energy – capacity to do work or transfer heat types of energy 1. kinetic – energy of motion

Thermodynamics

A. Energy – Capacity to do work or transfer heat

B. Types of Energy1. Kinetic – Energy of motion

a. KE = ½ mv2

b. Unit – Joules (kg –m2/s2)1 calorie = 4.184 J1000 cal = 1 Cal (nutr)

Types of Energy

Thermodynamics

F = ma

W = Fd

K = ½ mv2

U = mgh

Thermodynamics

c. Macroscale KE – energy

of movement (car, baseball)

d. Microscale KE – Temp.

measure of average KE of molecules

Types of Energy

Thermodynamics

2. Potential Energy – stored energy

a. Macroscale PE – energy of position (rollercoaster ex.)

b. Microscale PE – Energy in chemical bonds.

(Food example)

Types of Energy

Thermodynamics

System and Surroundings

1. System – What we are studying

2. Surroundings – rest of the universe

a. Open system – Car, earth

b. Closed system – Vacuum

Types of Energy

Thermodynamics

The First Law of Thermodynamics

Energy is neither created nor destroyed. It only changes form. Energy is conserved.

E = q + w

E = change in energy

q = heat

w = work

The First Law

Thermodynamics

B. Examples

1. Car

2. Vacuum Cleaner

The First Law

Thermodynamics

C. Transfer of Energy

1. Car – Chemical to heat to mechanical

D. Consequences of the First Law

1. No machine is 100% efficient

2. Always lose some energy to heat

The First Law

Thermodynamics

Exothermic Endothermic• Gives off heat• Surroundings get hotter

(transfer of heat from molecules)

• H –• Hot pack, burning gas,

exercise

• Absorbs heat from surroundings

• System gets hotter, surroundings get colder

• H +

• Cold Pack, cooking

Exo and Endothermic

Thermodynamics

A. Heat – Energy transferred because of a difference in temperature (Cooking a turkey example)

B. Extensive Property – depends on amount of materialGlass of water vs. iceberg“Which has more heat”

Enthalpy – Heat Energy

Thermodynamics

A. Calorimetry – Measuring heat by measuring a temperature change

B. Specific Heat – Amount of heat to raise temperature of one gram of a substance one degree Celsius or Kelvin1. Unit – J/goC2. Higher the sp. heat, more energy needed to raise temp

Calorimetry

Thermodynamics

3. Example – Wooden Spoon (~2 J/goC) vs. metal spoon (~0.50 J/goC)

4. Water

a. Very high specific heat - 4.18 J/goC

b. Oceans

c. Our bodies

Calorimetry

Thermodynamics

C. Heat for one substance

1. Formula

q = mCpT

q = heat

m = mass (grams)

Cp = specific heat (J/goC)

T = Tfinal – Tinitial

Calorimetry

Thermodynamics

1. How much heat is needed to warm 250.0 grams of water from 22.0oC to 98.0oC to make tea? (Ans: 79.5 kJ)

2. Calculate the specific heat of copper if 12.0 grams of copper cools from 98.0oC to 96.2 oC and tranfers 8.32 J of heat. (0.385 J/goC)

Calorimetry

Thermodynamics

a. Calculate how much heat is absorbed by 50.0 kg of rocks if their temperature increases by 12.0 oC? Assume the specific heat is 0.82 J/g oC. (492 kJ)

b. Calculate how many calories of heat they absorbed. (117 kcal)

c. Calculate the final temperature of the rocks if they absorbed 450 kJ of heat and began at a temperature of 26.9 oC. (37.9 oC)

d. 50 kg of Copper (0.385 J/goC)absorbs the same amount of heat as the rocks in part c. Would the temperature change be greater or less for copper?

Calorimetry

Thermodynamics

A student heats 60.0 grams of iron to 98.0oC in a hot water bath. They then submerge the metal in 150.0 g of water (Cp = 4.18 J/goC) at 20.0oC. After a few minutes, the cup and metal reach a final temperature of 23.2oC.

a.Calculate the heat gained by the water.b.Calculate the specific heat (Cp) of the metal.

Thermodynamics

3. Heat Capacity – heat to raise temperature of a particular object by 1K; unit of J/K

a. Used only for one object

b. Car part

c. heat capacity= mCp

Calorimetry

Thermodynamics

a. How many Joules of heat are required to raise 1 gram of water by 1 degree Celsius?

b. Calculate the molar heat capacity of water? (Ans: 75.2 J/ oC)

c. Calculate the heat capacity of 250.0 grams of water? (Ans: 1050 J/K)

d. Calculate the heat capacity of 6.20 mole of iron? (Ans: 156 J/oC)

Calorimetry

Thermodynamics

1. Calorimeter – Coffee Cup2. Used to measure heats of a chemical

reaction3 H = -q

H = -mCpTm = mass of BOTH reactants

Cp = often 4.18 J/goC for aqueous solns

Heat of Reaction

Thermodynamics

Example 1.

