17.4
Calculating Heats of Reaction
CALCULATING HEATS OF
REACTION
OBJECTIVES
•Apply Hess’s law of heat summation to find enthalpy
changes for chemical and physical processes
•Calculate
formation
enthalpy changes using standard heats of
Requisite Skills

Algebra I & II: System of Equations

VOCABULARY:
oHess’s law of heat of summation
oStandard heat of formation
. . . manipulate and combine algebraic equations.
In algebra I, you learned that you can combine the like terms from
two equations to make a third equation.
2x + 6y = 16
y = 2x + 12
+
If you do this right, this allows you
to solve for one of the variables by
eliminating the other one.
2x + 7y = 2x + 28
7y = 28
y=4
4 = 2x + 12
x = -4
Once you know of the variables,
you can plug it in and solve for the
other variable.
4 - 12 = 2x + 12 - 12
-8 = 2x
. . . manipulate and combine algebraic equations.
Sometimes, you need to flip one of the equations before adding
them to get one of the terms to disappear.
9y = 3x - 57
9y = 3x - 57
2y + 22 = 3x
3x = 2y + 22
+
If we flip the second equation,
3x + 9y = 3x + 2y - 35
and THEN add the two equations,
the “x” terms will disappear.
9y = 2y - 35
2(-5) + 22 = 3x
7y = -35
-10 + 22 = 3x
y = -5
12 = 3x
x=4
. . . manipulate and combine algebraic equations.
Other times, you need to multiply an equation by an integer to get
what you want.
4y = x + 5
4x + 7y = 49
Even if we flip one of the equations,
nobody disappears after adding the
equations, but if we multiply the first
equation by four first . . .
4(3) = x + 5
12 = x + 5
x=7
16y = 4x + 20
+
4x + 7y = 49
4x + 23y = 4x + 69
23y = 69
y=3
. . . manipulate and combine algebraic equations.
Sometimes, you can use these old algebra tricks
with chemical equations to solve for unknowns . . .
Which stone do you prefer? And
why?
Diamond
Graphite
Would you like Diamond to be
converted into Graphite?




Such conversion or reaction will take millions of
years to complete
And chemists are curious to know the enthalpy
changes for the conversion of diamond to
graphite.
Should they wait million of years when the
reaction will be completed to get the data? Or
What do you recommend they should do? So
that the enthalpy changes data would be available
ASAP
Hess’s law
Mr. Hess came up with a quick solution of
how to calculate heat of reaction, ΔHrxn
for:
 1. reactions that are too slow
 2. reactions that have intermediate steps
 3. reactions that are dangerous
 4. reactions that are not very useful
Hess’s law Heat of Summation

States that if you add to two or more
thermochemical equations to give final
equation, then you can also add the heat
of reaction to give the final heat of
reaction

Hess’s law allows us to determine the
heat of reaction indirectly.
Operations in Hess’s law
Any of these could be applied:
 Flipping a known chemical equation
 Changing the sign of the ΔH (-ve or +ve)
when you flipped
 Adding two or more chemical equations
 Subtracting two or more chemical equations
 Multiplying a chemical equation with a whole
number or fraction
 Dividing a chemical equation with a whole
number

C (s, diamond) = C (s, graphite) ΔH= ?kJ
C (s, graphite) + O2 (g) = CO2 (g) ΔH= -393.5kJ
b) C (s, diamond) + O2 (g) = CO2 (g) ΔH= -395.4kJ
a)


CO2 and O2 need to be cancelled out to achieve the desired
reaction we are looking for.
We need graphite to be on the product side, so we are going
to flip equation (a)
C (s, diamond) + O2 (g) CO2 (g)
 CO2 (g)
C(s, graphite) + O2 (g)

