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KVPY SX SB Exam 2015 Question Paper
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© examsnet.com
Question : 101
Total: 120
CHEMISTRY
For the reaction
N
2
+
3
X
2
→
2
N
X
3
where X = F, Cl (the average bond energies are F-F = 155 kJ mol
–
1
N-F = 272 kJ mol
–
1
, Cl-Cl = 242 kJ mol
–
1
, N-Cl = 200 kJ mol
–
1
and N ≡ N = 941 kJ mol
–
1
), the heats of formation of NF
3
and NCl
3
in kJ mol
–
1
, respectively, are closest to
– 226 and +467
+ 226 and – 467
– 151 and + 311
+ 151 and – 311
Validate
Solution:
N
2
+
3
F
2
⟶
2
N
F
3
Given,
B
E
N
≡
N
=
941
k
J
m
o
l
−
1
B
E
F
−
F
=
155
k
J
m
o
l
−
1
B
E
N
−
F
=
272
k
J
m
o
l
−
1
∆
H
f
=
Σ
B
E
reactants
−
Σ
B
E
products
=
B
E
N
≡
N
+
3
B
E
F
−
F
−
6
B
E
N
−
F
=
941
+
3
(
155
)
−
6
(
272
)
=
−
226
k
J
m
o
l
−
1
For the reaction,
N
2
+
3
C
l
2
⟶
2
N
C
l
3
∆
H
f
=
B
E
N
≡
N
+
3
B
E
C
l
−
C
l
−
6
(
B
E
N
−
C
l
)
=
941
+
3
(
242
)
−
6
(
200
)
=
+
467
k
J
m
o
l
−
1
© examsnet.com
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