Chapter 7: Hot & Cold Packs. Introductory Activity How many things can you think of in everday life...

Chapter 7: Hot & Cold Packs

Introductory Activity

How many things can you think of in everday life that either give off heat or absorb heat?

Which of these things are physical processes?

Which are chemical processes?

Hot & Cold Packs

This chapter will introduce the chemistry needed to understand how Hot & Cold Packs workSection 7.1: Endothermic & ExothermicSection 7.2: Calorimetry and heat capacitySection 7.3: Changes in StateSection 7.4: Heat of a Chemical ReactionSection 7.5: Hess’s Law

Hot/Cold Packs

Transfer of energy

Transfer of energy

Use

System & Surroundings


between

Materials ability to absorb energy

without noticeable temperature

change



change

Effect on temperature depends on

Physical change

Physical change

Is determined with

CalorimetryCalorimetry

Can be done in

Chemical change

Chemical change

Section 7.1—Endothermic and Exothermic

Why are hot packs “hot” and cold packs feel “cold”

Endothermic & Exothermic

When the system absorbs energy from the surroundings, it’s an endothermic process

When the system releases energy to the surroundings, it’s an exothermic process


It’s very important to define the system & surroundings correctly to use the endo- and exothermic definitions!

Most people define the system too broadly They include everything in the beaker or

container as the system

However, the system is only the molecules undergoing the change


The system is only made of the molecules undergoing the change

The water molecules & the container…

Your hand and the air…

Even the thermometer…

They are all the surroundings

Note that water is made up of water molecules—not a solid chunk of water…but for this picture, it’s best to represent water as one thing since it’s the surroundings and focus on the molecules reacting as the system.

Exothermic & You

You touch the beaker and it feels hot

Energy is being transferred TO YOU

You are the surroundings

When energy moves from system to surroundings, it’s exothermic

Exothermic & the Thermometer

The temperature (measured by the thermometer) is related to the average kinetic energy of the molecules in the container

The majority of the molecules in a solution are water

If the temperature is increasing, the energy of the water molecules is increasing

Since water is the surrounding (it’s not actually reacting), energy is being transferred to the surroundings

Exothermic shows an increase in the temperature within the container

Endothermic

The opposite is also trueIf the container feels cold to you, energy is

being transferred FROM YOU (the surroundings) into the system—endothermic

If the thermometer goes down, energy is being transferred FROM the water molecules (surroundings) into the system--endothermic

Let’s Practice

Example:Identify the system and surroundings when you hold an ice cube while it

melts. Is this endo- or exothermic?

Let’s Practice

Example:Identify the system and surroundings when you hold an ice cube while it

melts. Is this endo- or exothermic?

System: Water molecules in the form of iceSurroundings: You and the air

It feels cold to you…so energy is leaving you (surroundings)

When energy goes from surroundings to system it’s endothermic

Section 7.2—Calorimetry & Heat Capacity

Why do some things get hot more quickly than others?

Temperature

Temperature – proportional to the average kinetic energy of the molecules

Energy due to motion(Related to how fast the molecules are moving)

As temperature increases

Molecules move faster

Heat & Enthalpy

Heat (q)– The flow of energy from higher temperature particles to lower temperature particles

Under constant pressure (lab-top conditions), heat and enthalpy are the same…we’ll use the term “enthalpy”

Enthalpy (H)– Takes into account the internal energy of the sample along with pressure and volume

Energy Units

The most common energy units are Joules (J) and calories (cal)

4.18 J 1.00 cal

1000 J

1000 cal

1 kJ

1 Cal (food calorie)

=

=

=

Energy Equivalents

These equivalents can be used in dimensional analysis to convert units

Heat Capacity

Specific Heat Capacity (Cp) – The amount of energy that can be absorbed before 1 g of a substance’s temperature has increased by 1°C

Cp for liquid water = 1.00 cal/g°C or 4.18 J/g°C

Heat Capacity

High Heat Capacity Low Heat Capacity

Takes a large amount of energy to noticeably change temp

Small amount of energy can noticeably change temperature

Heats up slowly

Cools down slowly

Maintains temp better with small condition changes

Heats up quickly

Cools down quickly

Quickly readjusts to new conditions

A pool takes a long time to warm up and remains fairly warm over night.The air warms quickly on a sunny day, but cools quickly at night

A cast-iron pan stays hot for a long time after removing from oven.Aluminum foil can be grabbed by your hand from a hot oven because it cools so quickly

What things affect temperature change?

