Download 6.1. Define and calculate kinetic and potential energy.

Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
6.1. Define and calculate kinetic
and potential energy. [Reading 6.16.3 Problems 10, 11, 12, 34, & 36]
Energy
• Define Energy
• Energy is the capacity to do work or to
supply heat
Kinetic Energy
KE=1/2mv2
Units of energy
1
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Kinetic Energy
KE=1/2mv2
Units of energy
2
 m
2 −2
KE = kg *
= kgm s = J
s
Calculation
Calculate the kinetic energy of a jogger
weighing 180 lbs who runs at a speed of 20
mph.
180 lbs* 1 kg/2.2 lbs = 81.8 kg
Potential Energy
Potential energy is stored energy
Boulder on the side of a hill
Water behind a dam
A stretched spring
PE=mgh
Potential energy converts to kinetic energy
2
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Potential Energy
Chemical bonds store energy, thus contain
potential energy
CH4 (g) + 2 O2 (g) –––> CO2 (g) + 2 H2O (g)
+ heat
The potential energy contained in the C-H and
O=O bonds is converted to heat
Energy Associated with a Water
Molecule
KE - motion
1. Rotational
2. Vibrational
3. Translational
Energy Associated with a Water
Molecule
KE - motion
1. Rotational
2. Vibrational
3. Translational
PE - stored
1. Bonds holding molecule together
2. Forces holding electrons to nucleus
3. Forces holding nucleus together
3
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Kinetic Theory of Matter
• As temperature increases, the average
velocity of molecules increases
• KE = 1/2 mv2
• Temperature is related to KE
• As temperature increases, the average
velocity of the molecules increases
Kinetic Energy Distribution in Matter
Which curve represents a gas at a higher temperature?
For each curve what is the most probable kinetic energy?
Why is the energy distribution in 1 narrower than 2?
Heat Transfer
• Heat transfer is the transfer of kinetic
energy
• direction of
transfer
4
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Energy Convention
The system is defined in a chemical reaction
as the chemicals which are reacting
∆E = Ef - Ei
∆E = Eproducts - Ereactants
Energy Convention
∆E = Ef - Ei
If the system releases energy, i.e. energy
flows from the system to the surroundings,
the sign of ∆E is negative
Ef < Ei
∆E < 0
Exothermic Reaction - releases energy to the
surroundings.
Energy Convention
∆E = Ef - Ei
If the system absorbs energy, i.e. energy flows
from the surroundings to the system, the
sign of ∆E is positive
Ef > Ei
∆E > 0
Endothermic Reaction - consumes energy
from the surroundings.
5
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
6.2. Relate the Thermodynamic
quantities of heat, energy, work,
and enthalpy to the
Thermodynamic Equation.
[Reading 6.4 Problems 14, 15, 16,
17, 18, 19, 20, 21, 22, & 38]
Energy
Found in different forms; heat, light, electricity, &
mechanical.
Study of energy called thermodynamics.
3 laws of thermodynamics
1st law: amount of energy is constant.
It is neither created nor destroyed but merely
converted.
Energy is classified as being kinetic (motion) or
potential (stored).
E = KE + PE
Heat and work
Definition of terms
System - the mixture which is reacting
Surroundings - everything else in the universe
Closed vs. open system - An insulated system
is considered a closed system
6
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Closed System
In a closed system, Adiabatic, there is no heat
flow out of the system, ∆E = 0
∆E = ∆KE + ∆PE
Thus, ∆KE = – ∆PE
Open Systems
The increase in kinetic energy of the system
can be offset by the flow of heat energy to
the surroundings
An Exothermic process
The decrease in kinetic energy of the system
is offset by the flow of heat energy into the
system
An Endothermic process
State Functions
The value of the function depends only on the
initial and final state, not on the path
traveled
• Temperature?
• Displacement?
• Pressure?
• Work?
7
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
H - State Function
Copyright 2000 John Wiley and Sons, Inc.
State function - a function or variable whose value
depends only on the initial and final states of the system
and not on the path taken from initial to final state.
•Energy as a State Function
•
•
•
•
•
∆E = Efinal – Einitial
In a chemical reaction
∆E = Eproducts – Ereactants
E=q+w
Sign convention for q and w
First Law of Thermodynamics
E = q+w
Copyright 2000 John Wiley and Sons, Inc.
8
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Work = P V
Expansion Work
• Work is defined as a force (which moves an
object) x distance moved
• w=Fxd
• Pressure volume work is most common in
chemical systems (PV work)
• In a chemical reaction, if the number of
moles of gas increases, ∆V >0 thus P∆V > 0
and work will w < 0 . Work is done by the
system
Expansion Work
• In a chemical reaction, if the number of
moles of gas decrease
• ∆V is negative
• w = – P∆V thus work is positive.
• Work is done to the system
9
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
•Energy and Enthalpy
• Energy can be passed from a system to its
surroundings as either heat or work (or
both)
• ∆E = q + w
• Since w = – P∆V
∆E = q – P∆V
• q = ∆E + P∆V
• q, is defined as the heat energy gained or
lost
•q, at constant volume
•
•
•
•
In a closed reaction chamber
At constant volume, P∆V = 0
∆E = qv
The energy transferred is entirely heat
energy
Bomb Calorimeter
Copyright 2000 John Wiley and Sons, Inc.
