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MHT CET 2021 Chemistry Solved Paper
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© examsnet.com
Question : 12
Total: 100
What is enthalpy of formation of
N
H
3
, if bond enthalpies as
(
N
≡
N
)
=
−
941
k
J
∕
m
o
l
,
(
H
−
H
)
=
436
k
J
∕
m
o
l
and
(
N
−
H
)
=
389
k
J
∕
m
o
l
?
−
84.5
k
J
∕
m
o
l
−
63.45
k
J
∕
m
o
l
−
21.25
k
J
∕
m
o
l
−
42.5
k
J
∕
m
o
l
Validate
Solution:
N
2
+
3
H
2
⟶
2
N
H
3
∆
f
H
(
N
H
3
)
=
Bond energy of reactant - Bond energy of product
=
(
1
2
×
B
E
of
N
≡
N
bond
+
3
2
B
E
of
H
−
H
bond
)
- (3 BE of
N
−
H
bond)
=
(
1
2
×
941
+
3
2
×
436
)
−
(
3
×
389
)
=
470.5
+
654
−
1167
=
−
42.5
k
J
∕
m
o
l
© examsnet.com
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