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Byju's Answer
Standard XII
Chemistry
Hess' Law
2C4H10⟶ C8H18...
Question
2
C
4
H
10
⟶
C
8
H
18
+
H
2
. Given that bond energy of
C
−
C
,
C
−
H
are
347.3
and
414.2
kJ mol
−
1
and the heat of formation of hydrogen atom is
217.55
kJ mol
−
1
. Find
Δ
H
for the above reaction in kJ mol
−
1
.
(If
Δ
H
=
x
, input the answer to the nearest integer as
x
9
).
Open in App
Solution
For
2
C
4
H
10
⟶
C
8
H
18
+
H
2
;
Δ
H
=
?
∴
Δ
H
=
bond energy data used for the formation of bond + bond energy data used for dissociation of bond
Δ
H
=
−
[
7
×
E
(
C
−
−
C
)
+
18
×
E
(
C
−
−
H
)
+
E
(
H
−
−
H
)
]
+
2
[
3
×
E
(
C
−
−
C
)
+
10
×
E
(
C
−
−
H
)
]
Δ
H
=
−
e
(
C
−
−
C
)
+
2
×
e
(
C
−
−
H
)
−
e
(
H
−
−
H
)
∵
Given,
1
2
H
2
⟶
H
;
Δ
H
=
217.55
kJ
∴
e
(
H
−
−
H
)
=
2
×
217.55
=
435.10
kJ
∴
Δ
H
=
−
347.3
+
2
×
(
414.2
)
−
1
×
(
435.1
)
Δ
H
=
46.0
kJ mol
−
1
Now,
46
9
=
5.11
≡
5
kJ mol
−
1
Suggest Corrections
0
Similar questions
Q.
Using the data provided, calculate the multiple bond energy
(
k
J
m
o
l
−
1
)
of a
C
≡
C
bond in
C
2
H
2
.
Given that the heat of formation of
C
2
H
2
=
225
k
J
m
o
l
−
1
.
(take the bond energy of
C
−
H
bond as
350
k
J
m
o
l
−
1
):
2
C
(
s
)
→
2
C
(
g
)
△
H
=
1410
k
J
m
o
l
−
1
H
2
(
g
)
→
2
H
(
g
)
△
H
=
330
k
J
m
o
l
−
1
Q.
What will be
Δ
H
for the reaction,
C
H
2
C
l
2
→
C
+
2
H
+
2
C
l
?
(B.E. of C - H and C - Cl bonds are 416 kJ
m
o
l
−
1
and 325 kJ
m
o
l
−
1
respectively)
Q.
I
.
E
a
=
15
k
J
m
o
l
−
1
;
Δ
H
=
−
70
k
J
m
o
l
−
1
I
I
.
E
a
=
30
k
J
m
o
l
−
1
;
Δ
H
=
−
15
k
J
m
o
l
−
1
I
I
I
.
E
a
=
60
k
J
m
o
l
−
1
;
Δ
H
=
+
20
k
J
m
o
l
−
1
If above reactions are at same frequency factor and temperature, then fastest and slowest reactions are:
Q.
What is the heat of formation of liquid methyl alcohol in
k
J
m
o
l
−
1
,
use the following data.
Heat of vaporization of liquid methyl alcohol =
38
k
J
m
o
l
−
1
.
Heat of formation of gaseous atoms from the elements in their standard states:
H
=
218
k
J
m
o
l
−
1
,
C
=
715
k
J
m
o
l
−
1
,
O
=
249
k
J
m
o
l
−
1
Average bond energies :
C
−
H
=
415
k
J
m
o
l
−
1
C
−
O
=
356
k
J
m
o
l
−
1
O
−
H
=
463
k
J
m
o
l
−
1
Q.
Find the enthalpy of formation of Hydrogen flouride on the basis of following data :
Bond energy of
H
−
H
=
434
k
J
m
o
l
−
1
Bond energy of
F
−
F
=
158
k
J
m
o
l
−
1
Bond energy of
H
−
F
=
565
k
J
m
o
l
−
1
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