Transcript Chapter 5: Thermochemistry

Chapter 5: Thermochemistry
The Study of Energy and its Transformations
Chapter 5: Thermochemistry
The Study of Energy and its Transformations
The Nature of Energy
Energy is the capacity to do work or transfer heat
Work: energy used to make an object move
Heat: energy used to increase the temperature of an object
Chapter 5: Thermochemistry
(1) Kinetic Energy (Ek)
Energy of motion – a moving object has kinetic energy
• moving car
• molecules and atoms are constantly moving (thermal motion)
1
Ek  mv 2
2
mass
speed (velocity)
Chapter 5: Thermochemistry
(1) Potential Energy (Ep)
Energy an object has due to its position relative to
other objects
• potential energy due to gravity acting on the object
(diver jumping off a diving board)
• potential energy due to electrostatic attraction between
two charged objects
charge
Eel 
 Q1 Q2
d
distance between charges
• potential energy due to chemical energy stored in bonds of
molecules
Chapter 5: Thermochemistry
Different forms of energy are interconvertible:
Potential Energy
Kinetic Energy
Chemical Energy
Chapter 5: Thermochemistry
The Units of Energy:
A person (50kg) moving at a speed of 1m/s has a kinetic energy:
2
1 2 1
kg m 2
m

Ek  mv  (50 kg )  1   25
2
2
 s
s2
 25 J
1kJ = 1000 J
A calorie is an older energy unit:
1cal = 4.184 J
1Cal = 1kcal = 1000cal
Chapter 5: Thermochemistry
The System and its Surroundings:
Energy/Heat is transferred between regions of the universe
Surroundings
Region studied: system
Molecules cannot escape:
closed system
System
Everything else: surroundings
Chapter 5: Thermochemistry
The System and its Surroundings:
2 H2
+
O2
→
2 H 2O
Chapter 5: Thermochemistry
The System and its Surroundings:
2 H2
+
O2
→
2 H 2O
+ energy
The reaction releases energy
which can be transferred from
the system to the surroundings
Chapter 5: Thermochemistry
The System and its Surroundings:
2 H2
+
O2
→
2 H 2O
+ energy
The reaction releases energy
which can be transferred from
the system to the surroundings
… until the temperatures are equal
Chapter 5: Thermochemistry
The System and its Surroundings:
Energy can be transferred
OR
from the system to the
surroundings
Energy can also transferred
from the surroundings to
the system
Heat is transferred from the hotter to the colder object
… until their temperatures are equal
Chapter 5: Thermochemistry
Energy can be transferred as heat (q)
Energy can also be transferred as work (w)
Chapter 5: Thermochemistry
First law of Thermodynamics
Energy is conserved
• energy is neither created nor destroyed, it can only be
converted from one form into another
Chapter 5: Thermochemistry
Internal Energy
• The internal energy of a system (E) is the sum of all
kinetic and potential energies
• When a system undergoes change (D), the internal
energy of the system may change by DE
DE = Efinal - Einitial
Chapter 5: Thermochemistry
Internal Energy
Heat (q) transferred from
system to surroundings:
Heat (q) transferred from
surroundings to system:
Efinal < Einitial
Efinal > Einitial
DE < 0
DE > 0
Chapter 5: Thermochemistry
Internal Energy: sign conventions
DE = q + w
System
Heat (q) transferred TO surroundings
or work (w) done BY system:
negative sign
-q, -w => DE < 0
Chapter 5: Thermochemistry
Internal Energy: sign conventions
System
Heat (q) transferred TO system
or
work (w) done TO system:
positive sign
+q, +w => DE > 0
Chapter 5: Thermochemistry
DE = q + w
Can the sign of DE be predicted for the following experiment?
q transferred
to system
+q
work done
by system
-w
Chapter 5: Thermochemistry
What is the change in internal energy, DE, of a system that
gains 150 J of heat and does 432 J of work on the surroundings?
