Calendar 13-14 M 1/27 Sec 10.1-3:Thermodym, Phase B-1 1/29 Sec 10.3: Calorim, Hf o, Hess Law B-1 1/31 Lab M 2/3 Sec 10.4 - Entropy and S B-1 2/5.

Calendar 13-14 M 1/27 Sec 10.1-3:Thermodym, Phase B-1 1/29 Sec 10.3: Calorim, Hfo, Hess

Law B-1 1/31 Lab M 2/3 Sec 10.4 - Entropy and S B-1 2/5 Quiz H, S Sec 11.3: G, Equil

temp calcs B-2 2/7 Review M 2/10 Test

Calendar 14-15 B-2 1/23-4 Sec 10.1-3:Thermodym Intro M 1/26 Sec 10.2-3/11.3: Calorim, Phase

Demolab w/ phase thermodynamics B-1 1/27-8 Sec 10.3 Calcs of H Hfo, Hess

Law B-2 1/29-30 Quiz, Rev of H calcs, M = B-1 2/2-3 Lab Hess Wed 2/4 Sec 10.4 - Entropy and S, G B-2 2/5-6 Quiz S, G: Review M 2/9 Test

Driving ForcesWhat causes chemical reactions to occur?

Nature drives things toward lower energy

Enthalpy = total energy content of a systemNature likes things to be random, or more disorganized

Entropy = disorder in a system

Entropy = S A measure of the disorder in a system Which has a greater entropy

A chess board before the 1st move or in the middle of a game

A jigsaw puzzle that is put together, or the same puzzle unassembled in its box

Humpty Dumpty before or after his tragedy

Nature tends to move spontaneously to a more disorganized state

ThermodynamicsStudy of energy, heat, and its flow

Energy = ability to do workPush or pull a mass

Heat = q = a form of energy Heat content is a measure of total Ek in a substance

Temp = meas of average movement (Ek)Temp and Ek are directly relation

Ek = ½mv2

Temp diff determines direction of heat flow

System vs. Surrounding System = reaction or process you

are examining Surroundings = rest of the universe

Energy is exchanged between the system and the surroundings

Endothermic vs Exothermic reactions Diagrams follow

Enthalpy = H

The total energy content of a system Sum Ek + EpCannot be measured

Can meas Ek but not Ep Ek = ½mv2

Can determine the change in heat content of a system = HChange in Enthalpy = H = Hprod-Hreact

Thermochemical Equation Exothermic CH4(g) + 1½O2(g) CO2(g) + H2O(g) + 890 KJ

CH4(g) + 1½O2(g) CO2(g) + H2O(g) H = -890 KJ

Endothermic 483.6 KJ + H2O(g) H2(g) + O2(g)

H2O(g) H2(g)+ O2(g) H = 483.6 KJ

Thermo chemical equation show the reactants and product

with the energy change that occurs

Endothermic Rx Cure

Exothermic Rx Cure

Calorimetry The measurement of heat flow =

HWe can catch the energy going into or

out of a systemUse a calorimeter

Highly insulated device used to measure the amount of heat moving into or out of the system

Water is usually used in the cal is used to absorb energy from, or give off energy to the system

Heat Capacity = Cp To measure heat flow into or out of the

system we must know how much heat the material in the calorimeter can hold

Called heat capacity Unique for each substance

Specific Heat Capacity = Cp Amount heat (q) required to increase the temp

of 1 g of a subst 1˚C Water has a HUGE heat capacity

Units of J/g°C or J/g K

Coffee Cup Calorimeter qH2O = mH2O x TH2O x CpH2O

qH2O = H = energy absorbed or given off by the water in the calorimeter

mH2O = mass (g) of the water 1 mL H2O = 1 gram H2O

TH2O = change in temp Tfinal – Tinitial

Cp = specific ht capacity of the water in the calorimeter

CpH2O = 4.18 J/g˚C

1. A reaction occurs in a calorimeter that contains 45.0 g of water. If the temp of the water goes from 25.0°C to 55.0°C, calculate the energy exchanged.

Sample Problem

How much energy is needed to increase the temperature of 50.0 g of water 15.0⁰C?

