Answer:
I got
−
902 kJ
for the reaction as-written. How would you rewrite this into
kJ/mol NH
3
?
The standard molar enthalpy of formation is for the formation of
1 mol
of product from its elements as they exist in nature at
25
∘
C
and
1 atm
. For example...
1
2
N
2
(
g
)
+
3
2
H
2
(
g
)
→
NH
3
(
g
)
,
Δ
H
∘
f
(
NH
3
(
g
)
)
=
−
45.9 kJ/mol
But that also means formation of elements in their standard states must yield zero enthalpy change:
1
2
O
2
(
g
)
→
1
2
O
2
(
g
)
,
Δ
H
∘
f
(
O
2
(
g
)
)
=
0
That means we can simply derive from Hess's law and that fact to get:
Δ
H
∘
r
x
n
=
∑
P
n
P
Δ
H
∘
f
,
P
−
∑
R
n
R
Δ
H
∘
f
,
R
where
n
indicates the mols of product
P
or reactant
R
. Here we actually have just summed many formation reactions together that we know have
Δ
H
∘
r
x
n
=
Δ
H
∘
f
.
Here we have:
Δ
H
∘
r
x
n
=
[
4 mols NO
(
g
)
⋅
Δ
H
∘
f
,
N
O
(
g
)
+
6 mols H
2
O
(
g
)
⋅
Δ
H
∘
f
,
H
2
O
(
g
)
]
−
[
4 mols NH
3
(
g
)
⋅
Δ
H
∘
f
,
N
H
3
(
g
)
+
0 kJ/mol
]
where we have set a zero contribution
O
2
(
g
)
right off the bat.
=
[
4 mols NO
(
g
)
⋅
91.3 kJ/mol
+
6 mols H
2
O
(
g
)
⋅
−
241.8 kJ/mol
]
−
[
4 mols NH
3
(
g
)
⋅
−
45.9 kJ/mol
+
0 kJ/mol
]
=
[
−
1085.6 kJ
]
−
[
−
183.6 kJ
]
=
−
902 kJ
for the reaction as-written.
Step-by-step explanation: