T h er m oc h e mist r y - PowerPoint Presentation

Slide1

Thermochemistry Heat and Chemical Change

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TEMPERATURE VS. HEAT

Temperature is a measure of the average kinetic energy of the moleculesHeat is the sum total amount of energy of all the molecules

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Which is at a higher temperature?

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Which possesses more heat energy?

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heat energy = specific heat x mass x change in temperature

Q =

s∙m

∙T

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Specific Heat Capacity:

The amount of heat energyrequired to raise the temperature of 1 gram of a substance by 1 °C Energy is measured in Joules ( J )

Symbol = s Unit = J/g °C

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Specific Heat Capacities :

SubstanceH2O (l)J/g°C4.18Al0.89Fe0.45Hg0.14C0.71ethanol2.44

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Specific Heat Capacity of Water:

4.184 J/gºC

You must memorize this number!

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Specific Heat Capacities :

The higher the specific heat capacity, the more energy it takes to raise the Temperature of a substanceHow does water’s high specific heatcapacity make it such a valuable resource to living things?

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You place 10 gram samples of both Al and Fe in a hot oven for a short period of time….

0.89 J/gC 0.45 J/gC

Al

Fe

Which substance will heat up the fastest?

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You place 10 gram samples of both Al and Fe in a hot oven for a short period of time….

0.89 J/gC 0.45 J/gC

Al

Fe

Which substance has to absorb more heat energy to warm up?

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You now place both the Al and Fe in the same oven and let them sit overnight…..

Al

Fe

0.89 J/gC

0.45 J/gC

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You now place both the Al and Fe in the same oven and let them sit overnight…..

Al

Fe

0.89 J/gC

0.45 J/gC

Which substance will be at the higher temperature at the end of this time?

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You now place both the Al and Fe in a boiling water bath and let them sit overnight…..

Al

Fe

0.89 J/gC

0.45 J/gC

Which substance has absorbed more heat energy to get to this temperature?Let’s do a calculation!

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Al

Fe

0.89 J/gC

0.45 J/gC

Which substance has absorbed more heat energy to get to this temperature?

Q = s∙m∙T

QAl = 0.89∙10∙100 = 890

QFe = 0.45∙10∙100 = 450

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H2O (l)Ethanol

4.184 J/g°C2.44 J/g°C

Which substance would be a better coolant in a car’s radiator? Why?

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Calculations with Specific Heat

Q = Heat (J)s = specific heat (J/g·Cº)m = mass of sample (g)T = change in temperature (ºC) (Tfinal – Tinitial)J = Joules

Q = smT

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Calculations with Specific Heat

Example 1: How much heat is absorbed when the temperature of a 125 g piece of aluminum increases from 35ºC to 65ºC. The specific heat capacity of Aluminum is 0.90 J/g·ºC.

Q = smT

Q = 0.90 J g· oC

x 125g 1

x 30oC 1

= 3375 J 3400 J (2 sf)

Q = ? Js = 0.90 J/g·Cºm = 125 gT = change in temperature (ºC) (Tfinal – Tinitial)

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Calculations with Specific Heat

Example 2: The temperature of a 95.4g piece of copper increases from 25.0ºC to 48.0ºC when the copper absorbs 849 J of heat. What is the specific heat of copper?

Q = smT

s = Q mT

s = 849 J 95.4 g ∙ 23.0oC

s = .387 J g ∙ oC

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Q

= 849 J

s =

?

m =

95.4 g

T = 23.0ºC)

(

T

final

–

T

initial

)

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Calorimetry:

The measurement of heat energy transferred during a physical or chemical process

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Calorimetry:

The measurement of heat energy transferred during a physical or chemical processThe heat energy that is gained or lost is absorbed or released into water in a calorimeter

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Calorimetry:

The measurement of heat energy transferred during a physical or chemical processThe heat energy that is gained or lost is absorbed or released into water in a calorimeterHeat lost by process = heat gained by water

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Calorimetry:

Heat lost by metal = heat gained by water

smetalmmetalTmetal

= swatermwaterTwater

smmmTm

= swmwTw

MetalSpecific heat x mass x temp change

water = Specific heat x mass x temp change

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Calorimetry Experiment

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Heat lost = Heat Gained Calculations

smmmTm

= swmwTw

Example 1: A piece of metal with a mass of 678 g is heated to 99.8°C and dropped into 250.0 mL of water at 26.0°C. The final temperature of the system is 27.1°C. What is the specific heat of the metal?

