SCH4U Heats of Formation
And Multi-step Problems
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Formation Equations
The chemical equation showing the formation of 1
mole of a compound from the elements in their
standard state. The enthalpy change for the formation
is called the Standard Heat of Formation, DHof .
Elements in their standard state have a DHof = 0
Most compounds have a negative DHof , i.e. The
reaction is exothermic........the compound is lower in
energy than the elements from which it was formed.
Writing Formation Equations
PROBLEM:
Write balanced equations for the formation of 1 mol of the follo wing
compounds from their elements in their standard states and inclu
include
de
DH0f.
(a)Silver chloride,
PLAN:
AgCl , a solid at standard conditions.
Use the table of heats of formation for values.
SOLUTION:
(a) Ag( s)
+ 1/2Cl
2
( g)
AgCl(
s)
D H 0 f = - 127.0kJ
How do you use ÄHof with reactions without
elements?
Use Hess’s Law
Enthalpy is a state function
Make an intermediate reaction where all
reactants form elements in their standard
state
o
The heat for this reaction is - ÄH
f
The general process for determining DH 0 rxn from DH 0f values
Usin
HgessL'C
astow
alculate
DH
0
rxn
Now take all elements in their standard
states and react to form products
o
This is + ÄH
f
Adding these 2 reactions will eliminate the
intermediate reaction with elements
Adding these 2 reactions will give the
standard heat of reaction for the original
equation
.n Reactants Y m Products
ÄHoRXN = Ó m ÄHof (products) - Ó n ÄHof (reactants)
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DHorxn = [ cDHof (C) + dDHof (D) ] - [ aDHof (A) + bDHof (B) ]
DHorxn = S nDHof (products) - S mDHfo (reactants)
Hess's Law: When reactants are converted to products,
the change in enthalpy is the same whether the reaction
takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn't matter how you get
there, only where you start and end.)
The general process for determining DH 0 rxn from DH 0f values
Usin
HgessL'C
astow
alculate
DH
0
rxn
SamPprloeblem
Calculating
th e
HeaotR
f eaction
from
Heats
of
Formation
Nitric acid, whose worldwide annual production is about 8 billion kg
is used to make many products, including fertilizer, dyes, and
explosives. The first step in the industrial production process is the
oxidation of ammonia:
4NH
3
( g) +
Calculate
PLAN:
5 O 2 ( g)
DH
4 NO ( g) +
0
rxn
from
DH
0
f
6 H 2 O(
g)
values.
Look up the DH0f values and use Hess's Law to find DHrxn.
4NH
3
( g) +
5 O 2 ( g)
4 NO ( g) +
6 H 2 O(
g)
SOLUTION:
DH
rxn
= S
mD H 0 f
(products)
- S
nD H 0 f
(reactants)
DHrxn = [4 DH0f NO(g) + 6 DH0f H2O(g)] - [4 DH0f NH3(g) + 5 DH0f O2(g)]
= (4mol)(90.3kJ/mol) + (6mol)(
-241.8kJ/mol)
-
[(4mol)( -45.9kJ/mol) + (5mol)(0kJ/mol)]
DH
rxn
= - 906kJ
Benzene (C 6 H 6 ) 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.
2C
6
H
6
(l) + 15O
D H 0rxn =
2
(g)
12CO
2
(g) + 6H
2
O (l)
0
0
m
D
H
S
n
D
H
S
f (reactants)
f (products)
2C
6
H
6
(l) + 15O
2
(g)
12CO
2
(g) + 6H
2
O (l)
D H 0rxn = S n D H 0f (products) - S m D H f0 (reactants)
D H 0rxn =
[ 12 D H 0f (CO 2 ) + 6 D H 0f (H 2 O) ] - [ 2 D H 0f (C 6 H 6 ) ]
DH 0rxn =
[ 12x –393.5 + 6x –285.8 ] – [ 2x49.04 ] = -6534.88 kJ
Heat per mole of benzene (2 moles in equation)
- 6535kJ
= - 3267 kJ/mol C
2 mol
6
H
6
SamPprloeblemUsin
th
H
geeR
oaeftactio
(Dn
Ht)oFind
Amou
P
M
Sn
rtuo
e
ts
lb
pt:ilems
rxn
The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented b y
Al2 O3 (s)
2Al(s) +
3/2
O 2(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:
heat(kJ)
1676kJ=2mol Al
mol of Al
x
g of Al
M
SOLUTO
I N:
1.000x10 3kJ x
2mol Al
26.98g Al
1676kJ
1mol Al
= 32.20g Al
Breaking and Forming Chemical Bonds
During a chemical reaction, chemical bonds are
broken and new chemical bonds are formed.
Breaking each chemical bond requires energy, Endothermic
Forming each chemical bond releases energy, Exothermic
The DH for a reaction is the sum of the energy
required and released during the bond breaking and
forming process.
The enthalpy for any reverse process has the same
numerical value but a change in sign.
Enthalpy Changes and Bond Energies
Energy is absorbed when bonds break. The energy required to
break the bonds is absorbed from the surroundings.
O2(g) à2O(g) DHE=490.4 kJ
H2(g)à 2H(g) DHE =431.2 kJ
H2O(g)à2H(g) + O(g) DHE=915.6 kJ
We can estimate the bond enthalpies of O=O, H-H, and O-H as 490.4 kJ/mol, 431.2 kJ/mol,
and 457.7 kJ/mol, respectively.
2H2(g) + O2(g) à 2H2O(g)
DHE= ?
2H2(g) + O2(g)
moles of
bonds broken
Energy absorbed
à 2H2O(g)
moles of bonds
formed
Energy released
2 H-H @ 431.2 kJ each
862.4kJ
1 O=O @ 490.4 kJ each
490.4kJ
_____________________________________________
1352.7kJ
4 O-H @ 457.7 kJ each
1830.9kJ
1830.9kJ
DHE= 1352.7 - 1830.9 kJ = -478.2 kJ.
(Remember that the minus sign means "energy released", so you add the bond energies for
broken bonds and subtract energies for bonds formed to get the total energy.)
A calculation based on enthalpies of formation gave DHE = -483.7 kJ
Bonds in a molecule influence each other, which means that bond energies aren't really additive.
An O-H bond in a water molecule has a slightly different energy than an O-H bond in H2O2,
because it's in a slightly different environment.
Reaction enthalpies calculated from bond energies are very rough approximations!