SCH4U Hess’s Law
Additivity of Heats
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330 # 1 - 4
Herman
Hess
Potential energy of hiker 1 and hiker 2
is the same even though they took
different paths.
Hess's Law of Heat Summation
The enthalpy change of an overall process is the sum of
the enthalpy changes of its individual steps.
Example:
Problem: Calculate the energy involved in the oxidation of elemental
sulfur to sulfur trioxide from reactions:
1) S (s) + O2 (g)
SO2 (g)
H1 = -296.0 kJ
2) 2 SO2 (g) + O2 (g)
2 SO3 (g)
H2 = -198.2 kJ
SO3 (g)
H3 = ?
Target equation:
3) S (s) + 3/2 O2 (g)
If you add two or more
equations to get a new
equation, you must add the
H's to get the
the new equation.
HRXN for
Given:
1) S (s) + O2 (g)
+ ½[
2 SO 2 (g) + O2 (g)
S (s) + O 2 (g)
+ SO 2 (g) + ½ O2 (g)
SO2 (g)
2 SO3 (g)]
H1 = -296.0 kJ
[
SO2 (g)
+ SO 3
H2 = -198.2 kJ] *½
H1 = -296.0 kJ
+
H2 = -198.2/2 kJ
Find a substance that appears in only one place.
Make sure that the substance is on the correct side of the
targetequation. Add or subtract the equation.
Get the same coefficient as the target equation
Target equation:
S (s) + 3/2 O 2 (g)
SO3 (g)
H3 = ?
Thermochemical Equations
•
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
•
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
•
H2O (l)
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
Given:
1) S (s) + O2 (g)
+ 2)
2 SO2 (g) + O2 (g)
S (s) + O 2 (g)
+ SO 2 (g) + ½ O2 (g)
S (s) + 3/2 O 2 (g)
SO2 (g)
H1 = -296.0 kJ
2 SO3 (g)
H2 = -198.2 kJ
SO2 (g)
+ SO 3
SO3 (g)
H1 = -296.0 kJ
+
H2 = -198.2/2 kJ
H3 = -296 - 99.1 kJ
H3 = -395.1 kJ
Simplify to get the target equation and ÄH3.
Target equation:
S (s) + 3/2 O 2 (g)
SO3 (g)
H3 = ?
Thermochemical Equations
•
The physical states of all reactants and products must be
specified in thermochemical equations.
H 2O
(s )
H 2O
( l)
H 2O
H 2O
( l)
D H = 6.01 kJ
(g )
D H = 44.0 kJ
Determine the heat of reaction for the reaction:
4NH3(g) + 5O2(g) à
4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) à 2NO(g)
DH = 180.6 kJ
N2(g) + 3H2(g) à 2NH3(g)
DH = -91.8 kJ
2H2(g) + O2(g) à 2H2O(g)
DH = -483.7 kJ
The three reactions must be algebraically
manipulated to sum up to the desired reaction.
and.. the DH values must be treated accordingly.
Target:
4NH3(g) + 5O2(g) à
4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) à 2NO(g)
DH = 180.6 kJ
N2(g) + 3H2(g) à 2NH3(g)
DH = -91.8 kJ
2H2(g) +
O2(g) à 2H2O(g)
DH = -483.7 kJ
Find a substance that appears in only one place.
O2 : Found in more than one place, SKIP IT (its hard).
Make sure that the substance is on the correct side of the
targetequation.
Target:
4NH3(g) + 5O2(g) à
4NO(g) + 6H2O(g)
Using the following sets of reactions:
N2(g) + O2(g) à 2NO(g)
DH1 = 180.6 kJ
N2(g) + 3H2(g) à 2NH3(g)
DH2 = - 91.8 kJ
2H2(g) +
O2(g) à 2H2O(g)
DH3 = - 483.7 kJ
Make sure that the substance is on the correct side of the
targetequation.
Reverse equation 2 and change the sign ofÄH2
2NH3(g)
à
N2(g) + 3H2 (g)
DH2 = + 91.8 kJ
Get the same coefficient as the target equation
Target:
4NH3(g) + 5O2(g) à
2*[ N2(g) + O2(g) à 2NO(g)
2*[ 2NH3(g)
4NO(g) + 6H2O(g)
]
à N2(g) + 3H2 (g) ]
3*[2H2(g) + O2(g) à 2H2O(g)
]
2*[ DH1 = 180.6 kJ
]
2*[ DH2 = + 91.8 kJ ]
3*[ DH3 = - 483.7 kJ ]
DH1 = 2(180.6 kJ)
2N 2(g) + 2O 2 (g) à 4NO(g)
+4NH 3 (g)
+2N 2 (g) + 6H2 (g)
+ 2(91.8 kJ)
+3(- 483.7 kJ)
+ 6H2(g) + 3O2(g)
+6H 2 O(g)
4NH3(g) + 5O2(g) à
4NO(g) + 6H2O(g)
DH = - 906.3 kJ
Get the same coefficients as the target equation
Goal:
NH3:
O2 :
NO:
H2O:
4NH3(g) + 5O2(g) à
4NO(g) + 6H2O(g)
Reverse and x2 4NH3 à 2N2 + 6H2 DH =
Found in more than one place, SKIP IT.
x2
x3
2N2 + 2O2 à 4NO
6H2 + 3O2 à 6H2O
+183.6 kJ
DH = 361.2 kJ
DH = -1451.1 kJ
Cancel terms and take sum.
