chemistry 30 – unit 1 thermochemical changes

Post on 03-Jan-2016

47 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

To accompany Inquiry into Chemistry. PowerPoint Presentation prepared by Robert Schultz robert.schultz@ei.educ.ab.ca. Chemistry 30 – Unit 1 Thermochemical Changes. Preparation Info. Systems: Open, closed, and isolated - definitions - PowerPoint PPT Presentation

TRANSCRIPT

Chemistry 30 – Unit 1Thermochemical Changes

To accompany Inquiry into Chemistry

PowerPoint Presentation prepared by Robert Schultzrobert.schultz@ei.educ.ab.ca

Preparation Info

• Systems: Open, closed, and isolated - definitions

• First Law of Thermodynamics – Total energy of the universe is constant (energy can’t be created or destroyed)

• Second Law of Thermodynamics – In the absence of energy input, a system becomes more disordered (its entropy increases)

Preparation

• Meaning?

• A system at lower temperature will be more ordered as the particles have less average kinetic energy

• Two systems in thermal contact will transfer energy such that the more ordered (cooler) one gains energy and becomes more disordered

• Consequence: heat always flows from hotter systems to cooler ones

Preparation

Important Definitions:• Thermal Energy: the total kinetic

energy of all particles of a system

• Temperature: a measure of the average kinetic energy of the particles of a system

• Heat: a transfer of thermal energy between 2 systems

Chapter 9, Section 9.1Questions:

• Which has more thermal energy, a hot cup of coffee or an iceberg?

• Which has a larger average thermal energy, a hot cup of coffee or an iceberg?

• If an iceberg and a hot cup of coffee come into contact, in which direction will heat flow?

Preparation

• Heat energy transferred will be related to the temperature change of the system

• It takes different amounts of heat energy to change the temperature of

1 g of a substance by 1°C

• This number is called the specific heat capacity, c, and is measured in units of:

Jg C

• Water has a c value of

• This means that it takes 4.19 J of heat to raise the temperature of 1 g of water by 1°C

• Water has a very large c compared to most other common substances

4.19 Jg C

Preparation

• To determine the amount of heat transferred the formula used is

• Despite what your text says on page 337, I would always take ∆t as positive

• If heat is absorbed, temperature of surroundings will decrease; if heat is released temperature of surroundings will increase

• Examples: Practice Problems 1 and 4, page 337

Q mc t

Preparation

• Practice Problem 1, page 337

• Since 1 J is such a small amount of heat energy I start my questions in kJ as shown above

• If necessary I move into MJ or GJ

0.100 2.44 25 6.1JQ mc t kg C kJgk

k C

Preparation

• Practice Problem 4, page 337

• Putting kilo top and bottom cancels out and c stays the same

• The substance is granite

• Worksheet: WS 43 (Nelson) then BLM 9.1.1 (back only)

4.9370.790 0.790

0.25000 25.0

Q mc t

kJQ J Jc g Ck

kk g Cm t g C

Preparation

Chapter 9, Section 9.1

• Energy changes in chemical reactions crucial to life

• Not just in photosynthesis, fuels, and batteries, but in the very way that your body metabolizes food and makes the energy available for life processes

• Thermodynamics: the study of energy and energy changes

Chapter 9, Section 9.1

• Recall the first law of thermodynamics: ∆Euniverse= 0

• If a system loses energy, the surroundings gain energy (get warmer)

• If a system gains energy, the surroundings lose energy (get cooler)

∆Esystem = - ∆Esurroundings

Chapter 9, Section 9.1• Energy types:

• Kinetic energy, Ek, energy of motion of particles of a system

• Temperature is a measure of the average Ek of the particles of a system

• Potential energy, Ep, stored energy, usually in chemical bonds

Chapter 9, Section 9.1• Transfer of Ek: heat flows from hotter objects

to cooler ones (Preparation section of notes)

• Breaking bonds always requires energy (endothermic); forming bonds always releases energy (exothermic)

• Chemical reaction:breaking bonds + energy1 forming bonds + energy2

• If energy1 > energy2, reaction is endothermic

• If reverse is true, it is exothermic

• Worksheet BLM 9.1.3

input output

Chapter 9, Section 9.1• New term: enthalpy (not entropy)

• Enthalpy (change), ∆H: the difference in potential energy between reactants and products, measured at constant pressure – measured in kJ (or MJ, etc)

