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Chapter 11 - Thermochemistry Heat and Chemical Change. Milbank High School. Section 11.1 The Flow of Energy - Heat. OBJECTIVES: Explain the relationship between energy and heat . Section 11.1 The Flow of Energy - Heat. OBJECTIVES: Distinguish between heat capacity and specific heat .

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section 11 1 the flow of energy heat
Section 11.1The Flow of Energy - Heat
  • OBJECTIVES:
    • Explain the relationship between energy and heat.
section 11 1 the flow of energy heat3
Section 11.1The Flow of Energy - Heat
  • OBJECTIVES:
    • Distinguish between heat capacity and specific heat.
energy and heat
Energy and Heat
  • Thermochemistry - concerned with heat changes that occur during chemical reactions
  • Energy - capacity for doing work or supplying heat
energy and heat5
Energy and Heat
  • Heat - represented by “q”, is energy that transfers from one object to another, because of a temperature difference between them.
exothermic and endothermic processes
Exothermic and Endothermic Processes
  • In studying heat changes, think of defining these two parts:
    • the system
    • the surroundings
exothermic and endothermic processes7
Exothermic and Endothermic Processes
  • The Law of Conservation of Energy states that in any chemical or physical process, energy is neither created nor destroyed.
    • All the energy is accounted for as work, stored energy, or heat.
exothermic and endothermic processes8
Exothermic and Endothermic Processes
  • Fig. 11.3a, p.294 - heat flowing into a system from it’s surroundings:
    • defined as positive
    • q has a positive value
    • called endothermic
      • system gains heat as the surroundings cool down
exothermic and endothermic processes9
Exothermic and Endothermic Processes
  • Fig. 11.3b, p.294 - heat flowing out of a system into it’s surroundings:
    • defined as negative
    • q has a negative value
    • called exothermic
      • system loses heat as the surroundings heat up
exothemic and endothermic
Exothemic and Endothermic
  • Every reaction has an energy change associated with it
  • Exothermic reactions release energy, usually in the form of heat.
  • Endothermic reactions absorb energy
  • Energy is stored in bonds between atoms
heat capacity and specific heat
Heat Capacity and Specific Heat
  • A calorie is defined as the quantity of heat needed to raise the temperature of 1 g of pure water 1 oC.
    • Used except when referring to food
    • a Calorie, written with a capital C, always refers to the energy in food
    • 1 Calorie = 1 kilocalorie = 1000 cal.
heat capacity and specific heat12
Heat Capacity and Specific Heat
  • Joule-- the SI unit of heat and energy
    • 4.184 J = 1 cal
  • Specific Heat Capacity - the amount of heat it takes to raise the temperature of 1 gram of the substance by 1 oC (abbreviated “C”)
heat capacity and specific heat13
Heat Capacity and Specific Heat
  • For water, C = 4.18 J/(g oC), and also C = 1.00 cal/(g oC)
  • Thus, for water:
    • it takes a long time to heat up, and
    • it takes a long time to cool off!
  • Water is used as a coolant!
    • Note Figure 11.7, page 297
heat capacity and specific heat14
Heat Capacity and Specific Heat
  • To calculate, use the formula:
  • q = mass (g) x T x C
  • heat abbreviated as “q”
  • T = change in temperature
  • C = Specific Heat
  • Units are either J/(g oC) or cal/(g oC)
  • Sample problem 11-1, page 299
section 11 2 measuring and expressing heat changes
Section 11.2Measuring and Expressing Heat Changes
  • OBJECTIVES:
    • Construct equations that show theheat changes for chemical and physical processes.
section 11 2 measuring and expressing heat changes16
Section 11.2Measuring and Expressing Heat Changes
  • OBJECTIVES:
    • Calculate heat changes in chemical and physical processes.
calorimetry
Calorimetry
  • Calorimetry - the accurate and precise measurement of heat change for chemical and physical processes.
calorimetry18
Calorimetry
  • For systems at constant pressure, the heat content is the same as a property called Enthalpy (H) of the system
calorimetry19
Calorimetry
  • Changes in enthalpy = H
  • q = H These terms will be used interchangeably in this textbook
  • Thus, q = H = m x C x T
  • H is negative for an exothermic reaction
  • H is positive for an endothermic reaction (Note Table 11.3, p.301)
c o 2 co 2

C + O2

Energy

C O2

Reactants

®

Products

C + O2® CO2

+ 395 kJ

395kJ

in terms of bonds

O

C

O

O

C

O

O

C

O

In terms of bonds

O

C

O

Breaking this bond will require energy.

Making these bonds gives you energy.

In this case making the bonds gives you more energy than breaking them.

exothermic
Exothermic
  • The products are lower in energy than the reactants
  • Releases energy
slide23

CaO + CO2

Energy

CaCO3

Reactants

®

Products

CaCO3® CaO + CO2

CaCO3 + 176 kJ ® CaO + CO2

176 kJ

chemistry happens in
Chemistry Happens in

MOLES

  • An equation that includes energy is called a thermochemical equation
  • CH4 + 2O2® CO2 + 2H2O + 802.2 kJ
  • 1 mole of CH4 releases 802.2 kJ of energy.
  • When you make 802.2 kJ you also make 2 moles of water
thermochemical equations
Thermochemical Equations
  • A heat of reaction is the heat change for the equation, exactly as written
    • The physical state of reactants and products must also be given.
    • Standard conditions for the reaction is 101.3 kPa (1 atm.) and 25 oC
ch 4 2 o 2 co 2 2 h 2 o 802 2 kj
CH4 + 2 O2® CO2 + 2 H2O + 802.2 kJ
  • If 10. 3 grams of CH4 are burned completely, how much heat will be produced?