4NH3
+ 5O2
à
4NO
+ 6H2O
DH = -906.3 kJ
Is the reaction endothermic or exothermic?
Determine the heat of reaction for the reaction:
C2H4(g) + H2(g) à C2H6(g)
Use the following reactions:
C2H4(g) + 3O2(g) à 2CO2(g) + 2H2O(l)
DH = -1401 kJ
C2H6(g) + 7/2O2(g) à 2CO2(g) + 3H2O(l)
DH = -1550 kJ
H2(g) +
1/2O2(g) à H2O(l)
DH = -286 kJ
.
Determine the heat of reaction for the reaction:
Goal: C2H4(g) + H2(g) à C2H6(g)
DH = ?
Use the following reactions:
C 2H 4(g) + 3O 2(g) à 2CO 2(g) + 2H 2O(l)
DH = -1401 kJ
C 2H 6(g) + 7/2O 2(g) à 2CO 2 (g) + 3H2O(l)
DH = -1550 kJ
H 2(g) +
1/2O2(g) à H 2O(l)
DH = -286 kJ
C 2H 4(g) :use 1 as is C 2H 4(g) + 3O 2(g) à 2CO 2(g) + 2H 2O(l)
H 2(g) :# 3 as is
H 2(g) + 1/2O 2(g) à H 2O(l)
C 2H 6(g) : rev #2
2CO 2(g) + 3H 2O(l) à C 2H6(g) + 7/2O2(g)
C 2H 4(g) + H 2(g) à C 2H 6(g)
DH = -1401 kJ
DH = -286 kJ
DH = +1550 kJ
DH = -137 kJ
Calculate the standard enthalpy of formation of CS 2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
DHrxn
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
DHrxn
C(graphite) + 2S(rhombic)
CS2 (l)
DH0rxn= -393.5 + (2x-296.1) + 1072 = 86.3 kJ
Benzene (C 6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6
(l)
+ 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.88 kJ
-6535kJ
= - 3267 kJ/mol C6H6
2 mol
6.5
Sample Problem 6.6
PROBLEM:
Using the Heat of Reaction (DHrxn) to Find
Amounts
The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented by
Al2O3(s)
2Al(s) + 3/2O2(g)
DHrxn = 1676kJ
If aluminum is produced this way, how many grams of aluminum can
form when 1.000x10 3kJ of heat is transferred?
PLAN:
SOLUTION:
heat(kJ)
1676kJ=2molAl
mol of Al
xM
g of Al
1.000x10 3kJ x
2mol Al
26.98g Al
1676kJ
1mol Al
= 32.20g Al
Sample Problem 6.7
PROBLEM:
Using Hess's Law to Calculate an Unknown DH
Two gaseous pollutants that form auto exhaust are CO and NO.
An environmental chemist is studying ways to convert them to
less harmful gases through the following equation:
CO(g) + NO(g)
CO2(g) + 1/2N2(g)
DH = ?
Given the following information, calculate the unknown DH:
Equation A: CO(g) + 1/2O2(g)
Equation B: N2(g) + O2(g)
PLAN:
CO2(g) DHA = -283.0kJ
2NO(g) DHB = 180.6kJ
Equations A and B have to be manipulated by reversal and/or
multiplication by factors in order to sum to the first, or target, equation.
SOLUTION:
Multiply Equation B by 1/2 and reverse it.
CO(g) + 1/2O2(g)
NO(g)
CO(g) + NO(g)
CO2(g) DHA = -283.0kJ
1/2N2(g) + 1/2O2(g)
DHB = -90.6kJ
CO2(g) + 1/2N2(g) DHrxn = -373.6kJ
Dr. Schambaugh, of the University of Oklahoma
School of Chemical Engineering, Final Exam
question for May of 1997. Dr. Schambaugh is
known for asking questions such as, "why do
airplanes fly?" on his final exams. His one and
only final exam question in May 1997 for his
Momentum, Heat and Mass Transfer II class was:
"Is hell exothermic or endothermic? Support your
answer with proof."
Most of the students wrote proofs of their beliefs
using Boyle's Law or some variant. One student,
however, wrote the following:
"First, We postulate that if souls exist, then they
must have some mass. If they do, then a mole
of souls can also have a mass. So, at what rate
are souls moving into hell and at what rate are
souls leaving? I think we can safely assume
that once a soul gets to hell, it will not leave.
Therefore, no souls are leaving. As for souls
entering hell, let's look at the different
religions that exist in the world today. Some
of these religions state that if you are not a
member of their religion, then you will go to
hell. Since there are more than one of these
religions and people do not belong to more
than one religion, we can project that all
people and souls go to hell. With birth and
death rates as they are, we can expect the
number of souls in hell to increase
exponentially.
Now, we look at the rate of change in volume
in hell. Boyle's Law states that in order for
the temperature and pressure in hell to stay
the same, the ratio of the mass of souls and
volume needs to stay constant. Two options
exist:
If hell is expanding at a slower rate than the
rate at which souls enter hell, then the
temperature and pressure in hell will
increase until all hell breaks loose.
If hell is expanding at a rate faster than the increase of
souls in hell, then the temperature and pressure will
drop until hell freezes over.
So which is it? If we accept the quote given to me by
Theresa Manyan during Freshman year, "that it will be
a cold night in hell before I sleep with you" and take
into account the fact that I still have NOT succeeded
in having sexual relations with her, then Option 2
cannot be true...Thus, hell is exothermic."
The student, Tim Graham, got the only A.