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Standard XII
Chemistry
Heat of Formation
Question
The bond dissociation energy of gaseous
H
2
,
C
l
2
and
H
C
l
are 104, 58 and 103 kcal mol
−
1
respectively. The enthalpy of formation for
H
C
l
gas will be:
−
44.0
kcal
−
22.0
kcal
+
2.0
kcal
+
44.0
kcal
A
−
44.0
kcal
B
−
22.0
kcal
C
+
2.0
kcal
D
+
44.0
kcal
Open in App
Solution
Verified by Toppr
Reaction is
H
2
+
C
l
2
⇌
2
H
C
l
Δ
H
for the formation of
H
C
l
=
(
Δ
H
f
H
2
+
Δ
H
f
C
l
2
)
−
2
Δ
H
f
H
C
l
Δ
H
=
(
104
+
58
)
−
2
×
(
103
)
=
−
44
k
c
a
l
−
44
k
c
a
l
is the enthalpy for the formation of
2
moles
H
C
l
. So, for
1
mole
H
C
l
, it will be
44
/
2
=
−
22
k
c
a
l
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