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Chemistry
Chemistry
Problem
The enthalpy changes for two reactions are listed below. 1)
1
2
N
2
(
g
)
+
1
2
O
2
(
g
)
→
N
O
(
g
)
Δ
H
1
=
+
90.2
kJ
\tfrac12\,\mathrm{N_2}(g)+\tfrac12\,\mathrm{O_2}(g)\rightarrow\mathrm{NO}(g)\qquad \Delta H_1=+90.2\ \text{kJ}
2
1
N
2
(
g
)
+
2
1
O
2
(
g
)
→
NO
(
g
)
Δ
H
1
=
+
90.2
kJ
2)
N
O
(
g
)
+
1
2
O
2
(
g
)
→
N
O
2
(
g
)
Δ
H
2
=
−
57.0
kJ
\mathrm{NO}(g)+\tfrac12\,\mathrm{O_2}(g)\rightarrow\mathrm{NO_2}(g)\qquad \Delta H_2=-57.0\ \text{kJ}
NO
(
g
)
+
2
1
O
2
(
g
)
→
N
O
2
(
g
)
Δ
H
2
=
−
57.0
kJ
What is
Δ
H
\Delta H
Δ
H
for the reaction
N
2
(
g
)
+
2
O
2
(
g
)
→
2
N
O
2
(
g
)
\mathrm{N_2}(g)+2\,\mathrm{O_2}(g)\rightarrow2\,\mathrm{NO_2}(g)
N
2
(
g
)
+
2
O
2
(
g
)
→
2
N
O
2
(
g
)
? a)
+
123.4
kJ
+123.4\ \text{kJ}
+
123.4
kJ
b)
+
66.4
kJ
+66.4\ \text{kJ}
+
66.4
kJ
c)
−
66.4
kJ
-66.4\ \text{kJ}
−
66.4
kJ
d)
+
33.2
kJ
+33.2\ \text{kJ}
+
33.2
kJ
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