Heat Capacity of substanceThe higher the heat capacity, the slower the

temperature change

Mass of sampleThe larger the mass, the more molecules there are to

absorb energy, so the slower the temperature change

TCmH p

Energy added or removed

Mass of sample

Specific heat capacity of substance

Change in temperature

Positive & Negative T

Change in temperature (T) is always T2 – T1 (final temperature – initial temperature) If temperature increases, T will be positive

A substance goes from 15°C to 25°C. 25°C - 15°C = 10°CThis is an increase of 10°C

If temperature decreases, T will be negativeA substance goes from 50°C to 35°C35°C – 50°C = -15°CThis is a decrease of 15°C

Positive & Negative H

Energy must be put in for temperature to increaseA “+” T will have a “+” H

Energy must be removed for temperature to decreaseA “-” T will have a “-” H

Example

Example:If 285 J is added to 45 g of water at 25°C, what is the

final temperature? Cp water = 4.18 J/g°C

Let’s Practice #1

Example:How many joules must be

removed from 25 g of water at 75°C to drop the

temperature to 30°? Cp water = 4.18 J/g°C

Let’s Practice #2

Example:If the specific heat capacity of

aluminum is 0.900 J/g°C, what is the final temperature if 437 J is added to a 30.0 g

sample at 15°C

Calorimetry

1st Law of Thermodynamics – Energy cannot be created nor destroyed in physical or chemical changes

This is also referred to as the Law of Conservation of Energy

Conservation of Energy

If energy cannot be created nor destroyed, then energy lost by the system must be gained by the surroundings and vice versa

Calorimetry

Calorimetry – Uses the energy change measured in the surroundings to find energy change of the system

Hsurroundings = - Hsystem

Because of the Law of Conservation of Energy,The energy lost/gained by the surroundings is equal to but opposite of the energy lost/gained by the system.

(m×Cp×T)surroundings = - (m×Cp×T)system

Don’t forget the “-” sign on one sideMake sure to keep all information about surroundings together and all information about system together—you can’t mix and match!

Thermal Equilibrium – Two objects at different temperatures placed together will come to the same temperature

Two objects at different temperatures

So you know that T2 for the system is the same as T2 for the surroundings!

An example of CalorimetryExample:

A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water

at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of

the metal?

Metal:m = 23.8 gT1 = 100.0°CT2 = 32.5°CCp = ? Water:m = 50.0 gT1 = 24°CT2 = 32.5°CCp = 4.18 J/g°C Cp = 1.04 J/g°C

watermetal HH

waterpmetalp TCmTCm

CC

CgJgCCCg p

5.245.3218.40.500.1005.328.23

CCg

CCCg

JgC p

0.1005.328.23

5.245.3218.40.50

An example of CalorimetryExample:

A 23.8 g piece of unknown metal is heated to 100.0°C and is placed in 50.0 g of water

at 24°C water. If the final temperature of the water is 32.5°,what is the heat capacity of

the metal?

Let’s Practice #3Example:

A 10.0 g of aluminum (specific heat capacity is 0.900 J/g°C) at 95.0°C is placed in a

container of 100.0 g of water (specific heat capacity is 4.18 J/g°C) at 25.0°. What’s the

final temperature?

Section 7.3—Changes in State

What’s happening when a frozen ice pack melts?

The energy being put into the system is used for breaking IMF’s, not increasing motion (temperature)

Change in State

To melt or boil, intermolecular forces must be broken

Breaking intermolecular forces requires energy

A sample with solid & liquid will not rise above the melting point until all the solid is gone.

The same is true for a sample of liquid & gas

Melting

When going from a solid to a liquid, some of the intermolecular forces are broken

The Enthalpy of Fusion (Hfus) is the amount of energy needed to melt 1 gram of a substanceThe enthalpy of fusion of water is 80.87 cal/g or 334 J/g

All samples of a substance melt at the same temperature, but the more you have the longer it takes to melt (requires more energy).

fusHmH Energy needed to melt

Mass of the sample

Energy needed to melt 1 g

Example

Example:Find the enthalpy of

fusion of a substance if it takes 5175 J to melt 10.5 g of the

substance.

Vaporization

When going from a liquid to a gas, all of the rest of the intermolecular forces are broken

The Enthalpy of Vaporization (Hvap) is the amount of energy needed to boil 1 gram of a substanceThe Hvap of water is 547.2 cal/g or 2287 J/g

All samples of a substance boil at the same temperature, but the more you have the longer it takes to boil (requires more energy).

vapHmH Energy needed to boil

Mass of the sample

Energy needed to boil 1 g

Example

Example:If the enthalpy of

vaporization of water is 547.2 cal/g, how many calories are

needed to boil 25.0 g of water?