C(s, diamond)
ΔH= -395.4kJ
ΔH= 393.5kJ
C(s, graphite) ΔH= (-395.4 + 393.5)kJ
ΔH= (393.5 – 395.4)kJ
ΔH= - 1.9kJ
. . . use Hess’ law to determine DHrxn
Hess’s law allows us to figure out DHrxn for a reaction without
ever having to make the reaction happen in real life.
HLEx1: Imagine that experiments tell you the following:
Sn(s) + Cl2(g)  SnCl2(s)
SnCl2(s) + Cl2(g)  SnCl4(l)
DH = -325.1 kJ
DH = -186.2 kJ
If you add the two equations above, you get the following:
Sn(s) + SnCl2(s) + 2Cl2(g)  SnCl2(s) + SnCl4(l)
Just as with normal, algebraic equations, when the same term
appears on the left and right, it can be crossed out . . . yielding . . .
Sn(s) + 2Cl2(g)  SnCl4(l)
DH = ? kJ
. . . use Hess’ law to determine DHrxn
Now here’s the real magic. Hess’s law says that you can also add
the DH’s for the reactions to get the DH for the final reaction.
Sn(s) + Cl2(g)  SnCl2(s)
SnCl2(s) + Cl2(g)  SnCl4(l)
DH = -325.1 kJ
DH = -186.2 kJ
(-325.11 kJ) + (-186.2 kJ) = -511.3 kJ
The sneaky miracle here is that we figured this out without ever
having to make tin metal and chlorine gas react to form tin (IV)
chloride in real life.
Sn(s) + 2Cl2(g)  SnCl4(l)
DH==-511.3
? kJ kJ
DH
. . . use Hess’ law to determine DHrxn
That was a very simple usage of Hess’ law. We didn’t have to
manipulate any equations before adding them. Let’s try a harder
problem.
. . . use Hess’ law to determine DHrxn
HLEx2: Let’s say experiments have told us the following:
DH = -335 kJ
Os(cr) + 2O2(g)  OsO4(g)
OsO4(cr)  OsO4(g)
DH = 56.4 kJ
Use Hess’s Law to figure out DH for the following reaction:
Os(cr) + 2O2(g)  OsO4(cr)
DH = ? kJ
(The final equation looks a lot like the first equation, but notice that
OsO4 is a gas in the first equation and a crystal in the third.)
What would you have to do to figure this one out? Think a
moment . . .
. . . use Hess’ law to determine DHrxn
HLEx2:
DH = -335 kJ
Os(cr) + 2O2(g)  OsO4(g)
OsO4(cr)  OsO4(g)
DH = 56.4 kJ
Os(cr) + 2O2(g)  OsO4(cr)
DH = ? kJ
OsO4(g) is in both of the initial equations, but doesn’t appear in
the final equation, so it needs to be eliminated somewhow.
OsO4(g) is on the right on both equations, so they won’t cancel
each other out if you add the equations as they are.
One filthy little trick you can do is to flip the second equation to put
OsO4(g) on the left, and THEN add the two equations together.
. . . use Hess’ law to determine DHrxn
HLEx2:
DH = -335 kJ
Os(cr) + 2O2(g)  OsO4(g)
OsO4(cr)  OsO4(g)
DH = 56.4 kJ
Os(cr) + 2O2(g)  OsO4(cr)
Os(cr) + 2O2(g)  OsO4(g)
OsO4(g)  OsO4(cr)
DH = ? kJ
DH = -335 kJ
DH = -56.4 kJ
Notice how the sign of DH changed on the equation that we
flipped. Remember that DHsolid = -DHfus and DHcond = -DHvap
Freezing is the opposite of melting and condensation is the
opposite of vaporization, so their DH’s have opposite signs.
. . . use Hess’ law to determine DHrxn
HLEx2:
DH = -335 kJ
Os(cr) + 2O2(g)  OsO4(g)
OsO4(cr)  OsO4(g)
DH = 56.4 kJ
Os(cr) + 2O2(g)  OsO4(cr)
Os(cr) + 2O2(g)  OsO4(g)
OsO4(g)  OsO4(cr)
DH = ? kJ
DH = -335 kJ
DH = -56.4 kJ
Os(cr) + 2O2(g) + OsO4(g)  OsO4(g) + OsO4(cr)
Now we can add the equations . . . and simpify . . .
Now we have the equation we were looking for. Now what?
. . . use Hess’ law to determine DHrxn
HLEx2:
DH = -335 kJ
Os(cr) + 2O2(g)  OsO4(g)
DH = 56.4 kJ
OsO4(cr)  OsO4(g)
Os(cr) + 2O2(g)  OsO4(cr)
DH = -335 kJ
Os(cr) + 2O2(g)  OsO4(g)
OsO4(g)  OsO4(cr)
DH = ? kJ
DH = -56.4 kJ
Os(cr) + 2O2(g) + OsO4(g)  OsO4(g) + OsO4(cr)
Add the DH’s to get the DH for the final equation.
DH = -335 kJ + (-56.4 kJ) = -391.4 kJ = DH
. . . use standard heats of formation to determine DHrxn
[Section under construction]
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17.1
17.2
17.3
17.4
17.1 Temperature &
Heat
17.2 Calorimetry
17.3 Heating Curve
for Water
17.4 Alegbra
Review
17.1 Endothermic
& Exothermic
17.2
Thermochemical
Equations
17.3 State Change
Math Problems
17.4 Hess’ Law
17.1
Q = m DT Cp
Problems
17.2
Heat of
Combustion
17.3 Heat of
Solution Math
Problems
17.4 Standard Heat
of Formation