Transcript AP Ch 06 apchapt6r

Chapter 6
Energy
Thermodynamics
6.1 Nature of Energy
• The ability to do work or produce heat.
• Conserved - can be converted from one form to
another but can neither be created nor
destroyed.
• Work is a force acting over a distance.
• Potential: due to position or composition - can
be converted to work.
• Kinetic: due to motion of the object.
KE = 1/2 mv2
(m = mass, v = velocity)
Heat and Temperature
• Temperature reflects random motion of
particles in a substance.
• Heat is the measure of energy content.
• Heat is energy transferred between
objects because of temperature
difference.
• State Function - property of a system that
depends only on its present state.
• Independent of the path, or how you get
from point A to B.
The Universe
• Is divided into two halves, the system and the
surroundings.
• The system is the part you are concerned
with.
• The surroundings are the rest.
• Exothermic reactions release energy to the
surroundings.
• Endothermic reactions absorb energy from
the surroundings.
Potential energy
CH 4 + 2O 2  CO 2 + 2H 2 O + Heat
CH 4 + 2O 2
Heat
CO 2 + 2 H 2 O
N 2 + O 2 + heat  2NO
Potential energy
2NO
Heat
N2 + O2
Direction
•
Every energy measurement has three
parts.
1. A unit ( Joules or calories).
2. A number - how many.
3. A sign to tell direction.
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•
Negative - exothermic
Positive- endothermic
Surroundings
System
Energy
DE <0
Surroundings
System
Energy
DE >0
Same rules for heat and work
• Heat given off is negative.
• Heat absorbed is positive.
• Work done by system on surroundings
is negative.
• Work done on system by surroundings
is positive.
• Thermodynamics - The study of energy
and the changes it undergoes.
•
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First Law of Thermodynamics
• The energy of the universe is constant.
• Law of conservation of energy.
• q = heat
• w = work
 DE = q + w
• Take the systems point of view to
decide signs.
What is work?
•
•
•
•
•
Work is a force acting over a distance.
work = force  distance
since pressure = force / area,
work = pressure  volume
Work can be calculated by multiplying
pressure by the change in volume at
constant pressure.
• units of L•atm
Work needs a sign
• If the volume of a gas increases, the system
has done work on the surroundings.
• work is negative
• wsystem = PDV
• Expanding work is negative.
• Contracting, surroundings do work on the
system w is positive.
• 1 L•atm = 101.3 J
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Examples
• Calculate the DE for a system undergoing an
endothermic process in which 15.6 kJ of heat
flows and where 1.4 kJ of work is done on the
system. EX. 6.1
• What amount of work is done when 46L of
gas is expanded to 64 L at 15 atm pressure?
EX. 6.2
• If 2.36 J of heat are absorbed by the gas
above. What is the change in energy? 1 L•atm
= 101.3 J
6.2 Enthalpy
• Abbreviated H
• H = E + PV (that’s the definition), at constant
pressure.
 DH = DE + PDV
• the heat at constant pressure qp can be
calculated from:
DE = qp + w = qp - PDV
qp = DE + P DV = DH
• Where qP = DH at constant pressure.
 DH = energy flow as heat (at constant
pressure).
Examples
• When 1 mole of methane (CH4) is burned
at constant pressure, 890 kJ of energy is
released as heat. Calculate the DH for a
process in which 5.8 g sample of methane
is burned at a constant pressure.
• Consider the following reaction:
2H2(g) + O2(g)  2H2O(l) DH=-572kJ
How much heat is evolved when 2.56 g of
hydrogen is reacted with excess oxygen?
EX 6.4
#34b
Calorimetry
• Measuring heat. We use a calorimeter.
• The heat capacity for a material, C, is
calculated.
• C = heat absorbed/DT = DH/ DT
• specific heat capacity
heat capacity per gram = J/°C•g or J/K•g
• molar heat capacity
heat capacity per mole = J/°C•mol or
J/K•mol
Calorimetry
• Constant pressure calorimeter (coffee
cup calorimeter, used for solutions).
• heat = specific heat x m x DT
• heat = molar heat x moles x DT
• Make the units work and you’ve done
the problem right.
• A coffee cup calorimeter measures DH.
• The specific heat of water is 1 cal/gºC
(4.184 J/gºC)
• Heat of reaction= DH = s x mass x DT
Examples
• The specific heat of graphite is 0.71
J/gºC. Calculate the energy needed to
raise the temperature of 75 kg of
graphite from 294 K to 348 K.
• A 46.2 g sample of copper is heated to
95.4ºC and then placed in a calorimeter
containing 75.0 g of water at 19.6ºC.
The final temperature of both the water
and the copper is 21.8ºC. What is the
specific heat of copper?
Calorimetry
• Constant volume calorimeter is called a
bomb calorimeter.
• Material is put in a container with pure
oxygen. Wires are used to start the
combustion. The container is put into a
container of water.
