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Byju's Answer
Standard XII
Chemistry
Enthalpy
The enthalpy ...
Question
The enthalpy changes at
298
K
in successive breaking of
O
−
H
bonds of water are
H
2
O
(
g
)
→
H
(
g
)
+
O
H
(
g
)
;
Δ
H
=
498
k
J
m
o
l
−
1
O
H
(
g
)
→
H
(
g
)
+
O
(
g
)
;
Δ
H
=
428
k
J
m
o
l
−
1
The bond enthalpy of
O
−
H
bond is:
A
498
k
J
m
o
l
−
1
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B
428
k
J
m
o
l
−
1
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C
70
k
J
m
o
l
−
1
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D
463
k
J
m
o
l
−
1
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Solution
The correct option is
D
463
k
J
m
o
l
−
1
Suggest Corrections
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Similar questions
Q.
The enthalpy changes at
298
K in successive breaking of
O
−
H
bonds of
H
O
H
are:
H
2
O
(
g
)
→
H
(
g
)
+
O
H
(
g
)
,
Δ
H
=
498
kJ
m
o
l
−
1
O
H
(
g
)
→
H
(
g
)
+
O
(
g
)
,
Δ
H
=
428
kJ
m
o
l
−
1
The bond enthalpy of the
O
−
H
bond is:
Q.
The enthalpies of formation of
O
H
(
g
)
,
H
(
g
)
and
O
(
g
)
are 21
k
J
m
o
l
−
1
, 109
k
J
m
o
l
−
1
and 124
k
J
m
o
l
−
1
.
The value of bond enthalpy
O
−
H
is
Q.
The enthalpy of combustion of
H
2
(
g
)
to give
H
2
O
(
g
)
is
−
249
kJ mol
−
1
and bond enthalpies of
H
−
H
and
O
=
O
are
433
kJ mol
−
1
and
492
kJ mol
−
1
respectively. The bond enthalpy of
O
−
H
is:
Q.
The enthalpy of combustion of
H
2
(
g
)
at 298 K to give
H
2
O
(
g
)
is
−
249
k
J
m
o
l
−
1
and bond enthalpies of H - H and O = O are
433
k
J
m
o
l
−
1
and
492
k
J
m
o
l
−
1
,
respectively. The bond enthalpy of O - H is
Q.
Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g)
Bond energy
C
−
H
=
413.4
k
J
m
o
l
−
1
;
Bond energy
C
−
C
=
347.0
k
J
m
o
l
−
1
;
Bond energy
C
=
O
=
728.0
k
J
m
o
l
−
1
;
Bond energy
O
=
O
=
495.0
k
J
m
o
l
−
1
;
Bond energy
H
−
H
=
435.8
k
J
m
o
l
−
1
;
Δ
H
s
u
b
C
(
s
)
=
718.4
k
J
m
o
l
−
1
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