relate heat and work to the change in internal energy, DE
DE = q + w
q = +150J
w = -432J
DE = 150J - 432J = -282 J
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
Endothermic process:
• heat flows into the system
Exothermic process:
• heat flows out of the system
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
heat +
Water
Water vapor
H2O (l)
H2O (g)
Endothermic:
system absorbs heat
H2O (l)
H2O (s)
Exothermic:
system produces heat
+ heat
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
H2O (g)
H2O (l)
+ heat
Exothermic:
system produces heat
Water vapor
Water
Chapter 5: Thermochemistry
Heat released/absorbed during a reaction (Enthalpy, H)
is an extensive property
+ lots of O2
CO2
+ heat
H2O
+ lots of O2
twice as much
CO2
+ twice as much
H2O
heat
two logs
Chapter 5: Thermochemistry
Enthalpy (H)
Enthalpy measures the amount of heat exchanged
at constant pressure:
DH = qp
• Enthalpy is an EXTENSIVE property
depends on the amount of substance
Chapter 5: Thermochemistry
Exothermic / Endothermic Processes
heat +
Water
Water vapor
H2O (l)
H2O (g)
DH > 0
Endothermic:
system absorbs heat
H2O (l)
H2O (s)
+ heat
Exothermic:
system produces heat
DH < 0
Chapter 5: Thermochemistry
Enthalpies of Reaction
DH = Hproducts - Hreactants
2 H2 (g) + O2 (g)
→
2 H2O (l) + heat
DH = -483.6 kJ
exothermic !
How much heat (q) is given off by reacting 3.4 g of H2 gas?
1mol H 2
3.4 g H 2 
2g H 2

 483 .6kJ
2mol H 2
  411.1 kJ
Chapter 5: Thermochemistry
2 H2O (l)
DH = -483.6 kJ
2 H2 (g) + O2 (g)
DH = +483.6 kJ
2 H2 (g) + O2 (g)
2 H2O (l)
→
→
DH is equal in magnitude, but opposite in sign, to DH for
the reverse reaction
Chapter 5: Thermochemistry
Calorimetry
… is a way to measure the amount of heat given off or
absorbed in the course of a reaction
+ NaOH
dissolves
heat (q)
Chapter 5: Thermochemistry
Calorimetry
… when the reaction occurs in a container that is
thermally insulated …
… heat given off by reaction
cannot escape to the outside!
NaOH dissolves
heat
(q)
… and temperature of the
solution rises!
By how much?
Chapter 5: Thermochemistry
Calorimetry
“Coffee-cup Calorimeter”
Chapter 5: Thermochemistry
Calorimetry
The temperature change that a substance undergoes when it
absorbs heat depends on its specific heat (s)
1 g of H2O
+ 4.184 J of heat
1 g of H2O
11oC
+ 1 cal of heat
12oC
The specific heat (s) is the amount of heat required to heat
1 g of a substance by 1K (or 1oC)
Chapter 5: Thermochemistry
Calorimetry
The amount of heat a substance absorbs to undergo a
temperature change, DT, depends on its specific heat (s)
q  DT  m  s
heat absorbed
or released
mass of
substance
specific heat of substance
Similarly, DT a substance undergoes when it absorbs/releases
heat, q, depends on its specific heat (s)
q
DT 
ms
Chapter 5: Thermochemistry
The temperature of an aqueous NaOH solution (250g) was found
to have risen from room temperature (21oC) to 30oC. How much
heat was transferred? The specific heat of water is 4.184 J/g-K.
q  DT  m  s
DT= 30oC – 21oC
= 9 oC
= 9K
m = 250 g
J
s = 4.184
gK
DTK  (To C final  273)  (To C initial  273)
 To C final  273  To C initial  273
 To C final  To C initial
J
q  9 K  250 g  4.184
gK
 9414 J  9.41 kJ
Chapter 5: Thermochemistry
The temperature of an aqueous NaOH solution (250g) was found
to have risen from room temperature (21oC) to 30oC. How much
heat was transferred? The specific heat of water is 4.184 J/g-K.
9.41 kJ were transferred – was the reaction exo- or endothermic?
NaOH (s)
→
Na+(aq)
+
OH- (aq)
+ heat
The reaction gives off heat => exothermic
Chapter 5: Thermochemistry
Calorimetry
Where are the system and the surroundings in a thermally
insulated calorimeter?
- both are part of what is inside the calorimeter!
system
Heat transfer occurs from the warmer
water to the ice-cubes in bag:
system:
H2O (s) + heat →
H2O (l)
endothermic process: DH >0
surroundings
surroundings (water) will cool down!
Chapter 5: Thermochemistry
A 5g piece of iron with a temperature of 60oC is plunged into 100g
of room-temperature water (21oC). The temperature of the water
rises to 21.21oC. What is the specific heat of iron? sH2O=4.18 J/g-K
What is actually happening?