Solid

Liquid

Gas

Melting

Vaporizing or Evaporating

Condensing

Freezing

Incr

easi

ng m

olec

ular

mot

ion

(tem

pera

ture

)

Changes in State go in Both Directions

Going the other way

The energy needed to melt 1 gram (Hfus) is the same as the energy released when 1 gram freezes. If it takes 547 J to melt a sample, then 547 J would be

released when the sample freezes. H will = -547 J

The energy needed to boil 1 gram (Hvap) is the same as the energy released when 1 gram is condensed. If it takes 2798 J to boil a sample, then 2798 J will be

released when a sample is condensed. H will = -2798 J

Example

Example:How much energy is

released with 157.5 g of water is condensed?

Hvap water = 547.2 cal/g

Heating curve of water

-50

0

50

100

150

Energy input

Te

mp

era

ture

Heating Curves

Melting & Freezing

Point

Boiling & Condensing

Point

Heating curves show how the temperature changes as energy is added to the sample


-50

0

50

100

150

Energy input

Te

mp

era

ture

Going Up & Down

+H

-H

Moving up the curve requires energy, while moving down releases energy


-50

0

50

100

150

Energy input

Te

mp

era

ture

States of Matter on the Curve

Gas OnlyEnergy added

increases temp

Liquid & gasEnergy added breaks remaining IMF’s

Liquid OnlyEnergy added

increases temp

Solid & LiquidEnergy added breaks IMF’s

Solid OnlyEnergy added

increases temp


-50

0

50

100

150

Energy input

Te

mp

era

ture

Different Heat Capacities

Gas OnlyCp = 0.48 cal/g°C

Liquid OnlyCp = 1.00 cal/g°C

Solid OnlyCp = 0.51 cal/g°C

The solid, liquid and gas states absorb water differently—use the correct Cp!


-50

0

50

100

150

Energy input

Te

mp

era

ture

Changing States

Liquid & gasHvap = 547.2 cal/g

Solid & LiquidHfus = 80.87 cal/g


-50

0

50

100

150

Energy input

Te

mp

era

ture

Adding steps together

If you want to heat ice at -25°C to water at 75°C, you’d have to first warm the ice to zero before it could melt.

Then you’d melt the ice

Then you’d warm that water from 0°C to your final 75°

You can calculate the enthalpy needed for each step and then add them together

Example:How many calories are

needed to change 15.0 g of ice at -12.0°C to steam at

137.0°C?

ExampleUseful information:Cp ice = 0.51 cal/g°C

Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C

Hfus = 80.87 cal/gHvap = 547.2 cal/g

TCpmH

fusHmH

vapHmH

Example:How many needed to

change 40.5 g of water at 25°C to steam at 142°C?

Let’s PracticeUseful information:Cp ice = 0.51 cal/g°C

Cp water = 1.00 cal/g°CCp steam = 0.48 cal/g°C

Hfus = 80.87 cal/gHvap = 547.2 cal/g

TCpmH

fusHmH

vapHmH

Section 7.4—Energy of a Chemical Reaction

What’s happening in those hot/cold packs that contain chemical reactions?

Enthalpy of Reaction

Enthalpy of Reaction (Hrxn) – Net energy change during a chemical reaction

+Hrxn means energy is being added to the system—endothermic-Hrxn means energy is being released from the system—exothermic

Enthalpy of Formation

Enthalpy of Formation (Hf) – Energy change when 1 mole of a compound is formed from elemental states

Heat of formation equations: H2 (g) + ½ O2 (g) H2O (g) C (s) + O2 (g) CO2 (g)

A table with Enthalpy of Formation values can be found in the Appendix of your text

Be sure to look up the correct state of matter:H2O (g) and H2O (l) have different Hf values!

The overall enthalpy of reaction is the opposite of Hf for the reactants and the Hf for the products

Reactants are broken apart and Products are formed.

Breaking apart reactants is the opposite of Enthalpy of Formation.

Forming products is the Enthalpy of Formation.

reactHprodHH ffrxn

Hrxn = sum of Hf of all products – the sum of Hf reactants

Enthalpy of Formation & Enthalpy of Reaction

This is not the way a reaction occurs—reactants break apart and then rearrange…remember Collision Theory from Chpt 2! But for when discussing overall energy changes, this manner of thinking is acceptable.

