. Stronger bonds = more PE.

Thermodynamics Recall system and surroundings.

System = object(s) being studied- for us reactions Surroundings = everything outside of the object(s) being

studied- for us reaction container Signs always reflect what happens to the system!

Heat involved is given in terms of the system (reaction). Positive (+q) = system absorbs heat (takes in E). Negative (-q) = system releases heat (lets E out).

Same notation applies for work. +w = surroundings do work on system (system ‘takes in’

work). -w = system does work on surroundings (system ‘puts

out’ work).

Thermodynamics First Law of Thermodynamics: The E of the

universe is constant (law of conservation of E). ΔEuniverse = ΔEsystem + ΔEsurroundings = 0

KE + PE of all particles in a system = internal E of a system. Internal E can be changed by doing work, heat

flow, or both: ΔE = q + w (take signs into account!)

Thermodynamics & Work Work specification: V of system (gas!) can expand

during a reaction- the system is doing work on the surroundings. w can be replaced with –PΔV

Expansion: ΔV is positive, w is negative: gas (system) does work.

Compression: ΔV is negative, w is positive: gas (system) has work done on it.

ΔE = q – PΔV To convert units: -PΔV will give you Latm, but q will be

in J so you need both units to be in J. 101.3J = 1Latm use as a conversion factor!

AP Practice QuestionThe average ________ is the same for any ideal gas at a given temperature.

a)Free energyb)Lattice energyc)Kinetic energyd)Activation energy

AP Practice Question- ReviewWhat is the energy released when the gaseous ions combine to form an ionic solid?

a)Free energyb)Lattice energyc)Kinetic energyd)Activation energy

AP Practice QuestionWhen ammonium chloride dissolves in water, the temperature drops. Which of the following conclusions may be related to this?

a)Ammonium chloride is more soluble in hot water.b)Ammonium chloride produces an ideal solution.c)The heat of solution for ammonium chloride is exothermic.d)Ammonium chloride has a low lattice energy.

*Ideal solution = follows Raoult’s Law (assumptions are made for liquids like there were for gases)

6.1- The Nature of Energy HW: pg. 267 #6,10, 21, 24

Lesson Essential Questions What is enthalpy and how does it relate to

chemical reactions? How is calorimetry performed and what does

it study?

6.2- Enthalpy & Calorimetry Enthalpy is represented by H: heat involved in a reaction.

ΔH = q at constant pressure; the change in enthalpy of a system is equal to the E flow as heat.

For chemical reactions: ΔH = Hproducts – Hreactants Enthalpy of reaction shows the change in enthalpy

from reactants to products. This is the E flow as heat, what you feel as hot or cold!

Enthalpy and stoichiometry: ΔH for a reaction can be used to develop conversion factors. The given ∆H is often the enthalpy of reaction for one

mole of a substance involved in the reaction. But it could be for another amount (such as per gram)!

6.2- Enthalpy & Calorimetry Ex: When 1mol of CH4 burns at constant P, 890kJ

of E is released as heat. What is ΔH if 5.8g of CH4 is burned at constant P? *If not told whether P is constant or not in a problem,

assume P is constant.* Determine necessary conversion factor(s):

890kJ/molCH4

Convert to moles and solve; remember signs!!: 5.8gCH4 x (1molCH4/16.05gCH4) x (890kJ/molCH4) =

-320kJ Negative because you’re told E is released.

Calorimetry Lab technique to measure heat involved in a

chemical or physical change. Heat capacities: Cp = q/ΔT (heat capacity)

No specified amount. c = q/mΔT (specific heat capacity)

For 1 gram of a substance. C = q/nΔT (molar heat capacity)

For 1 mole of a substance. Two types of calorimeters: coffee cup and bomb.

Calorimetry at Constant Pressure Coffee cup calorimetry. At constant P: ΔH = c x m x ΔT = q

This is the formula for specific heat; now you’re solving for q since ΔH = q at constant P.

May need concentration, D, etc. to find mass in grams. Remember- AP questions like to combine various

topics/knowledge/calculations!

