(g)
CO
2
(g) + 2 H
2
O(l)
H = ?
Utilizing Hess’s law:
Reactants
Standard State Elements
H =
H
a
+
H
b
= 75 + 0 = 75 kJ
Standard State Elements
Products
H =
H
c
+
H
d
= –394 – 572 = –966 kJ
__________________________________________________________________
Reactants
Products
H = 75 – 966 = –891 kJ
b.
The standard enthalpy of formation for an element in its standard state is given a value of
zero.
To assign standard enthalpy of formation values for all other substances, there
needs to be a reference point from which all enthalpy changes are determined.
This
reference point is the elements in their standard state which is defined as the zero point.
So when using standard enthalpy values, a reaction is broken up into two steps.
The first
step is to calculate the enthalpy change necessary to convert the reactants to the elements
in their standard state.
The second step is to determine the enthalpy change that occurs
when the elements in their standard state go to form the products.
When these two steps
are added together, the reference point (the elements in their standard state) cancels out
and we are left with the enthalpy change for the reaction.

CHAPTER 9
ENERGY, ENTHALPY, AND THERMOCHEMISTRY
379
c.
This overall reaction is just the reverse of all the steps in the part a answer.
So
H
=
+966 – 75 = 891 kJ.
Products are first converted to the elements in their standard state
which requires 966 kJ of heat.
Next, the elements in the standard states go to form the
original reactants [CH
4
(g) + 2 O
2
(g)] which has an enthalpy change of
75 kJ.
All of the
signs are reversed because the entire process is reversed.
67.
In general:
Δ
H° =
o
products
,
f
p
H
Δ
n
,
Δ
H
n
o
reactants
f,
r
and all elements in their standard
state have
o
f
H
Δ
= 0 by definition.
a.
The balanced equation is:
2 NH
3
(g) + 3 O
2
(g) + 2 CH
4
(g)
2 HCN(g) + 6 H
2
O(g)
Δ
H° = (2 mol HCN ×
o
HCN
,
f
H
Δ
+ 6 mol H
2
O(g)
o
O
H
,
f
2
H
Δ
)
(2 mol NH
3
×
o
NH
,
f
3
H
Δ
+ 2 mol CH
4
×
o
CH
,
f
4
H
Δ
)
Δ
H° = [2(135.1) + 6(
242)]
[2(
46) + 2(
75)] =
940. kJ
b. Ca
3
(PO
4
)
2
(s) + 3 H
2
SO
4
(l)
3 CaSO
4
(s) + 2 H
3
PO
4
(l)
Δ
H° =
mol
kJ
1267
)
l
(
PO
H
mol
2
mol
kJ
1433
)
s
(
CaSO
mol
3
4
3
4
mol
kJ
814
)
l
(
SO
H
mol
3
mol
kJ
4126
)
s
(
)
PO
(
Ca
mol
1
4
2
2
4
3
Δ
H° =
6833 kJ
(-6568 kJ) =
265 kJ
c.
NH
3
(g) + HCl(g)
NH
4
Cl(s)
Δ
H° = (1 mol NH
4
Cl ×
o
Cl
NH
,
f
4
H
Δ
)
(1 mol NH
3
×
o
NH
,
f
3
H
Δ
+ 1 mol HCl ×
o
HCl
,
f
H
Δ
)
Δ
H°
=
mol
kJ
314
mol
1
mol
kJ
92
mol
1
mol
kJ
46
mol
1
Δ
H° =
314 kJ + 138 kJ =
176 kJ
d.
The balanced equation is:
C
2
H
5
OH(l) + 3 O
2
(g)
2 CO
2
(g) + 3 H
2
O(g)
Δ
H°
=
mol
kJ
242
mol
3
mol
kJ
5
.
393
mol
2
mol
kJ
278
mol
1
Δ
H°
=
1513 kJ
(
278 kJ) =
1235 kJ

380
CHAPTER 9


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