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# Chapter 15: Energy and Chemical Change

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## Sydney Sturgeon

on 10 February 2017

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#### Transcript of Chapter 15: Energy and Chemical Change

Chapter 15: Energy and Chemical Change
First Law of Thermodynamics
Law of Conservation of Energy:
States that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.
Energy
Chemical Potential Energy:
Energy that is stored in a substance because of its composition.

Heat:
Represented by "q"
A form of energy that transfers from 1 object to another because of a temperature difference.
Lancer Literacy Notebook #3: Energy
Take out your Lancer Literacy Notebook and write a paragraph describing your definition of ENERGY.
Energy
Energy: ability to do work or produce heat.
Feb. 3-10
2 basic forms of energy:
(1) Potential (PE) - energy of position or composition
(2) Kinetic (KE) - energy of motion
Chemical systems contain both KE and PE.
Measuring Heat
Units for measuring heat:

calorie (cal):
Amount of energy required to raise the temperature of one gram of pure water by one degree Celsius

Calorie:
Nutritional Calorie
1000 calories equal 1 nutritional Calorie

Joule (J):
SI unit of energy and heat
Measuring Heat
Energy Unit Conversion Factors (STAR THESE!!!):
1 J = 0.2390 cal

1 cal = 4.184 J

1 Calorie = 1 kcal = 1000 cal
Specific Heat
SPECIFIC HEAT: Amount of heat required to raise the temperature of one gram of a substance by one degree Celsius
Since different substances have different compositions, each substance will have their own specific heat value (c)

Large values for specific heat means that it takes a lot of energy to raise the temperature of that substance (STAR THIS!!!)
Water: 4.184 J/g C
Gold: 0.129 J/g C
Specific Heat/Heat Equation
q = mc T
q (or Q) = heat absorbed or released (units = J or kJ)
Positive q means that heat energy was gained (endothermic)
Negative q means heat energy was lost (exothermic)
m = mass of the sample (units = g)
c = specific heat of the substance (units = J/g C)
T = change in temperature in Celsius (T - T )
final
initial
Specific Heat Example #1
If the temperature of 34.4 g of ethanol increases from 25.0 C to 78.8 C, how much heat has been absorbed by the ethanol?
Specific Heat Example #2
The temperature of a sample of iron with a mass of 10.0 g changed from 50.4 C to 25.0 C with the release of 114 J. What is the specific heat of iron?
Chapter 15: Specific Heat WS
Specific Heat Example #1
q = mc T
q = ?
m = 34.4 g
c = 2.44 J/g C
T = 78.8 C - 25.0 C = 53.8 C
q = (34.4 g)(2.44 J/g C)(53.8 C)
q = 4,515.76 J
Specific Heat Example #2
q = mc T
q = -114 J (neg. - released)
m = 10.0 g
c = ?
T = 25.0 C - 50.4 C = -25.4 C
c =
q
_________
m T
c =
(-114 J)
(10.0 g)(-25.4 C)
_____________________
c = 0.45 J/g C
Calorimetry
Study of the HEAT and ENERGY transfer in a system.
CALORIMETER: An insulated device used for measuring the amount of heat absorbed or released during a chemical or physical process.
Calorimetry Practice Example
You heat up a 4.68 g piece of metal, put it into a foam cup calorimeter, and find that the temperature in the calorimeter increases by 182 C. You also calculate the amount of heat that's absorbed by the piece of metal to be 256 J. What is the specific heat of the metal?
Calorimetry Practice Example
q = mc T
q = 256 J
m = 4.68 g
c = ?
T = 182 C
Thermochemistry
Almost every chemical reaction and change of state either RELEASES or ABSORBS heat.
THERMOCHEMISTRY: Study of heat changes that accompany chemical reactions and phase changes.
3 Parts in Thermochemistry
(1) SYSTEM: Specific part of the universe that contains the reaction or process you wish to study.
(2) SURROUNDINGS: Everything in the universe other than the system.
(3) UNIVERSE: The system plus the surroundings.
Universe = System + Surroundings
When heat flows from the system to the surroundings = exothermic
When heat flows from the surroundings to the system = endothermic
Enthalpy (H)
ENTHALPY (H): Heat content of a system at constant pressure.
ENTHALPY (HEAT) OF REACTION: Change in enthalpy for a reaction
Thermochemical Equations
THERMOCHEMICAL EQUATION: Balanced chemical equation that includes the physical states of all reactants and products and the energy change (usually change in enthalpy, H)
Thermochemical Equations
Enthalpy (heat) of combustion: Enthalpy change for the complete burning of one mole of that substance. Units: kJ/mol
Standard enthalpy changes have the symbol H
Standard conditions are 1 atm and 298 K - don't confuse this with STP
Changes of State
Not only do chemical reactions release and absorb energy but changes of state also release and absorb energy.
Molar enthalpy (heat) of vaporization: Heat required to vaporize one mole of a liquid.
Units: kJ/mol

Molar enthalpy (heat) of fusion: Heat required to melt one mole of a solid.
Units: kJ/mol
H
H
vap
fus
Changes of State
When you reverse the process you switch the sign.
H = - H
H = - H
vap
fus
solid
cond
Change from a solid to a liquid to a gas the change in enthalpy is positive because it absorbs energy.