Calorimetry at Constant Pressure Example: Solid BaSO4 forms when 1.00L of 1.00M Ba(NO3)2

is mixed with 1.00L of 1.00M Na2SO4, and the temperature increases from 25.0°C to 28.1°C. If the amount of heat absorbed by the calorimeter is negligible, the specific heat of the mixed solutions is 4.18J/g°C, and the density of the mixed solutions is 1.0g/mL, what is the enthalpy change per mole of BaSO4 formed?

Answer: 26,000J/mol BaSO4

**Note stoichiometric relationships that can be made from reactions:

4Fe (s) + 3O2 (g) 2Fe2O3 (s) ΔH = -1652kJ

(1) 4mol Fe/1652kJ (3) 2molFe2O3/1652kJ

(2) 3mol O2/1652kJ

Calorimetry at Constant Volume Bomb calorimeter is used.

Reactants placed in steel container (constant V) and ignited.

Energy change determined by measuring T increase of water and other calorimeter parts.

ΔH = ΔT ccalorimeter + cwater mwater ΔT Need to account for all heat gained- by the water and the

calorimeter!

AP Practice QuestionA 1.5886g sample of glucose (C6H12O6) is ignited in a bomb calorimeter. The temperature increased by 3.682°C. The heat capacity of the calorimeter was 3.562kJ/°C and the calorimeter had 1.000kg of water inside. Find the molar heat of the reaction:

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

Answer: -3,234kJ/mol

6.2- Enthalpy & Calorimetry HW: pg. 267 #: 32, 34, 36, 42(a), 52, 54

Lesson Essential Question How is Hess’ Law used to determine the

change in enthalpy of a chemical reaction?

6.3- Hess’ Law Hess’ Law: for a chemical equation that can be

written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps. Add together ΔH’s of each reaction involved. This works because H is a state function! Ex: making nitrogen dioxide. N2 + 2O2 2NO2 ΔH1 = 68kJ

VS. N2 + O2 2NO ΔH2 = 180kJ

2NO + O2 2NO2 ΔH3 = -112kJ

N2 + 2O2 2NO2 ΔH = 180kJ – 112kJ = 68kJ

+

Add reactions together to obtain same reaction above; add ΔH together to get net ΔH.

Using Hess’ Law Two important characteristics about ΔH:

1. If a reaction is reversed, ΔH ’s sign is reversed. Ex: C(s) + O2(g) CO2(g) ΔH = -394kJ

CO2(g) C(s) + O2(g) ΔH = 394kJ 2. Size of ΔH is directly proportional to amounts of

reactants and products. If a reaction is multiplied by some integer, ΔH is also multiplied by the same integer. Ex: [C(s) + O2(g) CO2(g)]x2 ΔH = (-394kJ)x2

2C(s) + 2O2(g) 2CO2(g) ΔH = -788kJ Watch for ∆H given as kJ/mol of a substance- if the

balanced reaction has more than 1mol of the substance, multiply H by the appropriate coefficient!

Ex: Combustion of C4H4 has ∆H = -585.2kJ/mol CO2.

C4H4(g) + 5O2(g) 4CO2(g) + 2H2O(g) ∆H = -2341kJ

AP Practice QuestionGiven the following information:

C (s) + O2 (g) CO2 (g) ΔH = -393.5kJ

H2 (g) + 1/2O2 (g) H2O ΔH = -285.8kJ

C2H2 (g) + 5/2 O2 2CO2 (g) + H2O (l) ΔH = -1299.8kJ

Find the enthalpy change for: 2C (s) + H2 (g) C2H2 (g)

454.0kJ-227.0kJ0.0kJ227.0kJ

6.3- Hess’ Law HW: pg. 268 #: 57, 60, 62

AP Practice QuestionDetermine ΔH for the reaction if CH3OH (l) were formed instead of CH3OH (g). The ΔH of vaporization for CH3OH is 37 kJ/mol.

CO (g) + 2H2 (g) CH3OH (g) ΔH = -91kJ

-128kJ-54kJ128kJ54kJ

Lesson Essential Question How can standard enthalpies of formation be

used to calculate enthalpy changes in reactions?