s = ? J/g·Cºm = 678 gT = -72.7 ºC (Tfinal – Tinitial)

Q = s = 4.184 J/g·Cºm = 250.0 gT = 1.1 ºC (Tfinal – Tinitial)

First determine heat gained by water

Q = 4.184 ∙ 250.0g ∙ 1.1 = 1150J

Now determine specific heat of metal

s = Q mT

s = -1150 J 678 g ∙ -72.7oC

s = .0233 J g ∙ oC

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Q

=

-

1150J

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Heat lost = Heat Gained Calculations

smmmTm

= swmwTw

Example 2: A piece of metal is heated to 76.8°C and dropped into 75.0 mL of water at 22.0°C.The final temperature of the system is 27.1°C. The specific heat of the metal is .98 J/g°C. What is the mass of the metal piece?

MetalQ = -1600Js = .98 J/g·Cºm = ? gT = 27.1 - 76.8 = -49.7 ºC (Tfinal – Tinitial)m = -1600 (.98) · (-49.7) = 32.85 so 33g

Water – do first!Q = 4.184 · 75.0 · 5.1 = 1600Js = 4.184 J/g·Cºm = 75.0 gT = 27.1- 22.0 = 5.1 ºC (Tfinal – Tinitial)

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Changes in Heat Energy within a Chemical Reaction

Energy can be gained or lost in the form of heat during a chemical reactionWhen heat energy is gained (or absorbed), a reaction is called endothermicWhen heat energy is lost (or released), a reaction is called exothermicThe change in heat is represented by the symbol H

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Surroundings

E

n

e

rgy



H

=

negative

E

x

othermic

Sys

tem

Heat Released

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Exothermic Reactions

An exothermic reaction will feel warm or hot to the touch because you feel the heat being released from the reaction!

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+ 2H 2O + Heat

CH 4 + 2O2  CO2

C

H 4 + 2O2

E

nergy

Heat

CO

2

+

2 H 2O

EXOTHERMIC

 H = negative

Reactants

Products

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Surroundings



H

=

positive

Endothermic

Sys

t

e

m

E

n

ergy

Heat Absorbed

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Endothermic Reactions

An endothermic reaction will feel cool or cold to the touch because the heat from your hand is being absorbed by the reaction!

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N2 + O2

E

n

ergy

Heat

2

N

O

N

2

+ O 2 + heat 

 2 N O

ENDOTHERMIC

H = positive

Reactants

Products

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Change in Energy (Heat) H

The change in heat energy ( H) in a chemical reaction is called enthalpy H = negative (-) = exothermicH = positive (+) = endothermic H is expressed in kJ (kilojoules)

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Calculations Using Heat Energy

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2 S + 3 O2  2 SO3 H = -791.4 kJ

Example 1: How much heat will be released when 3.000 moles of Sulfur react with excess O2 in the reaction above?

-791.4

kJ

Is the reaction exothermic or endothermic? How do you know?

3 mol S

1

2 mol S

= -1187 kJ

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H

2 + Br2  2 HBr H = 72.80 kJ

How much heat will be absorbed when 38.2 g of bromine reacts with excess H2 in the reaction above?

Is the reaction exothermic or endothermic?

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H

2 + Br2  2 HBr H = 72.80 kJ

How much heat will be absorbed when 38.2 g of bromine reacts with excess H2 in the reaction above?

38.2 g Br2

159.80

g Br2

1 mol Br2

1

mol Br2

72.80 kJ

= 17.4 kJ of heat energy absorbed

Is the reaction exothermic or endothermic?

1

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HESS’S LAW

Sometimes reactions can occur in steps rather than as single reactionExample: 2C + H2 ---> C2H2This reaction actually occurs in 3 steps:C2H2 + 5/2 O2 ---> 2CO2 + H2OC + O2 ---> CO2H2 + ½ O2 ---> H2O

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HESS’S LAW

Regardless of the number of steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. 2C(s) + H2(g) ---> C2H2(g) H = ??