10
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
•q, at constant pressure
•
•
•
•
In an open flask on the bench top
P∆V ≠ 0
Thus, qp = ∆E + P∆V
q at constant pressure is defined as
Enthalpy (H) ==> qp = ∆H
• ∆H = Hfinal - H initial
• ∆H = Hproducts - Hreactants
6.3 Use thermal properties of
matter to calculate the energy
changes associated with a
calorimetry study [Reading 6.5
Problems 40, 42, 44, 48, 50, &
52]
Thermal Properties of Matter
Specific Heat (symbol is c)– the energy
required to raise the temperature of 1 g of a
substance by 1 °C
The unit calorie is defined as the amount of
energy required to raise the temperature of 1
g of water from 14.5 to 15.5 °C.
1 cal = 4.184 J/g°C
Specific heat of water is 4.184 J/g°C
11
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Thermal Properties of Matter
Heat capacity = specific heat x mass of object
Heat capacity depends on the size of the
sample
The amount of heat lost or gained by a system
is Q
Q = m c ∆T
Practice Problem
• Lead has a specific heat of 0.128J/g°C.
How much heat must be added to a 5.0 g
piece of lead in order to raise its
temperature by 5.0°C?
• The specific heat of copper is 0.387J/g°C.
What temperature change would the same
amount of heat as the lead example cause in
5.0 g of copper.
Thermal Properties of Matter
Molar heat capacity
Units of J/mol°C
1.0 g * 1 mol/18.0g = 0.0556 mol
12
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Thermal Properties of Matter
Molar heat capacity
Units of J/mol°C
1.0 g * 1 mol/18.0g = 0.0556 mol
4.18J
1g°C
=
4.18 J
0.0556mol°C
= 75.2
J
mol° C
Calorimetry
qmetal = – qH2O
Calorimetry Practice Problem
Suppose that the metal cylinder in the
preceding video released 3.55kJ of energy
into the water. If the water was at an initial
temperature of 25.0°C, what would be the
final temperature. (The specific heat of
water is 4.184J/g°C
13
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Calorimetry Practice Problem
When 2.50 g of NaOH is dissolved in 100.0
ml of water in a coffee cup calorimeter at
25.00°C, the temperature of the water
increases to 31.32°C. Assume that the
density of water is 1.00g/ml and that the
specific heat of the solution is 4.18 J/g°C.
Calculate ∆H for the dissolution reaction in
kJ/mol NaOH.
6.4. Given the thermochemical
equation for a reaction, calculate
the heat absorbed or released for a
specific amount of reactant or
product. [Reading 6.6 Problems 56,
58, 60, & 99]
•Standard State
•
•
•
•
Standard state is defined as:
Pressure = 1 atm
T = 25°C
1 M concentration
14
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
The Thermodynamic Equation
2Al(s) + Fe2O3(s) ––> Al2O3(s) + 2 Fe (s)
∆H° = – 852kJ
The Thermodynamic Equation
2Al(s) + Fe2O3(s) ––> Al2O3(s) + 2 Fe (s)
∆H° = – 852kJ
• Enthalpy
– The amount of heat energy gained or lost by
the system
• If ∆H<0, then the reaction is exothermic
The Thermodynamic Equation
2Al(s) + Fe2O3(s) ––> Al2O3(s) + 2 Fe (s)
∆H° = – 852kJ
15
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
The Thermodynamic Equation
C3H8(g) + 5O2(g) ––> 3CO2(g)+ 4H2O (g)
∆H° = – 2043kJ
The phase, or state, of each reactant must
be specified in the reaction
C3H8(g) + 5O2(g) –> 3 CO2(g) + 4 H2O (l)
∆H° = – 2219kJ
The Thermodynamic Equation
Ba(OH)2• 8 H2O (s) + 2 NH4Cl (s) ––>
BaCl2(aq) + 2 NH3 (l) + 10 H2O (l)
∆H° = + 80.3 kJ
If ∆H>0, then the reaction is endothermic
Heat is absorbed by the system from the
universe
Practice with the concept
According to the following reaction, how
much heat would be released if 3.52 g of
propane reacted with excess oxygen?
C3H8(g) + 5O2(g) ––> 3CO2(g)+ 4H2O (g)
∆H° = – 2043kJ
How much heat would be released if 3.65 g of
water were produced in the reaction?
16
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
6.5. Apply Hess’ Laws to
determine ∆H for reactions.
[Reading 6.7 Problems 64, 66, 68,
70, 72, & 74]
Hess’s Law
How does this diagram illustrate Hess’s law?
Hess’ Law(s)
1. Heat (Enthalpy)
amount of material reacting.
consider: A --> B + ∆
if 2g of A releases 50kJ of heat, 4g will release 100kJ.
