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
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)
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.”
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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
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• 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
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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
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• 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
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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
Tem
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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
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