Calculate the enthalpy of formation of benzene if the enthalpy of combustion of benzene is -3266KJ & enthalpy of formation of Co2&H2O are -393.1&-286
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2
Answer:
-491.8kJ/mol
Explanation:
Combustion reaction for above question is
C
2
H
4
O
2
+
2
O
2
→
2
C
O
2
+
2
H
2
O
Δ
H
f
(
c
o
m
b
u
s
t
i
o
n
)
=
2
[
Δ
H
C
O
2
+
Δ
H
H
2
O
]
−
Δ
H
C
2
H
4
O
2
Δ
H
C
2
H
4
O
2
=
2
[
Δ
H
C
O
2
+
Δ
H
H
2
O
]
−
Δ
H
c
o
m
b
u
s
t
i
o
n
Δ
H
C
2
H
4
O
2
=
867
+
(
2
×
−
679.4
)
=
−
491.8
k
J
m
o
l
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