2. The reverse reaction has the same amount of heat
(enthalpy) with opposite sign.
if A --> B has ∆H = -15kJ
then B --> A has ∆H = +15kJ
3. Heats (Enthalpies) are additive.
if A --> B ∆H = x
and B --> C ∆H = y
then A --> C ∆H = x + y
17
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Hess’ Law(s)
Allows one to calculate ∆Hrxn for an
unknown reaction from known reactions.
Uses the properties of multiplication (1),
reversing reactions (2), and the additive
propetries of reactions (3)
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Want CO2 as a product thus reverse rxn 2
18
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Want CO2 as a product thus reverse rxn 2
CO(g) + 1/2 O2(g) --> CO2(g)
∆Hº f = –283.0 kJ
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Want 1 C as a reactant thus divide rxn 1 by 2
CO(g) + 1/2 O2(g) --> CO2(g)
∆Hº f = –283.0 kJ
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Want 1 C as a reactant thus divide rxn 1 by 2
CO(g) + 1/2 O2(g) --> CO2(g)
C(s) + 1/2O2(g) --> CO(g)
∆Hº f = –283.0 kJ
Hºf = -110.5 kJ
19
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Add rxns to get desired rxn
CO(g) + 1/2 O2(g) --> CO2(g)
C(s) + 1/2O2(g) --> CO(g)
∆Hº f = –283.0 kJ
Hºf = -110.5 kJ
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Add rxns to get desired rxn
CO(g) + 1/2 O2(g) --> CO2(g)
1/2O2(g) --> CO(g)
C (s) + O2)g) ––> CO 2
∆Hº f = –283.0 kJC(s)
Hºf = -110.5 kJ
H° = H 1 + H2
C (s) + O2(g) --> CO2(g)
Find ∆H° for the above reaction given the following reactions.
2 C(s) + O2(g) --> 2 CO(g)
∆Hº f = -221.0 kJ
CO2(g) --> CO(g) + 1/2 O2(g)
∆Hºf = +283.0 kJ
Add rxns to get desired rxn
CO(g) + 1/2 O2(g) --> CO2(g)
C(s) + 1/2O2(g) --> CO(g)
C (s) + O2)g) ––> CO 2
∆Hº f = –283.0 kJ
Hºf = -110.5 kJ
H° = – 393.5 kJ
20
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
6.6 Apply standard ∆Hf to
determine ∆Hrxn of reactions.
[Readings 6.8 Problems76, 78, 80,
82, & 89]
Standard Heats of Formation
• The amount of energy absorbed or evolved
when one mole of substance is formed from
elements in their standard states
2 H2 (g) + O2 (g) –––> 2 H2O (l) ∆H° = – 571.8 kJ
Is ∆H° = ∆Hf°?
Standard Heats of Formation
• The amount of energy absorbed or evolved
when one mole of substance is formed from
elements in their standard states
2 H2 (g) + O2 (g) –––> 2 H2O (l)
∆H° ≠ ∆Hf°
21
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Standard Heats of Formation
• The amount of energy absorbed or evolved
when one mole of substance is formed from
elements in their standard states
2 H2 (g) + O2 (g) –––> 2 H2O (l)
∆H° ≠ ∆Hf°
H2 (g) + 1/2 O2 (g) –––> H2O (l)
Standard Heats of Formation
• The amount of energy absorbed or evolved
when one mole of substance is formed from
elements in their standard states
2 H2 (g) + O2 (g) –––> 2 H2O (l)
∆H° ≠ ∆Hf°
H2 (g) + 1/2 O2 (g) –––> H2O (l)
∆H° = – 285.9 kJ = ∆Hf°
Standard Heats of Formation
• The amount of energy absorbed or evolved
when one mole of substance is formed from
elements in their standard states
• NO (g) + 1/2 O2 (g) –––> NO2 (g)
• ∆H° = – 56.57 kJ
• ∆H° ≠ ∆Hf°
• Not formed from elements in std. state
22
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Standard Enthalpies – Table
6.2 & Appendix E
∆Hºrxn = Σ∆Hºf(prod) - Σ∆Hºf(react)
CO(g) + NO(g) --> CO 2(g) + 1/2 N2(g)
∆Hºf = -110.5 kJ
+90 kJ
mol
mol
x 1 mol
x 1 mol
-110.5 kJ
+90 kJ
|__________|
-20.5 kJ
-393.5 kJ
0 kJ
mol
mol
x 1 mol
x 1/2 mol
-393.5 kJ
0 kJ
|__________|
-393.5 kJ
∆Hº rxn = (-393.5 kJ) - (-20.5 kJ) = -373 kJ
Practice with the concept
Calculate the enthalpy of the following
reaction.
2 NOCl (g) –––> 2 NO (g) + Cl2 (g)
∆Hf°(kJ/mol)
NOCl
51.7
NO
90.3
Is this reaction exothermic?
23
Chapter 6 Notes
CHM 1045
Dr. Palmer Graves, Instructor
Practice
Calculate ∆H° for the following reaction
given that the heats of formation of CH4(g),
CH3Cl(g) and HCl(g) are respectively: 74.85kJ/mol, -82.0kJ/mol, and -92.30
kJ/mol.
CH4 (g) + Cl2(g) ––> CH3Cl (g) + HCl (gP
24