• The heat capacity of the calorimeter is
known and tested.
• Since DV = 0, PDV = 0, DE = q
Bomb Calorimeter
• thermometer
• stirrer
• full of water
• ignition wire
• Steel bomb
• sample
Properties
• Intensive properties - not related to the
amount of substance.
• Ex. density, specific heat, temperature.
• Extensive property - does depend on
the amount of stuff.
• Ex. heat capacity, mass, heat from a
reaction.
6.3 Hess’s Law
•
•
Enthalpy is a state function.
The change in enthalpy is the same whether
the reaction takes place in one step or a
series of steps.
• We can add equations to to come up with
the desired final product, and add the DH.
• Two rules:
1. If the reaction is reversed the sign of DH is
changed.
2. If the reaction is multiplied, so is DH.
Rules
1. If a reaction is reversed, DH is also reversed.
N2(g) + O2(g)  2NO(g)
2NO(g)  N2(g) + O2(g)
DH = 180 kJ
DH = 180 kJ
2. If the coefficients of a reaction are multiplied
by an integer, DH is multiplied by that same
integer.
6NO(g)  3N2(g) + 3O2(g)
DH = 540 kJ
Examples
• When using Hess’s Law, work by adding
the equations up to make it look like the
answer.
• Make the other compounds cancel out.
• N2(g) + 2O2(g)  2NO2(g) DH1 =68kJ
• Above reaction is carried out in two steps
below:
• N2(g) + O2(g)  2NO(g) DH2 =180kJ
• 2NO(g) + O2(g)  2NO2(g) DH3 =-112kJ
H (kJ)
2NO, O2
-112 kJ
180 kJ
N2 O2
2NO2
68 kJ
Practice
• Given(BDVD)
5
C 2 H 2 (g) + O 2 (g)  2CO 2 (g) + H 2 O( l)
2
DHº= -1300. kJ
C(s) + O 2 (g)  CO 2 (g)
DHº= -394 kJ
1
H 2 (g) + O 2 (g)  H 2 O(l)
2
DHº= -286 kJ
Calculate DHº for this reaction
2C(s) + H 2 (g)  C 2 H 2 (g)
Practice
Given
O 2 (g) + H 2 (g)  2OH(g) DHº= +77.9kJ
O 2 (g)  2O(g) DHº= +495 kJ
H 2 (g)  2H(g) DHº= +435.9kJ
Calculate DHº for this reaction
O(g) + H(g)  OH(g)
Practice
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P4(s) + 6Cl2(g)  4PCl3(g)
P4(s) + 5O2(g)  P4O10(s)
PCl3(g) + Cl2(g)  PCl5(g)
PCl3(g) + 1/2O2(g)  Cl3PO(g)
Calculate the DH for the reaction #58
P4O10(s) + 6PCl5(g)  10Cl3PO(g)
DH = -1225.6kJ
DH = -2967.3kJ
DH = - 84.2kJ
DH = -285.7 kJ
6.4 Standard Enthalpies of Formation
Standard States
• Compound
 For
a gas, pressure is exactly 1 atmosphere.
 For a solution, concentration is exactly 1 molar.
 Pure substance (liquid or solid), it is the pure
liquid or solid.
• Element
 The
form [N2(g), K(s)] in which it exists at 1 atm
and 25°C.
Standard Enthalpy of Formation
• The enthalpy change that occurs in the formation of
one mole of a compound for a reaction at standard
conditions (25ºC, 1 atm, 1 M solutions).
• Symbol DHºf
• There is a table in Appendix 4 (pg A21) It is a table
of standard heats of formation. The amount of heat
needed to for 1 mole of a compound from its
elements in their standard states.
Standard Enthalpies of Formation
• Need to be able to write the equations.
• What is the equation for the formation of
NO2 ?
• ½N2 (g) + O2 (g) ® NO2 (g)
• Have to make one mole to meet the
definition.
• Write the equation for the combustion of
methanol CH3OH.
• What is the equation for the formation of
solid aluminum oxide?
Since we can manipulate the equations
• We can use heats of formation to figure
out the heat of reaction.
• Lets do it with this equation. Ex.6.11
• CH3OH(l) + 3O2(g) ® 2CO2 (g) + 3H2O(l)
DH rxn  (DH products) -(DH reactants)
o
o
f
o
f
Thermite Reaction
• Using enthalpies of formation, calculate the
standard change in enthalpy for the thermite
reaction.Ex 6.10
2Al(s) + Fe2O3(s) --> Al2O3(s) + 2Fe(s)
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It’s a gas!
• Methanol (CH3OH) is often used as a
fuel in high performance engines in race
cars. Using the data in table 6.2,
compare the standard enthalpy of
combustion per gram of methanol with
that per gram of gasoline. Gasoline is
actually a mixture of compounds, but
assume for this problem that gasoline is
pure liquid octane (C8H18)