Change from a gas to a liquid to a solid the change in enthalpy is negative because it releases energy.
Energy Calculation Example #1
How much heat is released in the combustion of 54.0 g of glucose (C H O ) in a calorimeter? ( H = -2808 kJ/mol)
Energy Calculation Example #1
Molar mass glucose = 180.18 g/mol
Energy Calculation Example #2
How much energy is released by the combustion of 206 g of hydrogen gas? ( H = -286 kJ/mol)
Before you sit down pick up the following from the green demo table:

Specific Heat LAB WS (white)
c =
______
q
m T
c =
______________
(256 J)
(4.68 g)(182 C)
c = 0.30 J/g C
H
rxn
H = H - H = H - H
rxn
final
initial
products
reactants
Heat of Fusion and Heat of Vaporization
comb
6
12
6
comb
Energy Calculation Example #2
54.0 g glucose
0.30 mol glucose
= 0.30 mol glucose
= - 842.40 kJ
1 mol glucose
X
180.18 g glucose
______________________
-2808 kJ
1 mol glucose
X
_______________________
Molar mass hydrogen = 2 mol H (1.01 g/mol) = 2.02 g/mol
206 g hydrogen
1 mol hydrogen
X
2.02 g hydrogen
= 101.98 mol hydrogen
__________
101.98 mol hydrogen
X
-286 kJ
1 mol hydrogen
= -29,166.28 kJ
___________
SPECIFIC HEAT POP QUIZ!
Take out a new sheet of paper and write your name and "Specific Heat Pop Quiz" at the top. You may use your notes but not your neighbor.
Specific Heat Pop Quiz
1. When 10.2 g of canola oil at 25.0 C is placed in a wok, 3.34 kJ of heat is required to heat it to a temperature of 196.4 C. What is the specific heat of the canola oil (pay close attention to your units)?