How much energy (joules) is needed to increase the temp of 150.0 mL water from 40.0⁰C to 60.0⁰C?

Haw many grams of water can be heated from 20.0⁰C to 75⁰C using 12500.0 Joules?

A 120.0 mL sample of water, at 20.5 ⁰C, absorbs 9985 J of energy. Determine the final temp of the water.

q(H2O) = m(H2O) x t(H2O) x 4.18 J/g⁰C

A 120.0 mL sample of water, at 20.5 ⁰C, absorbs 9985 J of energy. Determine the final temp of the water.

q = m x (tf – ti) x 4.18 J/g⁰C

q = (tf – ti)

m x 4.18 J/g⁰C q

m x 4.18 J/g⁰C + ti = tf

q of the Reaction

The heat lost by the reaction is gained by the H2O in the calorimeterCalc q for the water in the calorimeter.

The value of Hrx for the reaction is = to –qH2O-qH2O = qrx = Hrx

Sample Problem A reaction is carried out in a coffee cup

calorimeter that contains 110.0 g of water at 23.5˚C. If the final temp of the water becomes 20.0˚C, determine the value of H for the reaction.

qH2O = mH2O x TH2O x CpH2O

= 110.0 g x (20.0˚C – 23.5˚C) x 4.18 J/g˚C = -1.61 x 103J

Hrx = -qH2O = +1.61 x 103 J or +1.61 KJ

Sample Problem An acid and a base are mixed in a coffee-

cup calorimeter. The calorimeter contains 125.0 mL of water at 22.0˚C. A neutralization reaction occurs that increased the temp to 28.0˚C. What is H for the reaction?

qH2O = mH2O x TH2O x CpH2O

= 125.0 g x (28.0˚C – 22.0˚C) x 4.18 J/g˚C = 3.34 x 103 J or 3.34 KJ

Hrx = -qH2O = -3.34 x 103 J or -3.34KJ

Calculation of C

Can calc the value of C for any substance through calorimetry

Heat the metal in a hot water bath and drop it in calorimeter: the heat picked up from the metal goes into the water Use the calorimetry equation to determine the

amount of energy transferred by the substance qH2O = -qPb = mPb x TPb x CpPb

A 90.0 g sample of metal is placed in a calorimeter that contains 105.0 g of water at 25.5˚C. If the temp of the water increases to 37.5 ˚C, determine the Amount of energy transferred in the process.

Hint: q(H2O) = m (H2O) x t (H2O) x C (H2O)

and t (H2O) = (tf – ti)

Phase Changes Chap 11.3

Solid GasLiquidfusion

crystallization

vaporization

condensation

sublimation

deposition

Each phase change requires a specific amount of energy

Molar Heat of fusion water = 6.01 KJ/mol

Molar Heat vaporization water = 40.7 KJ/mol

What are the molar heats of condensation and crystallization???

Blue = endoRed = exo

Lab Demo Heat Fusion Ice

How much energy does it take to melt a gram of ice? A mole of ice?

Place ice in warm water in a coffee-cup calorimeter

Measure 1- temp change for water in calorimeter (t) 2- how many grams (or mole)

Phase Change Sample Problem

How much energy is required to melt 27.0 g of ice at 0°C? q = molesH2O x Molar Ht FusionH2O

q = 1.50 mol water x 6.01 KJ = 9.01 KJ of is needed

1 mol water

Mol Htfus = 6.01 KJ/mol

Hint: 1 - calc the moles of water (mol=g/mm) 2 - multiply moles times 6.01 KJ/mol

27.0 g H2O = 1.50 mol H2O

18.0 g/mol H2O

How much energy is required to increase the temp of 27.0 g of water from zero to 100.0 degrees C? q(H2O) = m (H2O) x t (H2O) x C (H2O)

= 27.0 g x 100.0°C x 4.18 J/g°C = 11286 J = 11.3 KJ

Phase Change Sample Problem

Phase Change Sample Problem

How much energy is required to vaporize 27.0 g of water to steam, at 100.0°C? q = molesH2O x Molar Ht VapH2O

q = 1.50 mol water x 40.7 KJ = 61.1 KJ 1 mol water

Mol Htvap = 40.7 KJ/mol

Hint: - 1 calculate the moles of water (mol=g/mm) - 2 multiply moles times 40.7 KJ/mol 27.0 g H2O = 1.50 mol H2O

