The thermal or internal energy of a sample is the sum of all the kinetic and potential energies of all the atoms and molecules in a sample

Phase Changes

Solid

Liquid

Gas

Melting Vaporization

CondensationFreezing

Heating Curves

A plot of temperature vs. time that represents the process in which energy is added at a constant rate

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Heating Curves Animation

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Melting - PE

Solid - KE

Liquid - KE

Boiling - PE

Gas - KE

Heating Curves

A plot of temperature vs. time that represents the process in which energy is added at a

constant rate

Temperature and Phase Change

• The temperature doesn’t change during a phase change.

• If you have a mixture of ice and water, the temperature is 0ºC

• At 1 atm, boiling water is 100ºC• You can’t get the temperature higher until

it boils

Chemical Energy2 parts of the universe as it relates to a

chemical reaction:• System

– the reactants and the products  

• Surroundings – everything else in the universe

(such as container, the room, etc.)

Law of Conservation of Energy: the total energy of the universe is constant and can neither be created nor destroyed; it can only be transformed.

The First Law of Thermodynamics: The total energy content of the

universe is constant

Chemical energy lost by combustion = Energy gained by the surroundings

system

surroundingsSigns (+/-) will tell you if energy is entering or leaving a system

+ indicates energy enters a system

- indicates energy leaves a system

Two types of processes based on energy flow:

• Exothermic– produces energy (heat flows out of the system)

• Endothermic– absorbs energy (heat flows into the system)

 

Chemical Energy

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

System

Ene

rgy

In this example, the energy of the reactants and products decreases, while the energy of the surroundings increases.

In every case, however, the total energy does not change.

Myers, Oldham, Tocci, Chemistry, 2004, page 41

Exothermic Reaction

Reactant Product + Energy

Before reaction After reaction

Conservation of Energy in a Chemical Reaction

Surroundings

System

Surroundings

SystemEne

rgy

Before reaction After reaction

In this example, the energy of the reactants and products increases, while the energy of the surroundings decreases.

In every case, however, the total energy does not change.

Myers, Oldham, Tocci, Chemistry, 2004, page 41

Endothermic Reaction

Reactant + Energy Product

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Thermochemistry• Every reaction has an energy change

associated with it• Energy is stored in bonds between atoms• Making bonds gives energy• Breaking bonds takes energy

Enthalpy and enthalpy changes

• To more easily measure and study the energy changes that accompany chemical reactions, chemists have defined a property called enthalpy.

• Enthalpy (H) is the heat content of a system at constant pressure.

Enthalpy and enthalpy changes• Although you cannot measure the actual

energy or enthalpy of a substance, you can measure the change in enthalpy, which is the heat absorbed or released in a chemical reaction.

• The change in enthalpy for a reaction is called the enthalpy (heat) of reaction (∆Hrxn).

• You have already learned that a symbol preceded by the Greek letter ∆ means a change in the property.

Enthalpy and enthalpy changes

• Thus, ∆Hrxn is the difference between the enthalpy of the substances that exist at the end of the reaction and the enthalpy of the substances present at the start.

• Because the reactants are present at the beginning of the reaction and the products are present at the end, ∆Hrxn is defined by this equation.

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.

DH = H (products) – H (reactants)

DH = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

DH < 0

Hproducts > Hreactants

DH > 06.4

reaction

reaction

Exothermic, heat given off & temperature of water rises

Endothermic, heat taken in & temperature of water drops

Energy Change in Chemical Processes

Exothermic process: H < 0 (at constant pressure)

Endothermic process: H > 0 (at constant pressure)

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

6.2

energy + H2O (s) H2O (l)

Enthalpy and enthalpy changes

Endothermic Reactions

Exothermic Reactions

Endothermic or Exothermic?

exothermic

endothermic

exothermic

endothermic

endothermic

Effect of Catalyst on Reaction Rate

reactants

products

Ene

rgy

activation energy for catalyzed reaction

Reaction Progress

No catalyst

Catalyst lowers the activation energy for the reaction.