• Molar Enthalpy (change), ∆rH: the enthalpy change for 1 mole of a specified substance – measured inkJ/mol (or MJ/mol etc)

• In common usage the word change gets left out

Chapter 9, Section 9.1• Negative ∆H’s are exothermic

(think lose heat) and temperature of surroundings increases

• Positive ∆H’s are endothermic (think gain heat) and temperature of the surroundings decreases

• Note: this increase → negative, and decrease → positive is a stumbling block for many students

Chapter 9, Section 9.1• Chemical reactions can be written using

∆H notation:

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) ∆H=-2802.5 kJ

4 NO(g) + 6 H2O(g) 4 NH3(g) + 5 O2(g) ∆H=+906 kJ

• They can also be written with the heat as a term in the equation:

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2802.5 kJ

4 NO(g) + 6 H2O(g) + 906 kJ 4 NH3(g) + 5 O2(g)

Do ∆H Worksheet!

value for the reaction as written

Chapter 9, Section 9.1• Potential energy diagrams for the

same 2 reactions are shown below:

∆H = -2802.5 kJ

H (

kJ)

C6H12O6(s) + 6 O2(g)

6 CO2(g) + 6 H2O(l)

reactants

products

H (

kJ)

4 NO(g) + 6 H2O(g)

4 NH3(g) + 5 O2(g)reactant

s

products

∆H = +906 kJ

Chapter 9, Section 9.2• Recalling that breaking bonds always

endothermic and forming new bonds is always exothermic, more complete Ep diagrams might be shown as follows:

Endothermic Exothermic

reactants

intermediate

products

ΔH

Ep (

kJ)

reactants

intermediate

products

ΔH

Ep (

kJ)

Chapter 9, Section 9.1• Alternate forms of potential energy diagram

(from Chemistry 30 Diploma Exam Bulletin)

Chapter 9, Section 9.1• Example: Practice Problem 3, page 346a) C(s) + 2 H2(g) CH4(g) + 74.6 kJ

b) C(s) + 2 H2(g) CH4(g) ∆H = -74.6 kJ

c)

H (

kJ)

C(s) + 2 H2(g)

CH4(g)

products

reactants

∆H = -74.6 kJ

Do Ep diagrams for formation of Cr2O3(s), simple decomp* of AgI(s), and formation of SO2(g)

Chapter 9, Section 9.2

Formation of Cr2O3(s)

Ep (kJ)

reaction coordinate

2 Cr(s) + 3/2 O2(g)

Cr2O3(s)

ΔH=ˉ1139.7 kJ

Ep (kJ)

reaction coordinate

ΔH=+61.8 kJ

simple decomposition of AgI(s)

AgI(s)

Ag(s) + ½ I2(s) Ep (kJ)

reaction coordinate

ΔH=ˉ296.8 kJ

1/8 S8(s) + O2(g)

SO2(g)

formation of SO2(g)

Chapter 9, Section 9.1• Molar enthalpy of combustion: the

enthalpy change for the complete combustion of 1 mol of a substance

• Complete combustions of fossil fuels always yields CO2(g) and H2O

• Open systems – constant pressure – gases escape – H2O(g)

• Isolated systems – H2O(l)

• Human body – cellular respiration - H2O(l)

Chapter 9, Section 9.1• Table of Molar Enthalpies of Combustions

of alkanes, page 347

• Practice Problem 5b, page 347 (open system)

OR: note change in units!

• In thermodynamics it is acceptable to write equations with fractional coefficients – don’t do this elsewhere

• Try question 5a, page 347

C4H10(g) + 13/2 O2(g) 4 CO2(g) + 5 H2O(g) ∆H = -2657.3 kJ

2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(g) ∆H = -5314.6 kJ

Chapter 9, Section 9.1• Question 5a page 347

• Note that the value of ∆H varies directly as the number of moles of reacting substances

• This formula gets used to calculate enthalpy changes for ∆Ep like phase changes, chemical reactions, and nuclear reactions