2. What does it mean when a substance has a high specific heat value?
Hess's Law
States that if you can add two or more thermochemical equations to produce a final equation for a reaction, then the sum of the enthalpy changes for the individual reactions is the enthalpy change for the final reaction.
STEP #1
Step 1: Chemical equations must include the desired substances and have known enthalpy changes.
EXAMPLE:
2S(s) + 3O (g) 2SO (g) H = ?
S(s) + O (g) SO (g) H = -297 kJ
2SO (g) 2SO (g) + O (g) H = 198 kJ
STEP #2
Step 2: Equations should be balanced individually as well as with the other equations. If equations aren't balanced with each other multiply each coefficient and also the enthalpy change by the number needed to balance all equations.
2S(s) + 2O (g) 2SO (g) H = 2(-297 kJ) = -594 kJ
EXAMPLE:
STEP #3
Step 3: In the desired equation, if the desired reactant is a product or the desired product is a reactant, reverse the chemical equation and reverse the sign of the enthalpy change.
EXAMPLE:
2SO (g) + O (g) 2SO (g) H = -198 kJ
STEP #4
Step 4: Add all the equations together as well as the changes in enthalpy. Cancel any terms that are common to both sides of the chemical equation.
EXAMPLE:
2S(s) + 2O (g) 2SO (g) H = -594 kJ
2SO (g) + O (g) 2SO (g) H = - 198 kJ
____________________________________
2S(s) + 3O (g) 2SO (g) H = -792 kJ
2S(s) + 3O (g) + 2SO (g) 2SO (g) + 2SO (g)
Hess's Law Video
Hess's Law Example
Use the two equations below to determine H for the decomposition of hydrogen peroxide: [2H O (l) 2H O(l) + O (g)]
2H (g) + O (g) 2H O(l) H = -572 kJ
H (g) + O (g) H O (l) H = -188 kJ
Hess's Law Example
Standard Enthalpy (Heat) of Formation
H
f
The change in enthalpy that accompanies the formation of a compound in its standard state from its elements in their standard states (units usually kJ/mol).
STEP #1
Step 1: Find the chemical equations that have the desired products and reactants and also the standard enthalpies of formation for those equations.
STEP #2
Step 2: Rewrite each equation to show the reactants and products desired. Reverse the sign of the enthalpies of formation if the chemical equation was reversed.
STEP #3
Step 3: Multiply any chemical equation and its standard enthalpy of formation by a number that will balance all the equations.
STEP #4
Step 4: Add all the equations together as well as their standard enthalpies of formation. Cancel out any terms that are the same on both sides of the chemical equation.
The Summation Equation
Summation Equation Example
Summation Equation Example
Chapter 15: Hess's Law and Standard Enthalpies of Formation WS
2
3
2
2
2
2
3
Multiply 1st equation by 2.
2
2
Reverse 2nd equation.
2
2
3
2
2
2
2
3
2
2
2
3
2
3
2
2
2
2
2
2
2
2
2
2
2
2H (g) + O (g) 2H O(l) H = -572 kJ
2H O (l) 2H (g) + 2O (l) H = 2(188 kJ) = 376 kJ
____________________________________________________________
Reverse 2nd equation and multiply it by 2.
2
2
2
2
2
2
2
2H (g) + O (g) + 2H O (l) 2H (g) + 2O (g) + 2H O(l)
2H O (l) 2H O(l) + O (g) H = -196 kJ
2
2
2
2
2
2
2
2
2
2
2
EXAMPLE:
H S(g) + 4F (g) 2HF(g) + SF (g) H = ?
rxn
2
2
6
H (g) + F (g) HF(g) H = -273 kJ
S(s) + 3F (g) SF (g) H = -1220 kJ
H (g) + S(s) H S(g) H = -21 kJ
f
f
f
2
2
6
2
2
2
1
1
_
_
2
2
EXAMPLE:
H S(g) H (g) + S(s) H = 21 kJ
Reverse the 3rd equation.
2
2
f
EXAMPLE:
EXAMPLE:
Multiply 1st equation and standard enthalpy of formation by 2.
H (g) + F (g) 2HF(g) H = 2(-273 kJ) = -546 kJ
2
2
f
H (g) + F (g) 2HF(g) H = -546 kJ
S(s) + 3F (g) SF (g) H = -1220 kJ
H S(g) H (g) + S(s) H = 21 kJ
__________________________
2
2
2
6
2
2
f
f
f
H (g) + F (g) + S(s) + 3F (g) + H S(g) 2HF(g) + SF (g) + H (g) + S(s)
2
2
2
2
6
2
H S(g) + 4F (g) 2HF(g) + SF (g) H = -1745 kJ
2
2
6
rxn
f
f
rxn
H = H (products) - H (reactants)
rxn
f
f
H = standard enthalpy of the reaction
= sum of the terms
H (products) = standard enthalpies of formation of all the products
H (reactants) = standard enthalpies of formation of all the reactants
Use the standard enthalpies of formation to calculate H for the combustion of methane [CH (g) + 2O (g) CO (g) + 2H O(l)]
rxn
4
2
2
2
f
f
f
f
2
2
2
4
H (CO ) = -394 kJ
H (H O) = -286 kJ
H (CH ) = -75 kJ
H (O ) = 0.0 kJ
H = [(1)(-394 kJ) + (2)(-286 kJ)]
- [(1)(-75 kJ) + (2)(0.0 kJ)]
rxn
= -394 kJ - 572 kJ + 75 kJ - 0.0 kJ
H = -891 kJ
rxn

Chemical Equations Pop Quiz
Take out a new sheet of paper and write your name and "Chemical Equations Pop Quiz" at the top. You may use your notes, the textbook, and your phone, but not your neighbor (or anybody else).
Chemical Equations Pop Quiz
Write and balance a chemical equation for each of these word equations:

1. Hydrogen reacts with nitrogen to form ammonia.

2. The combustion of glucose (C H O ).
6
12
6
Chemical Equations Pop Quiz
1. Hydrogen reacts with nitrogen to form ammonia.

2. The combustion of glucose (C H O ).

Answer: C H O + 6O 6CO + 6H O
2
2
3
6
12
6
2
2
2
12
6
6
Before you sit down pick up the following from the green demo table:

Chapter 15: Sec. 1 Specific Heat NOTES (white)
specific heat of iron
Before you sit down pick up the following from the green demo table:

Chapter 15: Sec. 2-5 Heat, Calorimetry, and Heating Curve NOTES (white)
ABSOLUTE ZERO
At absolute zero particles stop moving - 0 K, -273.15 C, or -459.58 F
No pressure because no particles would be moving.

Pressure is created by particles hitting into the side of a container.

The higher the temperature, the higher the KE, the greater the movement of particles, and so the greater the pressure created.
Endothermic vs. Exothermic Rxns
Endothermic:
+ H
A + B + heat C + D
Feels colder
Exothermic:
- H
A + B C + D +heat
Feels warmer
Hot and Cold Pack Demo
HEATING CURVE OF WATER
pg. 531 Problem Solving Lab - heating curve of water
pg 531 - Problem Solving LAB
SPECIFIC HEAT LAB
THERMOCHEMISTRY
Study of the energy and heat associated with chemical reactions and physical changes.
o
Specific Heat Examples #1 & #2 on WHITEBOARD
o
o
TAKE OUT A CALCULATOR AND A PERIODIC TABLE ALSO
o
o
o
o
o
o
HEATING CURVE: simple line graph that shows the phase changes a given substance undergoes with increasing and decreasing temperature.
Heating Curve Calculation Example on WHITEBOARD
Full transcript