18.0 g/mol H2O

Phase Change Sample Problem

How much energy is required to take 27.0 g of ice at 0°C and change it to steam at 100.0°C?

q = molesH2O x Molar Ht FusionH2O

qH2O = mH2O x TH2O x CpH2O

q = molesH2O x Molar Ht VapH2O

= 9.1 KJ + 11.3 KJ + 61.1 KJ = 81.5 KJ

Phase Change Sample Problem

How much energy is required to take 63.0 g of ice at 0°C and change it to steam at 100.0°C?

q = molesH2O x Molar Ht FusionH2O

(63.0/18.0) mol x 6.01 KJ/mol = q1

qH2O = mH2O x TH2O x CpH2O

63.0 g x 100.0°C x 4.18 J/g°C = q2

q = molesH2O x Molar Ht VapH2O

(63.0/18.0) mol x 40.7 KJ/mol = q3

Phase Change Math

Molar heats of Fusion (etc) are used to calc energy change for phase change, (at a constant temp).q = moles x Heat Fusion (etc)

Calorimetry is used to calc energy exchanged went a substance temp increasesqH2O = mH2O x TH2O x CpH2O

qice = molice x Hfusice

qH2O = mH2O x TH2O x CpH2O

qsteam = msteam x Tsteam x Cpsteam

qH2O = molH2O x HvapH2O

qice = mice x Tice x Cpice

Phase Change Sample Problem

How much energy is given off when 27.0 g of steam condenses to water at 100.0°C? q = molesH2O x Molar Ht VapH2O

q = 1.50 mol water x 40.7 KJ = 61.1 KJ is given off

1 mol water

Mol Htvap = 40.7 KJ/mol

Hint: - 1 calculate the moles of water (mol=g/mm) - 2 multiply moles times 40.7 KJ/mol 27.0 g H2O = 1.50 mol H2O

18.0 g/mol H2O

3 Ways to Determine H

1. Calorimetry . H° = -qH2O

2. H°f tables 3. Hess’s Law of Heat Summation

Rules of thermodynamics

Rules of Thermodynamics

H is proportional to the amount of reactant or product

Reverse reactions have opposite sign Hess’s Law of Heat Summation

H is proportional to the amount of Reactant or

Product Based on stoichiometric coefficients

Must have a balanced thermochemical equation

CH4 + 2O2 CO2 + 2H2O H = -890 KJ How many KJ are given off by the complete

combustion of 2 moles of CH4?2 mol CH4 x -890 KJ = -1780 KJ

1 mol CH4

Sample ProblemHow many KJ of energy are produced by burning 1.25 g of CH4 according to the rx:

CH4 + 2O2 CO2 + 2H2O H = -890 KJ

How many KJ of energy are produced by burning 1.25 g of O2?

Reverse Reactions• H for reverse reactions have the

same value, but the opposite sign• Reactants Products H = (-)

• Products Reactants H = (+)

Reverse Reactions Have Opposite Signs on H

CH4 + 2O2 CO2 + 2H2O H = -890 KJ

CO2 + 2H2O CH4 + 2O2 H = +890 KJ

Hess's Law of Heat Summation

In a multi-step reaction the sum of the H values for all the steps equals H for the overall reaction

H1 + H2 = H3

A + B --> C H1 = -20 KJ B + C ---> D H2 = -40 KJ A + 2B --> D H3 = ? = -60 KJ

Hess's Law of Heat Summation

Can use Hess's Law to measure H for a reaction that is too hard to run in the lab

C + ½O2 ---> CO H1 = ?

