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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 / mole respectively . Calculate the enthalpy of formation of HCl gas.
-75 Kcal
-22 Kcal
-45 Kcal
none of these
A
-22 Kcal
B
none of these
C
-75 Kcal
D
-45 Kcal
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Solution
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1
2
+
H
2
+
1
2
C
l
2
→
H
C
l
Δ
h
+
=
(Bond energy of reactant)- (Bond energy of Product)
=
(
1
2
Δ
H
B
(
H
2
)
+
1
2
Δ
H
B
(
C
l
2
)
)
−
(
Δ
H
B
(
H
C
l
)
)
=
(
1
2
(
104
)
+
1
2
(
58
)
)
−
(
103
)
=
52
+
29
−
103
=
−
22
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Q1
The bond dissociation energy of gaseous
H
2
,
C
l
2
and
H
C
l
are 104 , 58 and 103 kcal / mole respectively . Calculate the enthalpy of formation of HCl gas.
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Q2
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