C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(g) ∆H = -3244.8 kJ

rH n H

Chapter 9, Section 9.1• Example Practice Problem 3a, page

349

Find for 56.78 g of pentane

56.783244.8 2553

72.17 /

r

r

H n H

H

g kJH n H kJmolg mol

Note: from table, page 347 - comment

mol of pentane

5 123244.8 kJ

molr C HH

Chapter 9, Section 9.1• Example Practice Problem 6, page

349

• molar enthalpy change for?• a) ammonia

• b) oxygen

• c) nitrogen monoxide

• d) water

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) ΔH = -906 kJ

3

906227

4r NH

kJ kJH molmol

2

906181

5r O

kJ kJH molmol

906227

4r NO

kJ kJH molmol

2

906151

6r H O

kJ kJH molmol

Chapter 9, Section 9.1• Do Worksheet BLM 9.1.6

Chapter 9, Section 9.2• Finding the value of energy changes

experimentally: calorimetry

• Device: calorimeter

• The following diagrams show the principle behind calorimetry – note arrow directions

Chapter 9, Section 9.2• A simple calorimeter like the one you

will use

2 nested styrofoam cups containing a measured volume of water

sitting in a beaker so that it doesn’t fall over

3rd styrofoam cup inverted on top with hole for thermometer (stirrer)

Chapter 9, Section 9.2• Assumptions in styrofoam cup

calorimetry:• Amount of energy transferred to

cups and thermometer is small and can be ignored

• The system is isolated• The solution produced has the

same density and specific heat capacity as water

• The process occurs at constant pressure

Chapter 9, Section 9.2• The enthalpy change of a chemical

reaction = energy lost or gained, and is indicated by the symbol ΔH

• Energy gained or lost by the water causes a temperature change and is indicated by the symbol Q

• In an ideal calorimeter ΔH = Q• But recall:

• Therefore

rH n H Q mc t and

rn H mc t calorimetry equation

system calorimeter “water”

Chapter 9, Section 9.2• I will redo the example on page

354 using this formula

r

mc t mc tH

n cv

limiting reagent, if not stated, or substance question asks about

• remember m c Δt is for the “water” and n (c v) for the CuSO4(aq) 2 0.05000 4.19 24.60 21.40

89.40.300 0.05000

kJkg C kJ

molr molL

kg CH

L

Since the temperature has gone up the process is exothermic

Correct answer: 89.4kJmol

Chapter 9, Section 9.2• Practice Problem 9, page 355• Note that question asks for molar

enthalpy of reaction for sodium• n will be moles of sodium (question asks)

20.175 4.19 25.70 19.302.9 10

0.3722.99

r

kJkg C kJ

molr

gmol

n H mc t

kg Cmc tH

gn

• Since temperature increases, answer is correctly expressed as

22.9 10 or 0.29kJ MJmol mol

Do Practice Problems 7, 10, 12, page 355

Chapter 9, Section 9.2• Investigation 9.A page 356 (goes

with the questions you’ve been doing)

• Molar enthalpy of combustion: Investigation 9.B, page 357

Chapter 9, Section 9.2• Bomb Calorimetry: a bomb

calorimeter is used to make accurate and precise measurements

Chapter 9, Section 9.2• Reaction takes place inside an

inner container called the “bomb” that contains pure oxygen

• Chemicals are electrically ignited and heat is released to or absorbed from calorimeter water

• Calorimeter materials: stirrer, thermometer, containers are not ignored

• With calorimeter filled to a set level with water, all of their heat capacities are combined as shown:

Chapter 9, Section 9.2

• Note that C contains the mass and specific heat capacity of each component of the calorimeter

• How do you know when to use

2 2

2 2

r H O H O ther ther stir stir contains contains

r H O H O ther ther stir stir contains contains

r

n H m c t m c t m c t m c t

n H m c m c m c m c t

n H C t

bomb calorimeter equation

versus ?r rn H C t n H mc t

Heat capacity of calorimeter

Chapter 9, Section 9.2• Look for:

- words “bomb calorimeter”- no mention of the mass or volume of water- words “heat capacity” rather than “specific heat capacity”- units J/°C rather than J/g°C

• Question 2, Worksheet 46

• Since temperature increases, answer is -286 kJ/mol

• Do rest of Worksheet 46

40.00 3.54286

1.002.02

r

kJC kJ

molr

gmol

n H C t

CC tH

gn

Chapter 9, Section 9.2• More practice with

• WS 9.1.5

Q mc t

Chapter 9, Section 9.2• Review: page 366-7 good

questions: 1, 3, 4 (no actual calculation needed), 5c (data page 347), 6a (data page 347), 8, 10, 13, 15, 16, 17, 18, 19, 21

Chapter 9, Section 9.2

top related