6.4- Standard Enthalpies of Formation Standard enthalpy of formation: enthalpy change when

one mole of a substance is formed from its elements when all substances are in their standard states. Represented by ∆H°f. Degree sign indicates standard

conditions, f indicates ‘formation’. Standard states for compounds:

Gases- at a pressure of 1atm. Solutions- concentration of 1M. Pure substances (elements & compounds)- most

stable form (how they’re typically found) at 1atm & 25°C. Ex: oxygen- O2 (g), mercury- Hg (l), sodium

chloride- NaCl (s)

Standard Enthalpies of Formation Always given per mole of substance in its standard

state; formed from elements in standard states: Ex: C(s) + 2H2(g) + ½O2(g) CH3OH(l) The enthalpy of formation of methanol is -239kJ/mol.

Enthalpy of formation for an element in its standard state is set to be zero. The element is found this way; no energy is involved to

put it into this form! Besides Hess’ Law, Enthalpy changes for reactions

can also be calculated by subtracting the enthalpies of formation of reactants from the enthalpies of formation of products:

∆H°rxn = ∑n∆H°products – ∑n∆H°reactants Note coefficients/moles (n) are taken into account!

∆H° = -239kJ/molf

f f

AP Practice QuestionThe decomposition of CaCO3 is shown in the equation below. Using the data below, which of the following values is closest to the ∆Hrxn of the decomposition of CaCO3?

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

∆H°f (kJ/mol): -1,207.1 -635.5 -393.5

-2,240 kJ/mol-180 kJ/mol180 kJ/mol1,207 kJ/mol2,240 kJ/mol

Bond Energies & Enthalpy Heats of reaction can also be calculated using

bond energies. Similar formula is used as seen with standard

enthalpy of formations:∆H°rxn = ∑nbonds broken– ∑nbonds formed

Bonds broken are the reactants, and bonds formed are the products.

Take into account ALL bonds broken and formed- in other words multiply each bond type by how many there are. Drawing out molecules can help! Watch out for single, double, and triple bonds!

Ex: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

∆H°rxn = [4(C-H) + 2(O-O)] – [2(C=O) + 4(H-O)]

AP Practice Question

Use the values of average bond energies to find the enthalpy change for the following reaction:

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

Average bond energies (in kJ/mol): H-H: 436, O=O: 499, and H-O: 464.

a)0kJb)485kJc)-485kJd)464kJ

Note: the actual enthalpy of reaction for this reaction is -572kJ. Why such a big difference?

Bond Energies & Enthalpy Cont. Note that the same bond in different molecules

impacts bond energy too. Thus ∆H°rxn is approximate here.

Ex: O-H bond in H2O does not have the exact same E as the O-H bond in CH3OH.

Note also that using bond energies to calculate enthalpies of reaction are only sufficient if all species are in the gaseous state!

Why might this be? Only in the gaseous state can we assume interactions

between particles are negligible (ideal gases), and so only bond energies need to be considered for energy changes.

In other states the interactions between particles are not negligible and cannot be ignored since these too contribute to energy changes.

6.4- Standard Enthalpies of Formation HW: pg. 269 #: 65, 67

Three Laws of Thermodynamics Three laws of thermodynamics:

1. Total E of the universe is constant (law of conservation!):

ΔEuniverise = ΔEsystem + ΔEsurroundings = 0 2. All processes that occur spontaneously move in

the direction of an increase in entropy of the universe (system + surroundings): ΔSuniverse = ΔSsystem + ΔSsurroundings > 0

3. The entropy of a pure, perfect crystal at 0K is zero. Note: the third law has more to it, but this is all

you need to take away. There is also a zeroth law that we will not cover.

We’ve already investigated #1, let’s look at #2!

Entropy Changes List all of the examples you can think of that

would result in an increase in entropy.

1)Entropy increases when the number of molecules increases from reactants to products.

2)Entropy increases when T increases.3)Entropy increases when a gas is formed from

a liquid or a solid.4)Entropy increases when a liquid is formed

from a solid.

