Transcript thermochemistry chapter 5x

THERMOCHEMISTRY
Energy
• The ability to do work or transfer heat.
– Work: Energy used to cause an object
that has mass to move.
– Heat: Energy used to cause the
temperature of an object to rise.
Definitions #1
Energy: The capacity to do work or produce heat
Potential Energy: Energy due to
position or composition
Kinetic Energy: Energy due to the
motion of the object
1 2
KE  mv
2
Definitions #2
Law of Conservation of Energy: Energy
can neither be created nor destroyed, but
can be converted between forms
The First Law of Thermodynamics: The
total energy content of the universe is
constant
Definitions #3
Internal energy:
The internal energy of a system is identified with
the random, disordered motion of molecules; the
total (internal) energy in a system includes
potential and kinetic energy.
The sample is at a height from the ground
etc). Symbol for Internal Energy Change: ΔU
1) The internal energy of a system
__________.
• A) is the sum of the kinetic energy of all of its
components
• B) is the sum of the rotational, vibrational, and
translational energies of all of its components
• C) refers only to the energies of the nuclei of the
atoms of the component molecules
• D) is the sum of the potential and kinetic energies
of the components
• E) none of the above
Definitions #4
Enthalpy:
Enthalpy is the amount of heat
content used or released in a
system at constant pressure
E = q + w
E = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving to the surroundings
+q if energy is entering from the surroundings
w = work done by, or on, the system
-w if work is done by the system on the
surroundings
+w if work is done on the system by the
surroundings
2) Which one of the following
conditions would always result in an
decrease in the internal energy of a
system?
• A) The system loses heat and does work
on the surroundings.
• B) The system gains heat and does work
on the surroundings.
• C) The system loses heat and has work
done on it by the surroundings.
• D) The system gains heat and has work
done on it by the surroundings.
• E) None of the above is correct.
• 3) The change in the internal
energy of a system that releases
2,500 J of heat and that does
7,655 J of work on the
surroundings is __________ J.
• A) -10,155
B) -5,155
C) 1.9x107
D) 10,155
E) 5,155
Work problems
Chapter 5
• 5.25
• 5.27 A and B
• 5.31 All
Calorimetry
The 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
The Joule
The unit of heat used in modern thermochemistry
is the Joule
1 kg  m
1 joule1 newton m eter
2
s
1 joule = 4.184 calories
2
A Bomb
Calorimeter
A Cheaper
Calorimeter
Specific Heat
The 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
Calculations Involving Specific Heat
q  s  m  T
OR
q
s
m  T
s = Specific Heat Capacity
q = Heat lost or gained
T = Temperature change
4) The temperature of a 15-g
sample of lead metal increases
from 22 °C to 37 °C upon the
addition of 29.0 J of heat. The
specific heat capacity of the lead
is __________ J/g-K.
A) 7.8
B) 1.9
C) 29
D) 0.13
Problems
• 5.53 a and B
State Functions depend ONLY on the
present state of the system
ENERGY IS A
STATE FUNCTION
A person standing at the
top of Mt. Everest has the
same potential energy
whether they got there by
hiking up, or by falling
down from a plane!
WORK IS NOT A
STATE FUNCTION
WHY NOT???
State Functions
Usually we have no way of knowing the
internal energy of a system; finding that
value is simply too complex a problem.
State Functions
• However, we do know that the internal
energy of a system is independent of the
path by which the system achieved that
state.
– In the system below, the water could have
reached room temperature from either
direction.
State Functions
• Therefore, internal energy is a state
function.
• It depends only on the present state of the
system, not on the path by which the
system arrived at that state.
• And so, E depends only on Einitial and Efinal.
State Functions
• However, q and w
are not state
functions.
• Whether the battery
is shorted out or is
discharged by
running the fan, its
E is the same.
– But q and w are
different in the two
cases.
Work
When a process
occurs in an open
container, commonly
the only work done
is a change in volume
of a gas pushing on
the surroundings (or
being pushed on by
the surroundings).
Work
We can measure the work done by the
gas if the reaction is done in a vessel
that has been fitted with a piston.
w = −PV
Work, Pressure, and Volume
w   PV
Expansion
+V (increase)
-w results
Esystem decreases
Work has been done
by the system on the
surroundings
Compression
-V (decrease)
+w results
Esystem increases
Work has been done
on the system by the
surroundings
Energy Change in Chemical Processes
Endothermic:
Reactions in which energy flows into the
system as the reaction proceeds.
+ qsystem
- qsurroundings
Exothermic:
Reactions in which energy flows out of
the system as the reaction proceeds.
- qsystem
+ qsurroundings
Endothermic Reactions
Exothermic Reactions
• 5) Which one of the following is an
exothermic process?
• A) ice melting
B) water evaporating
C) boiling soup
D) condensation of water vapor
• E) Ammonium thiocyanate and
barium hydroxide are mixed at 25
°C: the temperature drops
Enthalpy
H = E + PV
At constant pressure and volume the change
in enthalpy is the heat gained or lost
H = q
Enthalpies of Reaction
The change in
enthalpy, H, is
the enthalpy of the
products minus the
enthalpy of the
reactants:
H = Hproducts −
Hreactants
Enthalpies of Reaction
This quantity, H, is called the enthalpy
of reaction, or the heat of reaction.
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
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
Step #1: CH4 must appear on the reactant
side, so we reverse reaction #1 and change the
sign on H.
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
C + O2  CO2
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
-393.50 kJ
Step #2: Keep reaction #2 unchanged,
because CO2 belongs on the product side
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
C + O2  CO2
2H2 + O2  2 H2O
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
-393.50 kJ
-571.66 kJ
Step #3: Multiply reaction #3 by 2
Hess’s Law Example Problem
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Reaction
C + 2H2  CH4
C + O2  CO2
H2 + ½ O2  H2O
CH4  C + 2H2
C + O2  CO2
2H2 + O2  2 H2O
CH4 + 2O2  CO2 + 2H2O
Ho
-74.80 kJ
-393.50 kJ
-285.83 kJ
+74.80 kJ
-393.50 kJ
-571.66 kJ
-890.36 kJ
Step #4: Sum up reaction and H
Calculation of Heat of Reaction
Calculate H for the combustion of methane, CH4:
CH4 + 2O2  CO2 + 2H2O
Hrxn =  Hf(products) -   Hf(reactants)
Substance
CH4
O2
CO2
H2O
Hf
-74.80
0
-393.50
-285.83
kJ
kJ
kJ
kJ
Hrxn = [-393.50kJ + 2(-285.83kJ)] – [-74.80kJ]
Hrxn = -890.36 kJ
13) Given the following reactions the enthalpy of the
reaction in which sulfur dioxide is oxidized to sulfur
trioxide 2SO2 (g) +O2 (g)  2SO3(g) is
__________ kJ.
2SO2  2S + 2O2
ΔH = 594 kJ
2SO3  2S + 3 O2 ΔH = 790 kJ
6.
2SO2 
keep
2SO3 
2S +
2S
2S +
3 O2 
Ans: -196kJ
+ 2O2
ΔH = 594 kJ
3 O2 ΔH = 790 kJ reverse
2SO3
ΔH = -790 kJ
Problems
5.63