Chemistry 30 – Thermo 1

Chemistry 30 – Unit 1

Thermochemical Changes

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

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

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

Chapter 9, Section 9.1

Questions:

• 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?

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:

J

• 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 J _{g C}



• 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



• 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 J 25 6.1

Q  mc t  kg k _{kg C}  C kJ



• 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)

4.937

0.790 0.790

0.25000 25.0 Q mc t

kJ

Q _{J J}

c k _{kg C}

k g C

m t g C

 

   _  _

    

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

Chapter 9, Section 9.1

• Recall the first law of thermodynamics:

∆E_universe= 0

• If the system loses energy, the

surroundings gain energy (get warmer)

• If the system gains energy, the

surroundings lose energy (get cooler)

Chapter 9, Section 9.1

• Energy types:

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

• Temperature is a measure of the

average E_k of the particles of a system

Chapter 9, Section 9.1

• Transfer of E_k: 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 > energy_2, reaction is endothermic • If reverse is true, it is exothermic

• Worksheet BLM 9.1.3

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), ∆H: the enthalpy change for 1 mole of a specified substance – measured in kJ/mol (or MJ/mol etc)

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

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!

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)

reactants products

Chapter 9, Section 9.2

• Recalling that breaking bonds always endothermic and forming new bonds is always exothermic, more complete E_p diagrams might be shown as follows:

Chapter 9, Section 9.1

• Alternate forms of potential energy diagram (from

Chapter 9, Section 9.1

• Example: Practice Problem 3, page 346 a) 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

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)

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 Practice Problem 5a, page 347

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

Chapter 9, Section 9.1

• Practice Problem 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 ∆E_p like phase changes, chemical reactions, and nuclear reactions

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

r

H n H

Chapter 9, Section 9.1

• Example Practice Problem 3a, page 349





Find for 56.78 g of pentane 56.78

3244.8 2553

72.17 /

r

r

H n H H

g _kJ

H n H _mol kJ

g mol

   

       

Note: from table, page 347 - comment

mol of pentane

5 12 3244.8

kJ mol rHC H

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

3 906 227 4 r NH kJ _kJ H _mol mol      2 906 181 5 r O kJ _kJ H _mol mol      906 227 4 r NO kJ _kJ H _mol mol      2 906 151 6

r H O

kJ _kJ

H _mol

mol



Chapter 9, Section 9.1

Chapter 9, Section 9.2

• Finding the value of energy changes experimentally: calorimetry

• Device: calorimeter

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}

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

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

r

H n H

   and _{Q mc t} 

r

n H mc t   calorimetry _equation

Chapter 9, Section 9.2

• I will redo the example on page 354 using this formula

r

mc t mc t H

n c v

 

   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

89.4 0.300 0.05000

kJ

kg C _kJ

mol r _mol L kg C H L          

Since the temperature has gone up the process is exothermic

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)

  2

0.175 4.19 25.70 19.30

2.9 10 0.37

22.99

r

kJ

kg C _kJ

mol r

g mol n H mc t

kg C mc t H g n              

• Since temperature increases, answer is correctly expressed as

2

2.9 10 kJ or 0.29MJ

mol mol

  

Chapter 9, Section 9.2

• Investigation 9.A page 356 (goes with the questions you’ve been doing)

Chapter 9, Section 9.2

• Bomb Calorimetry: a bomb calorimeter is used to make accurate and precise

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

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





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 r

n H C t   n H mc t  

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 3.54 286 1.00 2.02 r kJ C _kJ mol r g mol

n H C t

Chapter 9, Section 9.2

• More practice with

• WS 9.1.5