What is a catalyst? What does it do during a chemical reaction?

Calculating Energy Changes - Heating Curve for Water

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DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp, liquid

Heat = mass x Dt x Cp, gas

Heat = mass x Dt x Cp, solid

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Heat of Fusion and Solidification• The heat that is absorbed by one mole of a

substance in melting at a constant temperature is the molar heat of fusion DHfus

• The heat lost when one mole of a liquid solidifies at a constant temperature is the molar heat of solidification DHsol

H2O (s) H2O (l) DHfus = 6.01 kJ/mol

H2O (l) H2O (g) DHsol = - 6.01 kJ/mol

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Heat of Vaporization&Condensation• The molar heat of vaporization:

– Heat needed to change one mol of a liquid to gas DHvap

• The molar heat of condensation:– Heat needed to change one mol of a gas to

liquid DHcon

H2O (l) H2O (g) DHvap = 40.7 kJ/mol

H2O (g) H2O (l) DHcon = - 40.7 kJ/mol

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Heat of Reaction• The heat that is released or absorbed in a

chemical reaction is equivalent to DH

C + O2(g) CO2(g) +394 kJ

C + O2(g) CO2(g) DH = -394 kJ

• In thermochemical equation it is important to say what state

H2O(g) H2(g) + ½ O2 (g) DH = 241.8 kJ

H2O(l) H2(g) + ½ O2 (g) DH = 285.8 kJ

Difference = 44.0 kJ

Hess’s Law

“In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one step or a series of steps.”

Hess’s Law

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Standard Heat of Formation• The change in heat that accompanies the formation of

a mole of a compound from its elements at standard conditions

• Standard conditions 25°C and 1 atm.

• Symbol is H◦f

The standard heat of formation of an element at its most stable form is 0

This includes the diatomics

39

• There are tables of heats of formations (pg. 316)

• For most compounds it is negative– Because you are making bonds– Making bonds is exothermic

• The heat of a reaction can be calculated by subtracting the heats of formation of the reactants from the products

H = H◦f (products) - H◦f (reactants)

Hess’s Law

Rules

1. If a reaction is reversed, the sign of ∆H must be reversed as well.

– because the sign tells us the direction of heat flow as constant P

2. The magnitude of ∆H is directly proportional to quantities of reactants and products in reaction.

If coefficients are multiplied by an integer, the ∆H must be multiplied in the same way.

– because ∆H is an extensive property

41

If H2(g) + 1/2 O2(g) H2O(l) H=-285.5 kJ/mol then

H2O(l) H2(g) + 1/2 O2(g) H =+285.5 kJ/mol

If you turn an equation around, you change the sign

2 H2O(l) 2 H2(g) + O2(g) H =+571.0 kJ/mol

If you multiply the equation by a number, you multiply the heat by that number.

– Twice the moles, twice the heat

Hess’s Law

42

• You make the products, so you need their heats of formation

• You “unmake” the reactants so you have to subtract their heats.

Hess’s Law

s)f(reactantr)f(productspreaction ΔHΣnΔHΣnΔH

https://www.youtube.com/watch?v=_NLAgSnqNOE&noredirect=1

Calculate the heat of combustion of methane, CH4

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

H◦f CH4 (g) = -74.86 kJ/mol

H◦f O2(g) = 0 kJ/mol

H◦f CO2(g) = -393.5 kJ/mol

H◦ fH2O(g) = - 241.8 kJ/mol

43

pg. 316

Hess’s Law Example Problem

Step #1: since 2 moles of water are produced by each mole of methane, we multiply the H◦ f. of water by 2.