Example

Example:Find the Hrxn for:

CH4 (g) + 2 O2 (g) 2 H2O (g) + CO2 (g)

Hf (kJ/mole)

CH4 (g) -74.81

O2 (g) 0

H2O (g) -241.8

CO2 (g) -393.5

Let’s Practice #1

Hf (kJ/mole)

CH3OH (l) -238.7

O2 (g) 0

H2O (l) -285.8

CO2 (g) -393.5

Example:Find the Hrxn for:

2 CH3OH (l) + 3 O2 (g) 2 CO2 (g) + 4 H2O (l)

Enthalpy & Stoichiometry

The Enthalpy of Reaction can be used along with the molar ratio in the balanced chemical equation

This allows Enthalpy of Reaction to be used in stoichiometry equalities

Example:If 1275 kJ is released, how many grams of B2O3 is

produced?B2H6 (g) + 3 O2 (g) B2O3 (s) + 2 H2O (g)

H = -2035 kJ

Example

Let’s Practice #2

If you need to produce 47.8 g B2O3, how many kilojoules will be released?

B2H6 (g) + 3 O2 (g) B2O3 (s) + 2 H2O (g) H = -2035 kJ

Section 7.5—Hess’s Law

How can we find the enthalpy of a reaction using step-wise reactions?

Hess’s Law

Hess’ Law – The sum of the energy changes during a series of reactions is equal to the sum of the reaction.

In other words…if you go from Reactant A to Product Z all in one step, you will have the same total energy change as someone that went from A to Z in 7 step—the energy from each of their 7 steps would add up to your 1 step energy change.

Example

1Label each step-wise equations with letters (“a”, “b”, “c”) if not already done for you.

Calculate the enthalpy of the reaction 2N2 (g) + 5O2 (g) 2N2O5 (g) Hrxn = ?

Use: 2 H2(g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ

abc

For the first reactant in the overall reaction, find the step-wise reaction that has the same chemical and the same state of matter. It doesn’t have to be on the reactants side of the step-wise equation

If it is on the correct “side” write it as is. Write it’s label beside it, too (“a”, “b”)

If it’s on the wrong “side”, flip the equation. If you flip it, write it’s label as “-a” or “-b”.

2



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

Example



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one.

3

Example



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

Repeat Step 2 for each reactant & product in the overall equation. If a reactant appears in more than one step-wise reaction, skip that reactant or product and move onto the next one.

3

-b 2HNO3 (l) N2O5 (g) + H2O (l)

Example

-b 2 HNO3 (l) N2O5 (g) + H2O (l)


Use: 2 H2 (g) + O2 (g) 2 H2O (l) Hrxn = -571.6 kJ N2O5 (g) + H2O (l) 2 HNO3 (l) Hrxn = -76.6 kJ N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l) Hrxn = -74.1 kJ

abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

Use any un-used step-wise equations to get rid of unwanted things. Putting them on opposite sides will allow them to cancel.

4

-a 2 H2O (l) 2 H2 (g) + O2 (g)

Example

-b 2 HNO3 (l) N2O5 (g) + H2O (l)



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

-a 2 H2O (l) 2 H2 (g) + O2 (g)

Begin to cancel things out that appear on both the reactants and products side. Your goal is to add up all the step-wise equations to equal the overall equation.

Multiply equations by a whole number if you need more of something to match the overall reaction or to fully cancel something out that you don’t want in the overall equation.

5

2 2 6 2 4

Example

-b 2 HNO3 (l) N2O5 (g) + H2O (l)



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

-a 2 H2O (l) 2 H2 (g) + O2 (g)



5

2 2 6 2 42 4 2 2

Example

-b 2 HNO3 (l) N2O5 (g) + H2O (l)



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

-a 2 H2O (l) 2 H2 (g) + O2 (g)



5

2 2 6 2 42 4 2 2

5

2 N2 (g) + 5 O2 (g) 2 N2O5 (g)

Example

-b 2 HNO3 (l) N2O5 (g) + H2O (l)



abc

c N2 (g) + 3 O2 (g) + H2 (g) 2 HNO3 (l)

-a 2 H2O (l) 2 H2 (g) + O2 (g)

2 2 6 2 42 4 2 2

5

2 N2 (g) + 5 O2 (g) 2 N2O5 (g)

Use the step-wise “labels” as a math expression for solving for Hrxn.

6

2 × (-74.1 kJ)2 × -(76.6 kJ)1 × -(-571.6 kJ)

270.2 kJ

Example

What did you learn about hot/cold packs?

Hot/Cold Packs

Transfer of energy

Transfer of energy

Use



between



change



change

Effect on temperature depends on

Physical change

Physical change

Is determined with

CalorimetryCalorimetry

Can be done in

Chemical change

Chemical change

Chapter 7: Hot & Cold Packs. Introductory Activity How many things can you think of in everday life...

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Transcript of Chapter 7: Hot & Cold Packs. Introductory Activity How many things can you think of in everday life...