1 mol CH4

802.2 kJ

10. 3 g CH4

16.05 g CH4

1 mol CH4

= 514 kJ

ch 4 2 o 2 co 2 2 h 2 o 802 2 kj27
CH4 + 2 O2® CO2 + 2 H2O + 802.2 kJ
  • How many liters of O2 at STP would be required to produce 23 kJ of heat?
  • How many grams of water would be produced with 506 kJ of heat?
enthalpy
Enthalpy
  • The heat content a substance has at a given temperature and pressure
  • Can’t be measured directly because there is no set starting point
  • The reactants start with a heat content
  • The products end up with a heat content
  • So we can measure how much enthalpy changes
enthalpy30
Enthalpy
  • Symbol is H
  • Change in enthalpy is DH (delta H)
  • If heat is released, the heat content of the products is lower

DH is negative (exothermic)

  • If heat is absorbed, the heat content of the products is higher

DH is positive (endothermic)

slide31

Energy

Change is down

DH is <0

Reactants

®

Products

slide32

Energy

Change is up

DH is > 0

Reactants

®

Products

heat of reaction
Heat of Reaction
  • The heat that is released or absorbed in a chemical reaction
  • Equivalent to DH
  • C + O2(g) ® CO2(g) + 393.5 kJ
  • C + O2(g) ® CO2(g) DH = -393.5 kJ
  • In thermochemical equation, it is important to indicate the physical state
  • H2(g) + 1/2O2 (g)® H2O(g) DH = -241.8 kJ
  • H2(g) + 1/2O2 (g)® H2O(l) DH = -285.8 kJ
heat of combustion
Heat of Combustion
  • The heat from the reaction that completely burns 1 mole of a substance
  • Note Table 11.4, page 305
section 11 3 heat in changes of state
Section 11.3Heat in Changes of State
  • OBJECTIVES:
    • Classify, by type, the heat changes that occur during melting, freezing, boiling, and condensing.
section 11 3 heat in changes of state36
Section 11.3Heat in Changes of State
  • OBJECTIVES:
    • Calculate heat changes that occur during melting, freezing, boiling, and condensing.
heats of fusion and solidification
Heats of Fusion and Solidification
  • Molar Heat of Fusion (Hfus) - the heat absorbed by one mole of a substance in melting from a solid to a liquid
  • Molar Heat of Solidification(Hsolid) - heat lost when one mole of liquid solidifies
heats of vaporization and condensation
Heats of Vaporization and Condensation
  • Molar Heat of Vaporization(Hvap) - the amount of heat necessary to vaporize one mole of a given liquid.
  • Table 11.5, page 308
heats of vaporization and condensation39
Heats of Vaporization and Condensation
  • Molar Heat of Condensation(Hcond) - amount of heat released when one mole of vapor condenses
  • Hvap = - Hcond
heats of vaporization and condensation40
Heats of Vaporization and Condensation
  • Note Figure 11.5, page 310
  • The large values for Hvapand Hcondare the reason hot vapors such as steam is very dangerous
    • You can receive a scalding burn from steam when the heat of condensation is released!
heats of vaporization and condensation41
Heats of Vaporization and Condensation
  • H20(g) H20(l) Hcond = - 40.7kJ/mol
  • Sample Problem 11-5, page 311
heat of solution
Heat of Solution
  • Heat changes can also occur when a solute dissolves in a solvent.
  • Molar Heat of Solution (Hsoln) - heat change caused by dissolution of one mole of substance
section 11 4 calculating heat changes
Section 11.4Calculating Heat Changes
  • OBJECTIVES:
    • Apply Hess’s law of heat summation to find heat changes for chemical and physical processes.
section 11 4 calculating heat changes44
Section 11.4Calculating Heat Changes
  • OBJECTIVES:
    • Calculate heat changes using standard heats of formation.
hess s law
Hess’s Law
  • If you add two or more thermochemical equations to give a final equation, then you can also add the heats of reaction to give the final heat of reaction.

Called Hess’s law of heat summation

  • Example shown on page 314 for graphite and diamonds
why does it work
Why Does It Work?
  • If you turn an equation around, you change the sign:
  • If H2(g)+ 1/2 O2(g)® H2O(g) DH=-285.5 kJ
  • then, H2O(g) ® H2(g)+ 1/2 O2(g) DH =+285.5 kJ
  • also,
  • If you multiply the equation by a number, you multiply the heat by that number:
  • 2 H2O(g) ® 2 H2(g)+ O2(g) DH =+571.0 kJ
standard heats of formation
Standard Heats of Formation
  • The DH for a reaction that produces 1 mol of a compound from its elements at standard conditions
  • Standard conditions: 25°C and 1 atm.
  • Symbol is
  • The standard heat of formation of an element = 0
  • This includes the diatomics
what good are they
What good are they?
  • Table 11.6, page 316 has standard heats of formation
  • The heat of a reaction can be calculated by:
    • subtracting the heats of formation of the reactants from the products

DHo

= (

Products) -

(

Reactants)

examples

CH4 (g) = - 74.86 kJ/mol

O2(g) = 0 kJ/mol

CO2(g) = - 393.5 kJ/mol

H2O(g) = - 241.8 kJ/mol

Examples
  • CH4(g) +2O2(g)® CO2(g) + 2 H2O(g)
  • DH= [-393.5 + 2(-241.8)] - [-74.68 +2 (0)]

DH= - 802.4 kJ