AP Practice QuestionChoose the reaction expected to have the greatest increase in entropy.

a)H2O (g) H2O (l)

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

c)Ca (s) + H2 (g) CaH2 (s)

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

AP Practice QuestionWhich of the following combinations is true when sodium chloride melts?

a)ΔH > 0 and ΔS > 0b)ΔH = 0 and ΔS > 0c)ΔH > 0 and ΔS < 0d)ΔH < 0 and ΔS < 0

AP Practice QuestionWhich of the following reactions would be accompanied by the greatest decrease in entropy?

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

b)C (s) + O2 (g) CO2 (g)

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

d)2Na (s) + Cl2 (g) 2NaCl (s)

Note: according to the practice book the reaction that produces the most gas will have the greatest increase in S and the one losing the most gas will have the greatest decrease in S.

Standard Molar Entropies Standard molar entropies: entropy associated

with 1mol of a substance in its standard state. Represented by S°. The change in entropy of

a reaction can be calculated like ΔH° was: ∆S° = ∑nS° products – ∑nS° reactants

Standard states are the same as previously discussed.

BUT entropies of elements in standard states are not zero! Everything will have entropy values!

AP Practice QuestionCalculate ΔS° for the following given the following S° values.

a. H2 (g) + ½ O2 (g) H2O (g) *Do you expect a positive or negative ΔS° value? Why?S° (J/molK): H2: 131.0 O2: 205.0 H2O: 188.7

ΔS° = -44.8J/molK

b. CaCO3 (s) + H2SO4 (l) CaSO4 (s) + H2O (g) + CO2 (g)

*Do you expect a positive or negative ΔS° value? Why?S° (J/molK): CaCO3: 92.9 H2SO4: 157 CaSO4: 107

H2O: 188.7 CO2 213.6

ΔS° = 259J/molK

Gibbs Free Energy Are ΔH and ΔS the only thermodynamic quantities

that determine if a reaction will occur? What if one is favorable and the other is not? Recall favorable signs for each: -ΔH and +ΔS.

What do you recall about ΔG? It combines ΔH and ΔS, and determines whether

or not a reaction will occur or not. Favorable sign: -ΔG. In terms of spontaneity: -ΔG = spontaneous; +ΔG = nonspontaneous

(energy needs to be added for the reaction to occur); ΔG = 0 the reaction is at equilibrium.

AP Practice QuestionA sample of Ga metal is sealed inside a well-insulated, rigid container. The T inside the container is at the melting point of the Ga metal. What can be said about the E and the S of the system after equilibrium has been established? Assume the insulation prevents any E change with the surroundings.

a)ΔH > 0 and ΔS > 0b)ΔH = 0 and ΔS > 0c)ΔH > 0 and ΔS < 0d)ΔH < 0 and ΔS < 0

Recall ΔG Formula ΔG = ΔH -TΔS ΔG°rxn = ΔH°rxn -TΔS°rxn Remember- you can use these formulas to

calculate values of ΔG, but you can also use it with signs for ΔH and ΔS to see if a reaction will occur and the effect T has on the reaction.

Ex: A salt is dissolved in a beaker of water and the temperature drops by 4.3°C. What are the signs of ΔH, ΔS, and ΔG for this reaction?

+ΔH, +ΔS, -ΔG

AP Practice QuestionA certain reaction is nonspontaneous under standard conditions, but becomes spontaneous at higher temperatures. What conclusions may be drawn under standard conditions?

a)ΔH < 0, ΔS > 0, ΔG > 0b)ΔH > 0, ΔS < 0, ΔG > 0c)ΔH > 0, ΔS > 0, ΔG > 0d)ΔH < 0, ΔS < 0, ΔG > 0

Gibbs Free Energies of Formation Just like with ΔH and S, ΔG also has energies of

formation: ΔG°f

ΔG°f for elements in their standard states are zero, just like they were for ΔH°f.

Same equation is used for ΔG°rxn as was used for ΔH°rxn:

∆G°rxn = ∑n∆G°f products – ∑n∆G°f reactants This is another way to calculate ΔG, in addition

to using the formula previously discussed.

