Read:
HW:
Thermochemistry Part 2
HESS’S LAW
BLB 5.6–7; 8.8
BLB 5:63, 67a,b, 69, 73, 75, 83, 85;
BLB 8:65a, 67a,c, 92a
BLB 18:72ab, 74
The sum of the !H values for each step
is the same as !H for the overall
process.
Supplemental 5:1–7;
Supplemental 8:11–13
This is true because !H is a state
function.
Know:
• Hess!s Law
• heats of formation
• enthalpy of reactions
• estimating reaction enthalpy from bond energies
C
!H2
!Hrxn
B
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Dr. L. S. Van Der Sluys
Page 1
Ch. 5 Part 2
!H1
A
A
B
B
C
A+B "# B + C
Dr. L. S. Van Der Sluys
!H1
!H2
!H1 + !H2= !Hrxn
Page 2
Ch. 5 Part 2
Example: Given the following
information
A
B
H2(g) + F2(g) # 2HF(g)
2H2(g) + O2(g) # 2 H2O(g)
Given the following information:
!HA = –537kJ
!HB = –572kJ
2SO2(g)+ O2(g)
2S(s) + 3 O2(g)
Determine !H for the reaction:
C
# 2SO3(g)
!H
$196kJ
# 2SO3(g)
$790kJ
What is !Hrxn for the following reaction?
2F2(g) + 2H2O(g)# 4HF(g) + O2(g) !HC =?
IDEA: find combinations of reactions
such that
nA+mB=C
then
S(s) + O2(g)
A.
B.
C.
D.
# SO2(g)
– 986 kJ
– 594 kJ
+ 594 kJ
– 297 kJ
n !HA + m !HB = !HC
Here:
2xA
–1xB
2H2(g) + 2F2(g) # 4HF(g) !H = 2(–537kJ)
2 H2O(g) # 2H2(g) + O2(g) !H = –(–572kJ)
_
2F2(g) + 2H2O(g)# 4HF(g) + O2(g)
!HC = 2(–537kJ) –1(–572kJ) = –502kJ
Dr. L. S. Van Der Sluys
Page 3
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 4
Ch. 5 Part 2
Standard States of the Elements
Heat of Formation
!Hf
enthalpy of formation
heat given off (or absorbed) when
elements combine to give one mole or
one molecule of a compound
combine
Elements
Compounds
!Hf
!H° f standard enthalpy of formation
enthalpy of formation when all
substances are in their Standard State
DEFINITION OF STANDARD STATE
1.
P = 1 atm
2.
T = 25°C
(298K)
3.
element is in its most stable
state (gas/liquid/solid)
For an element in its standard state:
!H° f = ___
Dr. L. S. Van Der Sluys
(by definition)
Page 5
Ch. 5 Part 2
(Most stable phase under standard
conditions of 298K and 1 atm)
1. Metals: all are _______ at 298K
and 1 atm except one (Which
one?)
2. Semi metals (metalloids): all are
_________ at 298K and 1 atm
3. Nonmetals at 298K and 1 atm
i: Noble gases (Group 8):
atomic ___________
ii: Diatomics: H2, N2, O2, Group 7
(F2, Cl2, Br2, I2)
H2, N2, O2, F2, Cl2, are ______
Br2 is a __________
I2 is a ___________
iii: all other non-metals are _____
C (graphite), S8, P(s), Se
Dr. L. S. Van Der Sluys
Page 6
Ch. 5 Part 2
Are these Elements in their
standard states?
YES
NO
O2(g)
$
N3 (s)
F2(g)
N2(g)
O3(g)
C(diamond)
C(graphite)
Fe(s)
Br2(g)
H(g)
Dr. L. S. Van Der Sluys
Page 7
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 8
Ch. 5 Part 2
For which of the following reactions (at
25° C and 1 atm) is !Hrxn = !H° f ?