H2 (g) + ½ O2 (g) H2O(g) 2x(- 241.8)= - 483.6kJ/mol

Calculate the heat of combustion of methane, CH4

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

H◦f CH4 (g) = +74.86 kJ/mol

H◦f O2(g) = 0 kJ/mol

H◦f CO2(g) = -393.5 kJ/mol

H◦ fH2O(g) = -483.6 kJ/mol

44

pg. 316

Hess’s Law Example Problem

Step #2: sum up all the H◦ f. :

H◦f = [-393.5 kJ/mol + (-483.6 kJ/mol)]- [-74.86 kJ/mol + (0 kJ/mol )]

H◦f = [-393.5 -483.6] + 74.86 = -877.1 + 74.86 = -802.2 kJ/mol

Hrxn = Hf(products) - Hf(reactants)

How do we relate change in temperature to the energy transferred?

Specific Heat capacity (J/oC) = heat supplied (J)

temperature (oC)

Specific Heat Capacity = heat required to raise the temperature of 1 gram of a substance object by 1oC

Affected by −What the substance is−Mass of the object

CalorimetryThe amount of heat absorbed or released during a physical or chemical change can be measured…

…usually by the change in temperature of a known quantity of water

1 calorie is the heat required to raise the temperature of 1 gram of water by 1 C

1 BTU is the heat required to raise the temperature of 1 pound of water by 1 F

– A device used to experimentally determine the amount of heat released or absorbed during a physical or chemical change

heat gained = heat lost

Calorimeter

Units of energy

Most common units of energy

1. S unit of energy is the joule (J), defined as 1 (kilogram•meter2)/second2, energy is also expressed in kilojoules (1 kJ = 103J).

2. Non-S unit of energy is the calorie.

One cal = 4.184 J or 1J = 0.2390 cal.

Units of energy are the same, regardless of the form of energy

Specific Heat Capacity aka Specific HeatThe amount of heat required to raise the temperature of one gram of

substance by one degree Celsius.

Substance Specific Heat (J/g·K)

Water (liquid) 4.18

Ethanol (liquid) 2.44

Water (solid) 2.06

Water (vapor) 1.87

Aluminum (solid) 0.897

Carbon (graphite,solid) 0.709

Iron (solid) 0.449

Copper (solid) 0.385

Mercury (liquid) 0.140

Lead (solid) 0.129

Gold (solid) 0.129

Page 296

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• The higher the specific heat the more energy it takes to change its temperature.

• Pizza burning the roof of your mouth

• The same amount of heat is released when an object cools down

Specific Heat Capacity

Calculations Involving Specific Heat

C = Specific Heat (J/ ºC.g)

Q = Heat lost or gained ( J)

T = Temperature change = Tf – Ti (ºC)

Q = m. C . T

How do we find the amount of heat energy?Change in energy =mass * specific heat * change in temp.

How much heat is need to raise 5 g of water 10 ̊C? (Water’s specific heat is 4.18 J/( ̊C.g)

1. Q = m * C * T

2. Q = m * C * T 5g * C * T 4.18 J/(g- ̊C) * T 10 ̊C

Q = m * C * T

Knownm= 5 KgT = 10 ̊CC = 4.18 J/(g- ̊C)

Unknownheat needed?

3. Q = 209 J

28,875 J of energy are added to a 5-kg piece of copper that has an initial temperature of 293 K. What will be the final temperature of this piece of copper? (Copper’s specific heat:385 J/(kg-K))

1. Q = m * C * T Knownm= 5 KgTi = 293 K

C = 385 J/(g- C)

UnknownFinal temperature?

1. Q = m * C * T

2. 28,875 J = m * C * T 5 kg * C * T 385 J/(kg-K) * T

3. T = 15 K

293 K + 15 K = 308 K

Calculating Energy Changes - Heating Curve for Water

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DH = mol x DHfus

DH = mol x DHvap

Heat = mass x Dt x Cp, liquid

Heat = mass x Dt x Cp, gas

Heat = mass x Dt x Cp, solid

Choose all that apply...C(s) + 2 S(g) CS2(l) H = 89.3 kJ

Which of the following are true?

A) This reaction is exothermic

B) It could also be written

C(s) + 2 S(g) + 89.3 kJ CS2(l)

C) The products have higher energy than the reactants

D) It would make the water in the calorimeter colder