AP Practice QuestionCalculate ΔG° for the following reaction.

2NH4Cl (s) + CaO (s) CaCl2 (s) + H2O (l) + 2NH3 (g)

ΔG°f (kJ/mol) : -203.9 -604.2 -750.2 -237.2 -16.6

Answer: -8.6kJ/mol

Ch. 17 Homework Pg. 783 # 8, 20, 24, 25, 27(a&b), 28(a), 33,

36, 37(a), 43, 45(a), 52

Note: problems in red are technically work for this information, but due to the large quantity these will be your class work for Friday when I’m out.

Thermodynamics & Equilibrium What if substances in a reaction are not under

standard conditions? Pressures and concentrations are no longer equal to one.

The new conditions must be taken into account:ΔG = ΔG° + RTlnQ

Q is the reaction quotient; used to indicate where a reaction is in terms of reaching equilibrium (we will do more with this later):

Q = [products]x/[reactants]y

Note: x and y superscripts come from coefficients! Ex: 2NO (g) + O2 (g) 2NO2 (g)

Q = [NO2]2/([NO]2[O2])

Example Calculate ΔG for the following reaction at 500. K.

2NO (g) + O2 (g) 2NO2 (g)

The concentrations of the species are as follows: NO: 2.00M, O2: 0.500M, and NO2: 1.00M. First find ΔG° using ΔGf° values and the equation

used before: ∆G°= ∑n∆G°f products – ∑n∆G°f reactants

ΔGf° values (kJ/mol): NO: 86.71, O2: 0.000, NO2: 51.84 ∆G° = -69.74 kJ/mol Now use the previous formula to find ΔG.

Note: be sure all units are kJ or J! ΔG = -7.262 x 104 J/mol

Example #2- Gases & Pressures Calculate ΔG at 25°C for the reaction where CO

gas at 5.0atm and H2 gas at 3.0atm form liquid methanol (CH3OH).

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

ΔGf° values (kJ/mol): CH3OH: -166, CO: -137, H2: 0 ΔG = ΔG° + RTlnQ Notes: (1) Reaction quotients are written the

same way for pressures as for concentrations. (2) Pure solids and liquids have a value of 1. Q = 1/(5.0 x 3.02) = 0.022 ΔG = -38kJ/mol, or -38,000 J/mol

Thermodynamics & Equilibrium Only use the previous equation if the reaction is

not under standard conditions and if you have not been told the reaction is at equilibrium.

If the reaction is not under standard conditions but IS at equilibrium:

ΔG° = - RTlnK K is the equilibrium constant for the reaction and

can only be used if the reaction is at equilibrium (again, we’ll do more with this later).

Recall that at equilibrium ΔG = 0, so it no longer appears in the equation used above.

Ex: Find ΔG° for 2O3 (g) 3O2 (g) Kp = 4.17 x 1014

ΔG° = -8.34 x 104 J/mol

A few more things about K & Q: Q is used when you are not told the reaction is at

equilibrium; if it is, use K. Recall that Q = [products]x/[reactants]y

Also true for K: Keq = [products]x/[reactants]y

K >1 means the reaction favors the products; spontaneous in the direction of the products, mostly all products present.

K < 1 means the reaction favors the reactants; spontaneous in the direction of the reactants (essentially nonspontaneous), mostly all reactants present.

K = 1 means the reaction is at equilibrium; both reactants and products are present.

Thermodynamics & Equilibrium

AP Practice Question Under standard conditions, calcium metal reacts readily with chlorine gas. What conclusions may be drawn from the fact?

a) Keq <1 and ΔG° > 0

b) Keq >1 and ΔG° = 0

c) Keq <1 and ΔG° < 0

d) Keq >1 and ΔG° < 0

AP Practice Question What is the minimum energy required to force a nonspontaneous reaction to occur?

a) free energyb) lattice energyc) kinetic energyd) activation energy

AP Practice Question Which of the following is the minimum energy required to initiate a reaction?

a) free energyb) lattice energyc) kinetic energyd) activation energy

Homework Pg. 786 #65