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AIEEE 2003 Chemistry Solved Paper
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© examsnet.com
Question : 19
Total: 76
If at
298
K
the bond energies of
C
−
H
,
C
−
C
,
C
=
C
and
H
−
H
bonds are respectively
414
,
347
,
615
and 435
kJ
∕
mol
, the value of enthalpy change for the reaction
H
2
C
=
CH
2
(
g
)
+
H
2
(
g
)
→
H
3
C
−
CH
3
(
g
)
at
298
K
will be
−
250
kJ
+
125
kJ
−
125
kJ
+
250
kJ
Validate
Solution:
CH
2
=
CH
2
(
g
)
+
H
2
(
g
)
→
CH
3
−
CH
3
Enthalpy change
=
Bond energy of reactants
−
Bond energy of products.
∆
H
=
1
(
C
=
C
)
+
4
(
C
−
H
)
+
1
(
H
−
H
)
−
1
(
C
−
C
)
−
6
(
C
−
H
)
=
1
(
C
=
C
)
+
1
(
H
−
H
)
−
1
(
C
−
C
)
−
2
(
C
−
H
)
=
615
+
435
−
347
−
2
×
414
=
1050
−
1175
=
−
125
kJ
.
© examsnet.com
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