C2H2 + 5/2 O2 ---> 2CO2 + H2O H = -1299.5 kJ C + O2 ---> CO2 H = -393.5 kJH2 + ½ O2 ---> H2O H = -285.8 kJ

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HESS’S LAW

Unfortunately, it is not as simple as just adding up all the Enthalpies. We must manipulate the various steps to the reaction until each matches the final reaction exactly.2C(s) + H2(g) ---> C2H2(g) H = ??

C2H2 + 5/2 O2 ---> 2CO2 + H2O H = -1299.5 kJ C + O2 ---> CO2 H = -393.5 kJH2 + ½ O2 ---> H2O H = -285.8 kJ

Which equation needs to be flipped

? What

is

its new H value?

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Steps to Solving Hess’ Law Problems

Circle the chemical formulas of interest in each step equationFlip the step equations so all reactants and products are in the same locations as in the reaction of interest. Remember to change the sign of H when you flip the equation.Multiply step equations so coefficients match the equation of interest. Remember to multiply H also.Cancel out chemical formulas that appear in both the reactant and product sides of the step equations.Sum all H values of the step equations.

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HESS’S LAW

Hess Law Video example

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Let’s do an example togetherGiven the following data:Calculate H for the reaction 2NO2  N2 + 2O2

N2 + 2O2 ---> N2O4 H = 9.6 kJ2NO2 ---> N2O4 H = -58.2 kJ

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Let’s do an example together:Given the following data:Calculate H for the reaction 2S + 2OF2  SO2 + SF4

OF2 + H2O ---> O2 + 2HF H = -277 kJSF4 + 2H2O ---> SO2 + 4HF H = 828 kJS + O2 ---> SO2 H = -297 kJ

2OF

2

+

2H

2O ---> 2O2 + 4HF H = -554 kJ (x2)SO2 + 4HF ---> SF4 + 2H2O H = -828 kJ2S + 2O2 ---> 2SO2 H = -594 kJ (x2)

flip

X 2

X 2

H =

-1976

kJ

H =

-1976

kJ

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States of Matter

Phase Changes, Heat of Fusion, and Heat of Vaporization

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Phases of Matter: (review)Solid matter that has definite volume and definite shapeLiquid matter that has definite volume but indefinite shapeGas matter that has indefinite volume and indefinite shape 

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Phase Change Descriptions:Melting the change from solid to liquid. Freezing the change from liquid to solid. Evaporation (vaporization) the change from liquid to gas . Condensation the change from gas to liquid . Sublimation the change from solid to gas . Deposition the change from gas to solid .

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Changes of State

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freezing

melting

condensing

evaporating

Phase Change Diagram (Heating/Cooling Curve)

Boiling Point

Melting

Point

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•Heat of fusion: (Hfus)the energy absorbed to change one mole of a substance from a solid to a liquid. (melting)

Heat of vaporization: (Hvap) the energy absorbed to change one mole of a substance from liquid to gas. (vaporizing)

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NO TEMPERATURE CHANGE OCCURS DURING A PHASE CHANGE!

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freezing

melting

condensing

evaporating



Hfus

Hvap

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When moving UP the curve, the Hfus and Hvap will be POSITIVE (+) because they are endothermic processes

When moving DOWN the curve, the Hfus and Hvap will be NEGATIVE (-) because they are exothermic processes

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freezing

melting

condensing

evaporating



Hfus

Hvap

Exothermic H = (-)

Endothermic H = (+)

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Calculations with Phase Changes

Example 1: How much energy would it take to completely melt a 7.20 mol sample of ice at 0°C? The heat of fusion (Hfus) of H2O is 6.02 kJ/mol.

7.20 mol H2O

6.02 kJ

1 mol H2O

1

= 43.3 kJ

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Calculations with Phase Changes

Example 2: How much energy would it take to completely melt a 15.0 g sample of ice at 0°C? The heat of fusion (Hfus) of H2O is 6.02 kJ/mol.What must we do first?

15.0g H20

18.20g H20

1 mol H20

1 mol H20

6.02kJ

= 4.96kJ

1

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Calculations with Phase Changes

Example 3: How much energy would be released when .456 g of water vapor (gas) at 100°C condense completely? The heat of vaporization (Hvap) of H2O is 40.7 kJ/mol.

.456g H20

18.20g H20

1 mol H20

1 mol H20

40.7kJ

= 1.02kJ

1

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