When a student mixes 50.0 mL of 1.00 M HCl and 50.0 mL of 1.00 M NaOH in a calorimeter, the temperature rises from 21.0oC to 27.5oC. Calculate the change in enthalpy for the reaction.

(Ans: -2.7 kJ and –54 kJ/mol NaOH)

Heat of Reaction

Thermodynamics

Example 2.

When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed, the temperature rises from 22.30oC to 23.11oC. Calculate the change in enthalpy for the reaction per mole of AgNO3.

(Ans: –68 kJ/mol AgNO3)

Heat of Reaction

Thermodynamics

A. Standard State – State of an element @25oC and 1 atm.

1. Standard Enthalpy of formation of an element in its st. state is zero

2. Takes no energy to make an element assume st. state.

“Exists in this form naturally”

Standard State

Thermodynamics

3. Examples Hfo

O2(g) 0H2(g) 0N2(g) 0F2(g), Cl2(g), Br2(l), I2(s) 0

C(graphite) 0Fe(s) 0

Standard State

Thermodynamics

4. Carbon Allotropes – different physical forms of the same element

a. Graphite

b. Diamond

c. Buckyball (Buckminster Fuller)

C(gr) 0

C(dia) 1.88 kJ/mol

C60(s) 2269 kJ/mol

Standard State

Thermodynamics

Thermodynamics

Thermodynamics

Thermodynamics

Thermodynamics

5. Writing Standard State Equations

Examples:

CO(g)

H2O(g)

KClO3(s)

Standard State

Thermodynamics

NaNO3(s)

NH4CN(s)

Al2(CO3)3(s)

Standard State

Thermodynamics

Fe(NO3)3(s)

CH3COOH(l)

NaOH (s)

Standard State

Thermodynamics

6. Using Standard Enthalpies of Formation

Hor = nHf

oprod – mHf

oreactants

n,m = coefficients

Hfo = values in text

H of Reaction

Thermodynamics

Example 1.

Calculate H for the combustion of liquid C6H6 to CO2 and water

(Ans:-3267 kJ/mol C6H6)

H of Reaction

Thermodynamics

Example 2.

Calculate H for the combustion of liquid C2H5OH to CO2 and water

(Ans:-1367 kJ/mol C2H5OH)

H of Reaction

Thermodynamics

Example 3.

Calculate Hfo for CaCO3(s) if you are given:

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

Hrxno = +178.1 kJ

(Ans:-1207.1 kJ/mol CaCO3)

H of Reaction

Thermodynamics

Example 4.

Calculate Hfo for CuO(s) if you are

given:

CuO(s) + H2(g) Cu(s) + H2O(l)

Hrxno = -129.7 kJ

(Ans:-156.1 kJ/mol CuO)

H of Reaction

Thermodynamics20.0 mL of 0.100 M NaOH(aq) is neutralized with 0.0950 M HNO3(aq).

a. Write the balanced chemical reaction for this process.

b.Calculate the volume of HNO3 required.(21.1 mL)

c. When these two volumes are mixed, the temperature rises from 22.3oC to 23.0 oC. Calculate the Hrxn/mole of NaOH. (-60.1 kJ/mol)

d. Calculate heat of reaction the using the heats of formation values found in the appendix (not the pancreas). (-55.83 kJ/mol)

e. How do your two answers compare?

Thermodynamics

A. Example

2H2(g) + O2(g) 2H2O(g) H = -483.6 kJ

1. Hindenberg

2. Heat is released to the surroundings.

3. -483.6 kJ released per 2 mole of H2, 1 mole O2, or 2 mole of H2O

Enthalpy – Heat Energy

Thermodynamics

B. Example 1.

How much heat is released from the combustion of 10.0 g of H2?

2H2(g) + O2(g) 2H2O(g) H = -483.6 kJ

(Ans: -1210 kJ)

Enthalpy – Heat Energy

Thermodynamics

Example 2.

How much heat is released when 5.00 g of H2O2 decomposes?

2H2O2 2H2O + O2 H= -196 kJ

Answer: -14.4 kJ

Enthalpy – Heat Energy

Thermodynamics

Example 3

How much heat is required to produce 25.0 g of H2O2?