Final temperatures:
100g H2O
21oC
Water: 21.21oC
Iron: 21.21oC
Iron
5g
60oC
Chapter 5: Thermochemistry
A 5g piece of iron with a temperature of 60oC is plunged into 100g
of room-temperature water (21oC). The temperature of the water
rises to 21.21oC. What is the specific heat of iron? sH2O=4.18 J/g-K
-q(Fe) = +q(H2O)
q(Fe) = DT(Fe) x m(Fe) x s(Fe)
q(H2O) = DT(H2O) x m(H2O) x s(H2O)
-DT(Fe) x m(Fe) x s(Fe) = DT(H2O x m(H2O x s(H2O)
s(Fe) = DT(H2O) x m(H2O) x s(H2O)
-DT(Fe) x m(Fe)
s(Fe) = 0.21K x 100g x 4.18 J/g-K
38.8K x 5g
s(Fe) = 0.45 J/g-K
DT(H2O) = (21.21-20)K
= +0.21K
DT(Fe) = (21.21-60)K
= -38.8K
Chapter 5: Thermochemistry
Hess’s law
H
H2O (g)
-44kJ
-50kJ
H2O (l)
-6 kJ
H2O (s)
Chapter 5: Thermochemistry
Hess’s law
Converting water vapor (steam) into ice at a constant temperature
can be done two steps: first, convert H2O into liquid water, then, in
a second step, into ice:
add:
H2O (g)
H2O (l)
DH1 = -44 kJ
H2O (l)
H2O (s)
DH2 = -6.0 kJ
H2O (g) + H2O (l)
net:
H2O (g)
H2O (l) + H2O (s)
H2O (s)
DH = DH1 + DH2
= -44kJ + (-6.0kJ)
= -50 kJ
Chapter 5: Thermochemistry
Hess’s law
If a reaction is carried out in a series of steps, DH for the overall
reaction will equal the sum of the enthalpy changes for the
individual steps
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
A
+
2B
→
D3
DH = ?
Given the following enthalpies of reaction:
a) A + B
b) D3
→
→
C
C + B
DH = -125 kJ
DH = -41 kJ
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
2B
→
D3
DH = ?
a) A + B
→
C
DH = -125 kJ
A
+
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
2B
→
D3
DH = ?
a) A + B
→
C
DH = -125 kJ
A
+
b) D3
→
C + B
DH = -41 kJ
→
DH = +41 kJ
invert:
C + B
D3
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
2B
→
D3
DH = ?
a) A + B
→
C
DH = -125 kJ
→
D3
DH = +41 kJ
+
A
add
C +
A
+
B
2B
→
D3
DHtot = -125 kJ + 41 kJ = -84 kJ
Chapter 5: Thermochemistry
From the enthalpies of reaction
a) 2 H2 (g)
+
b)
O2 (g)
→
2 H2O (g)
DH = -483.6 kJ
3 O2 (g)
→
2 O3 (g)
DH = +284.6 kJ
calculate the heat of the reaction
3 H2 (g)
+
O3 (g)
→
3 H2O (g)
Strategy:
Find two equations, the addition of which will give you
the final reaction
Chapter 5: Thermochemistry
a) 2 H2 (g)
+
O2 (g)
→
2 H2O (g)
3 H2 (g)
+
O3 (g)
→
3 H2O (g)
DH = -483.6 kJ
(1) start by considering the first reactant: find an equation where
H2 appears - reaction a). Then, multiply reaction a) so that
the coefficients match the final equation
O2 (g)
→
2 H2O (g)
DH = -483.6 kJ ) 
+ 3/2 O2 (g)
→
3 H2O (g)
DH = -725.4 kJ
( 2 H2 (g)
3 H2 (g)
+
3
2
Chapter 5: Thermochemistry
b)
3 O2 (g)
→
2 O3 (g)
DH = +284.6 kJ
b) inverted:
2 O3 (g)
→
3 O2 (g)
DH = -284.6 kJ
O3 (g)
→
3 H2O (g)
3 H2 (g)
+
(2) Consider the second reactant:
Find equation where O3 appears – rxn. b).
