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Byju's Answer
Standard XII
Chemistry
Enthalpy
Calculate C...
Question
Calculate
C
−
C
l
bond enthalpy from following reaction :
C
H
3
C
l
(
g
)
+
C
l
2
(
g
)
⟶
C
H
2
C
l
2
(
g
)
+
H
C
l
(
g
)
Δ
H
o
=
−
104
k
J
. If
C
−
H
,
C
l
−
C
l
and
H
−
C
l
bond enthalpies are
414
,
243
and
431
k
J
m
o
l
−
1
respectively.
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Solution
Δ
H
=
∑
H
r
e
a
c
t
a
n
t
b
o
n
d
−
∑
H
p
r
o
d
u
c
t
b
o
n
d
−
104
=
(
3
×
H
C
−
H
+
H
C
−
C
l
+
H
C
l
−
C
l
)
−
(
2
×
H
C
−
H
+
H
C
−
C
l
+
H
H
−
C
l
)
−
104
=
(
3
×
414
+
X
+
243
)
−
(
2
×
414
+
2
X
+
431
)
−
104
=
(
1242
+
X
+
243
)
−
(
828
+
2
X
+
431
)
−
104
=
226
−
X
∴
X
=
330
k
J
m
o
l
−
1
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0
Similar questions
Q.
Calculate the bond energy of
C
l
−
C
l
bond from the following data:
C
H
4
(
g
)
+
C
l
(
g
)
→
C
H
3
C
l
(
g
)
+
H
C
l
(
g
)
;
Δ
H
=
−
100.3
k
J
. Also the bond enthalpies of
C
−
H
,
C
−
C
l
,
H
−
C
l
bonds are
413
,
326
and
431
k
J
m
o
l
−
1
respectively.
Q.
The following reaction
has
Δ
H
as
−
25
k
c
a
l
.
C
H
4
(
g
)
+
C
l
2
(
g
)
→
C
H
3
C
l
(
g
)
+
H
C
l
(
g
)
Bond
Bond Energy kCal
ε
C
−
C
l
84
ε
H
−
C
l
109
ε
C
−
H
x
ε
C
l
−
C
l
y
x
:
y
=
9
:
5
From the given data, what is the bond enthalpy of
C
l
−
C
l
bond?
Q.
Calculate the enthalpy of the reaction,
H
2
(
g
)
+
C
l
2
(
g
)
⟶
2
H
C
l
(
g
)
given that, bond energies of
H
−
H
,
C
l
−
C
l
and
H
−
C
l
are
436
,
243
and
432
k
J
m
o
l
−
1
.
Q.
Calculate the enthalpy ,
Δ
H
, for the given reaction using the the given bond energies ?
C
2
H
4
+
C
l
2
→
C
l
H
2
C
−
C
H
2
C
l
Bond energies
kJ/ mol
C
−
C
347
C
=
C
612
C
−
C
l
341
C
−
H
414
C
l
−
C
l
243
Q.
The reaction
C
H
4
(
g
)
+
C
l
2
→
C
H
3
C
l
(
g
)
+
H
C
l
(
g
)
has
δ
H
=
−
25
k
c
a
l
.
From the data given below, what is the bond enthalpy of
C
l
−
C
l
bond?
Bond
Bond Energy (kcal)
C
−
C
84
C
−
C
l
81
C
−
H
x
C
l
−
C
l
y
H
−
C
l
103
Given:
x
:
y
=
9
:
5
.
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