CO + ½O2 ---> CO2 H2 = -283 KJ

C + O2 ---> CO2 H3 = -393 KJ

H1 + H2 = H3

H1 = H3 - H2

H1 = (-393 KJ) - (-283 KJ)

= -110 KJ

Use the following 2 equations to determine

H for the overall reaction:

Sn (s) + 2 Cl2 (g) --- > SnCl4 (aq)

Sn (s) + Cl2(g) ----> SnCl2(s) -325 kJ/mol

SnCl2 (s) + Cl2 (g) ----> SnCl4(aq) -186

kJ/mol

Mn (s) + 02 (g) ---> MnO2 (s)

MnO2 (s) + Mn (s) --- > 2 MnO (s) -240 kJ

2 MnO2 (s) --- > 2 MnO (s) + 02 (g) +264 kJ

SnO2 (s) + 2H2 (g) --- > Sn (s) + 2 H20 (l)

Sn (s) + 02 (g) --- > SnO2 (s) -580.7 kJ

H2 (g) + ½ 02 (g) --- > H20 (l) -285.8 kJ

2 Mg (s) + SiCl4 (1) --- > Si (s) + 2 MgCl2

(s)

SiCl4 (l) --- > Si (s) + 2 Cl2 (g) +687 kJ

Mg (s) + Cl2 (g) --- > MgCl2 (s) -641 kJ

N2 + 2 H2 --> N2H4

H2 + ½ 02 --> H20 -242 kJ

N2 + 2 H20 --> N2H4 + 02 +534 kJ

CH4 + 2 02 --- > C02 + 2 H20

C + 2 H2 ---> CH4 -74.8 kJ

C + O2 ----> CO2 -393.5 kJ

H2 + ½ O2 ----> H2O -235.8 kJ

Find H° for the reaction

2H2(g) + 2C(s) + O2(g) --> C2H5OH(l)

using the following thermochemical data.

C2H5OH (l) + 2 O2 (g) --> 2 CO2 (g) + 2 H2O (l) H = -875. kJ

C (s) + O2 (g) --> CO2 (g) H = -394.51

kJ

H2 (g) + ½ O2 (g) --> H2O (l) H = -285.8 kJ

H2(g) + Br2(g) → 2 HBr(g) ΔH = −72 kJ

H2(g) → 2 H(g) ΔH = 436 kJ

Br2(g) → 2 Br(g) ΔH = 224 kJ

Calculate the enthalpy change for

H(g) + Br(g) → HBr(g)

Enthalpies of Formation = Hf°

Cu(s) + Cl2(g) ---> CuCl2(s) H = -220 KJ The energy change H⁰, when 1 mol of a

substance is formed from its elements in their normal (standard) state (the ° means 25°C and normal pressure) Elements (including diatomic like H2, N2, Cl2) are

arbitrarily assigned Hf° = 0

Hf° = a relative measure of how stable compounds

are A(-)Hf° means compound is more stable than its elements

A(+)Hf° means compound is less stable than its elements

Enthalpies of Formation

H°rx = Hf°(products) - Hf°(reactants)

Hf° values are on table A-11 pg 833 They are used to calculate, NOT

measure, the change in enthalpy, H°, for a chemical reaction that is hard or impossible to carry out in the lab.

Sample prob E pg 356 Do pract prob #1,2 pg 356

Sample Problem Calculation H° for the reaction: C3H8(g) + 5O2 ---> 3CO2(g) + 4H2O(l)

-103.8 0 -393.5 -285.8 = Hf° KJ/mol

H°rx = Hf°(products) - Hf°(reactants)

H°rx = [3mol(-393.5)KJ/mol + 4

mol(-285.8)KJ/mol]- [1mol(-

103.8)KJ/mol + 0] = -2219.9 KJ

Entropy S

Randomness or disorder in a system Entropy is ever increasing Molar entropy is entropy possessed by 1 mole of a substance Nature favors increases in entropy

Reactions that increase entropy occur more easily

Elements have a value for entropy

Entropy S = Randomness As temp entropy Sol Liq Gas entropy goes up A solution has more entropy than its

separate components NaCl water solution vs water and solid NaCl

Reactions that produce more moles of gases increase entropy

Is S (+) or (-) for: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

Entropy Change = S°

Calculted in the same manner as H° S°rx = S°(products) - S°(reactants)

Sample prob F pg 361

Calc S° for: 2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

S°J/molK = 127.2 205.0 213.8 188.7

S° = [2 mol (213.8KJ/molK) + 4 mol (188.7KJ/molK)] - [2 mol (127.2 KJ/molK) + 3 mol (205.0 KJ/molK)]

= 313 KJ/K Does the entropy change, S, favor a spontaneous reaction or

not???