O3 appears on the product side – invert it to get O3 on the
reactant side
multiply “b) inverted” so that the coefficients match:
(2 O3 (g)
→
3 O2 (g)
DH = -284.6 kJ) x 1/2
O3 (g)
→
3/2 O2 (g)
DH = -142.3 kJ
Chapter 5: Thermochemistry
3) Now add the two partial reactions - and their DH’s - and check
whether you get the overall reaction:
3 H2 (g)
add:
+ 3/2 O2 (g)
→
3 H2O (g)
DH = -725.4 kJ
O3 (g)
→
3/2 O2 (g)
DH = -142.3 kJ
3 H2 (g) + O3 (g)
→
3 H2O (g)
DH = -867.7 kJ
Chapter 5: Thermochemistry
Enthalpy of Formation, DHof, …
... is the DH for the reaction that forms one mole of that compound
from its elements, with all substances (products and reactants) in their
standard states [25oC, 1atm]
this is what the o stands for
H2 (g)
½ N2 (g)
2 C (s)
+
½ O2 (g)
→
H2O (l)
+ ½ O2 (g)
→
NO (g)
+ 3 H2 (g) + ½ O2 (g) → C2H5OH (l)
DHof = -285.8 kJ
DHof = 90.4 kJ
DHof = -277.7 kJ
Chapter 5: Thermochemistry
Enthalpy of Formation, DHof
DHof of the most stable form of any element is zero
½ O2 (g) + ½ O2 (g)
→
O2 (g)
DHof = 0
DHof for O2 (g) , N2 (g) , H2 (g), Br2 (l) etc. = 0
Chapter 5: Thermochemistry
Enthalpies of formations can be used to calculate
enthalpies of reactions (under standard conditions)
DHorxn = sum of all DHof(products) - sum of all DHof(reactants)
DHorxn = Σ n DHof(products) – Σ m DHof(reactants)
“sum”
moles of reactant
moles of product
Chapter 5: Thermochemistry
Enthalpies of formations (DHof) can be used to calculate
enthalpies of reactions (DHorxn)
C3H8 (g)
+ 5 O2 (g)
→
3 CO2 (g) + 4 H2O (l)
DHorxn = [3 x DHof [CO2 (g)] + 4 x DHof [H2O (l)]) - (DHof [C3H8] (g) + 0)
= [(3 x -394 kJ) + (4 x -286 kJ)] - [-104 kJ]
= -1182 kJ - 1144 kJ + 104 kJ = -2222 kJ
DHof [CO2 (g)] = -394 kJ
DHof [H2O (l)] = -286 kJ
DHof [C3H8 (g)] = -104 kJ
DHof values are in Table 5.3 and
Appendix C
Chapter 5: Thermochemistry
For which of the following reactions would DH represent a
“standard enthalpy of formation”?
a. K (l)
+ ½ Cl2 (g)
b. CO (g)
+ ½ O2 (g)
c. 2 Na (s)
+ Cl2 (g)
d. 2 Na (s)
+
→
½ O2 (g)
KCl (s)
→
→
→
DHo = DHof
CO2 (g)
DHo = DHof
2 NaCl (s)
DHo = DHof
Na2O (s)
DHo = DHof
Chapter 5: Thermochemistry
Calculate the enthalpy change for the reaction
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = ?
Given the following enthalpies of reaction:
a) P4 (s) + 3 O2(g)
→
P4O6 (s)
DH = -1640.1 kJ
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
Chapter 5: Thermochemistry
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = ?
Where in the following reactions does P4O6 appear?
a) P4 (s) + 3 O2(g)
→
P4O6 (s)
DH = -1640.1 kJ
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
Chapter 5: Thermochemistry
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = ?
Where in the following reactions does P4O6 appear?
a) P4 (s) + 3 O2(g)
→
P4O6 (s)
DH = -1640.1 kJ
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
I) invert a)
a) inv.
P4O6 (s)
→
P4 (s) + 3 O2(g)
DH = +1640.1 kJ
Chapter 5: Thermochemistry
P4O6 (s)
→
P4 (s) + 3 O2(g)
P4O6 (s) + 2 O2(g)
→
P4O10 (s)
DH = +1640.1 kJ
DH = ?
II) O2 already appears in eq. a). P4O10 needs to be
introduced on the product side - use equation b)
b) P4 (s) + 5 O2(g)
→
P4O10 (s)
DH = -2940.1 kJ
Chapter 5: Thermochemistry
P4O6 (s)
→
add: P4 (s) + 5 O2(g)
P4 (s) + 3 O2(g)
→
P4O6 (s) + P4(s) + 5 O2(g)
P4O10 (s)
DH = +1640.1 kJ
DH = -2940.1 kJ
→ P4(s) + 3 O2(g) + P4O10 (g)
Chapter 5: Thermochemistry
P4O6 (s)
→
add: P4 (s) + 5 O2(g)
P4 (s) + 3 O2(g)
→
P4O6 (s) + P4(s) + 2 O2(g)
P4O6 (s) + 2 O2(g)
P4O10 (s)
DH = +1640.1 kJ
DH = -2940.1 kJ
→ P4(s) + 3 O2(g) + P4O10 (s)
→ P4O10 (s)
DH = -1300.0 kJ