Reverse above reaction

Answer: +72.1 kJ

Enthalpy – Heat Energy

Thermodynamics

Example 4

How much H2O2 is produced if 300.0 kJ of heat are absorbed?

2H2O + O2 2H2O2 H= +196 kJ

Answer: 104 grams

Enthalpy – Heat Energy

Thermodynamics

Example 5

How many grams of water are produced if 8437 kJ of heat are released?

C6H6 + 15/2O2 6CO2 +3H2O H= -3267 kJ

Answer: 139.5grams

Enthalpy – Heat Energy

Thermodynamics

A. Hess’s Law - If a reaction is carried out in a series of steps, you add the H’s for the individual steps

B. Can calculate H without having to do the experiment.

C. Rules

1. If you flip the reaction, flip the sign of H

2. If you multiply the reaction, multiply theH.

Hess’s Law

Thermodynamics

D.Example 1.

C(gr) C(diamond)

C(gr) + O2(g)CO2(g) H = -393.5 kJ

C(dia) + O2(g)CO2(g) H = -395.4 kJ

(Ans: +1.9 kJ)

Hess’s Law

Thermodynamics

Example 2.

NO (g) + O(g) NO2(g)

NO(g) + O3(g) NO2(g) + O2 (g) H=-198.9 kJ

O3(g) 3/2 O2 (g) H=-142.3kJ

O2(g) 2 O(g) H=+495.0kJ

(Ans: -304.1 kJ)

Hess’s Law

Thermodynamics

Example 3.

2C(s) + H2(g) C2H2(g)

C2H2(g) + 5/2O2(g) 2CO2(g) + H2O (l)

H= -1299.3 kJ

C(s) + O2(g) CO2(g) H= -393.5 kJ

H2(g) + ½O2(g) H2O(l) H= -285.8 kJ

(Ans: +226.5 kJ)

Hess’s Law

Calculate the enthalpy for the reaction:

1/2N2(g) + 2H20(l)NO2(g) + 2H2(g)

2NH3(g) N2(g) + 3H2(g) H=161KJ

NO2(g) + 7/2H2(g) 2H2O(l) + NH3(g)

H=-378KJ

(ANS: 297.5 kJ)

Thermodynamics

A. Food

1. Carbohydrates – break down into glucose, then CO2 and H2O

C6H12O6 + O2 CO2 + H2O

H = -2803 kJ/mol

Provide 17 kJ/g

About 4 Cal/g (4 per raisin)

Food

Thermodynamics

2. Fats – produce CO2 and H2O as waste

38 kJ/g

9 Cal/g (9 Calories per 2 M&M’s)

Food

Thermodynamics

3. Proteins – Produce CO2, H2O and (NH2)2CO as waste

a. Amino Acid structure responsible for the urea.

b. 17 kJ/g

4 Cal/g

Food

Thermodynamics

1. Fossil Fuels (coal, oil, natural gas, propane, diesel, gasoline)

2. Coal

a. All anthracite in PA

b. Fossilized plant matter

c. Sulfur often in coal (SO3)

SO3 + H2O H2SO4

Fuels

Thermodynamics

Acid rain affected this statue at the Field Museum in Chicago (botttom is after restoration)

Thermodynamics

3. Hydrogen

a. Can be produced by breaking down water or fossil fuels (Maybe solar powered)

b. Burns to produce water

c. 20 to 50 % more efficient than gasoline

d. Explosive

Fuels

Thermodynamics42.a) endo b) 127 kJ c) 1.15 g d) -165 kJ

44.a) -18.8 kJ b) -5.14 kJ c) not spontaneous

46.a) -630 kJ b) + 630 kJ c) reverse favored

50.a) Hg(l) b) 70 J

52. 4110 J or 4.11 kJ

54.a) 25 kJ/mol b) endothermic

62. -867.7kJ

64.155.7 kJ

Thermodynamics68.

½ H2(g) + ½ Br2(l) HBr(g)

Ag(s) + ½ N2(g) + 3/2 O2(g) AgNO3(s)

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

2C(gr) + 2H2(g) + O2(g) CH3COOH(l)

a) -36.23 kJ b) -124.4 kJ c) -822.2 kJ

d) -487.0 kJ

Thermodynamics72. a) -426.74 kJ b) -382.5 kJ c) -433.7 kJ

d) -150 kJ

74. -60.6 kJ

Mr. Fredericks’ old ID card was getting a bit dated.