What determines if a reaction or process occurs spontaneously? What energy change favors a

spontaneous reaction?(-) change in energy

What entropy change favors a spontaneous reaction?(+) positive change in entropy

Neither H or S alone, can predict if a rx will be spontaneous

Gibb's Free Energy = G

Determines whether a reaction is spontaneous or not spontaneous at a specific temperature

If G is (-) reaction is spontaneous If G is (+) reaction is NOT

spontaneous If G = 0 reaction is just beginning

Called equilibrium

Gibb's Free Energy = G Can be calculated in 2 ways 1-From free energy tables Must be at standard conditions to use:

25˚C and pressure at surface of the earthG°rx = G°(products) - G°(reactants)

2-From H and S values at temperatures other than 25˚CG = H – TS

Gibb's Free Energy = G

Calc G˚ for H2(g) + CO2(g) H2O(g) + CO(g)

0 -394.4 -228.6 -137.2 (KJ/mol)

G°rx = G°(products) - G°(reactants)

= [1 mol (-228.6KJ/mol) + 1 mol (-137.2KJ/mol] – [0 + 1mol (-394.4 KJ/mol)]

= 28.6 KJ

Is the rx spontaneous???

G at Temp other than 25˚C

Calc G at 100˚C for the reaction: H2(g) + CO2(g) H2O(g) + CO(g)

H 0 -393.5 -241.8 -110.5

S 130.7 213.8 188.7 197.2

H˚ = [(1 x -241.8) + (1 x -110.50)] – [(1 x -393.5) + 0] = 41.2 KJS˚ = [(1 x 188.7) + (1 x 197.2)] – [(1 x 130.7) + (1 x 213.8)] = 41.4 J/K

= 0.041 KJ/K(Move decimal 3 places to convert J/K to KJ/K)

G = H – TS

Is a reaction spontaneous at 100˚C if H = 41.2 KJ and S = 41.4 J/K

G = H – TS= 41.2KJ – (100 + 273 K) 0.0414 KJ/K

= 25.8 KJ is the reaction spontaneous?

Determination of Equilibrium Temp

The temp (K) at which the rx just beginsEquilibrium

T = H = 41.2 KJ = 995 K S 0.0414 KJ/K

ºC + 273 = K or ºC = K – 273 995 – 273 = 722 ºC

Gibb’s Free Energy

At what temp does H2O(l) H2O(g) becomes spontaneous

H = (-241.8 KJ/mol) – (-285.8 KJ/mol) = 44.0 KJ/mol S = (188.7 J/mol K) – (70.0 J/mol K) = 118.7 J/mol K

T = H = 44.0 KJ/mol = 370 K S .1187 KJ/mol K

ºC = K - 273 = 370 K – 273 = 97 ºC

Affects of Signs of H and G

If H is (-) and S is (+)Rx is always spontaneous

If H is (+) and S is (-)Rx is never spontaneous

If H is (+) and S is (+) Increase temp to make rx spontaneous

If H is (-) and S is (-)Decreases temp to make rx spontaneous

Initial water temp 42.3 ºC final water temp 8.1ºC Change in water temp ºC Final water volume 194.5 mL Initial water volume 130.0 mL Volume of melt mL mass ice melted g Heat released by cooling water J J/g ice melted (heat of fusion) J/g kJ/mol ice melted (molar heat of fusion) kJ/mol

How much energy is lost when 125.50 g of water in a calorimeter changes from 98.6°C

to 18.9°C?

CaO + CO2 --------> CaCO3 ΔHº = -178 kJ

CaO + H2O(l) --------> Ca(OH)2 ΔHº = -65 kJ

Ca(OH)2 + CO2 -----> CaCO3 + H2O(l)

SOCl2 + NiO --------> SO2 + NiCl2 150 kJ

SOCl2 + H2O --------> SO2 + 2HCl -27 kJ

NiO + 2HCl -----> NiCl2 + H2O(g)