Answer A: !Hrxn = !H° f
Answer B: !Hrxn ! !H° f
We know that !H° f is the Heat of formation;
formation of _____________(how much?)
from ______________ when all reactants
are in their __________________
1) H2(g) + F2(g) "# 2HF(g)
2) NO(g)+ 1/2 O2(g)
We know that !H° rxn is the Standard
Enthalpy of reaction, or Heat of reaction;
all elements must be in their
________________.
# NO2(g)
How are they related?
3) 2C(graphite)+3H2(g)+1/2O2(g)"#C2H5OH(l)
4) Pb(s) + Cl2(g) "#PbCl2(s)
Obtaining !H° rxn from !H° f:
!H° rxn = %n !H° f (prod) $ %m !H° f (react)
5) S(s) + O3(g) "# SO3(g)
where n and m are stoichiometric
coefficients of products and reactants.
6) Br2(l) "# Br2(g)
(This is an application of Hess’s Law.)
Dr. L. S. Van Der Sluys
Page 9
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 10
Ch. 5 Part 2
Using the following information, what is
!H° rxn for the reaction:
C2H5OH(l) + 3O2(g)#2CO2(g) + 3H2O(l)
!H° f (in kJ/mole):
-277.7
a)
b)
c)
d)
-393.5
-285.8
– 115.8 kJ
+ 115.8 kJ
– 1366.7 kJ
+ 13366.7 kJ
If a piece of fruit contains 16.0 g of
fructose, and !Hrxn = $2803 kJ, how many
food calories does it contribute to the
body?
1 cal = 4.184J
1kcal = 1 Cal (= 1 food calorie)
Combustion!
C6H12O6(s) +6____# 6____+ 6_____
NOTE: Appendix C in the text has a
more extensive table of !H° f
Dr. L. S. Van Der Sluys
Page 11
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 12
Ch. 5 Part 2
Changes in Enthalpy with Breaking and
Formation of Bonds
Bond Properties Review
COVALENT BOND LENGTHS and ENERGIES
Bond length: distance between nuclei
bond
Bond energy Bond length pm
kJ/mol
348
154
C"C
614
134
C=C
839
121
C&C
more electrons shared, shorter bond length
bond
!
Bond Enthalpies (bond energy)
can be used to estimate !Hrxn.
Bond energy
Bond length
pm
kJ/mol
413
110
C"H
328
176
C"Cl
276
196
C"Br
Shorter bond length, stronger the bond
!
The overall reaction has two steps;
breaking the original bonds, and forming
new ones.
Dr. L. S. Van Der Sluys
Page 13
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 14
Ch. 5 Part 2
Estimating !Hrxn
• Bond energies provide estimates of
reaction enthalpies:
!Hrxn ' %nDbroken $ %mDformed
n, m = # of bonds
Energy is given off ($) when bonds form.
Helps understand origins of !Hrxn
Dr. L. S. Van Der Sluys
Page 15
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 16
Ch. 5 Part 2
Example: Predict the heat of reaction for
the decomposition of hydrogen
peroxide.
2 H2O2 # 2 H2O + O2 !Hrxn = ?
Draw Lewis structures of reactants and products:
Peroxide Decomposition:
Elephant Toothpaste Demonstration
2 H2O2(l) # 2 H2O(l) + O2(g)
!Hrxn = ?
Add KI to catalyze the reaction:
I-(aq) + H2O2(l) # H2O(l) + IO-(aq)
IO-(aq) + H2O2(l) # H2O(l) + O2(g)+ I-(aq)
Reactants (broken)
Products
bond #
bond #
D
(formed)
D
2 H2O2(l) # 2 H2O(l) + O2(g)
Can also catalyze the reaction with MnO2
Compare to Actual value of
!H° rxn = -196 kJ/mole
Dr. L. S. Van Der Sluys
Page 17
Ch. 5 Part 2
Dr. L. S. Van Der Sluys
Page 18
Ch. 5 Part 2