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Byju's Answer
Standard XII
Chemistry
Enthalpy
The enthalpy ...
Question
The enthalpy change for the process
C
(
g
r
a
p
h
i
t
e
)
→
C
(
g
)
is called
A
heat of vaporisation
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B
heat of sublimation
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C
heat of allotropic change
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D
heat of atomisation
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Solution
The correct options are
B
heat of sublimation
D
heat of allotropic change
Sublimation is the process of changing a solid into a gas without passing through the liquid phase.
So this process is the heat of sublimation.
C
(
g
r
a
p
h
i
t
e
)
→
C
(
g
)
but here also an allotropic form of C changes so this is also the heat of allotropic change.
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Similar questions
Q.
For the allotropic change represented by the equation
C
(
g
r
a
p
h
i
t
e
)
→
C
(
d
i
a
m
o
n
d
)
,
δ
H
=
1.9
k
J
.
If 6 g of diamond and 6 g of graphite are separately burnt to yield
C
O
2
,
the enthalpy liberated in first case is:
Q.
For the allotropic change represented by the equation:
C
(
g
r
a
p
h
i
t
e
)
→
C
(
d
i
a
m
o
n
d
)
,
△
H
=
1.9
k
J
.
.
If
6
g of diamond and
6
g of graphite are separately burnt to yield
C
O
2
, the enthalpy liberated in first case is?
Q.
If the enthalpy of combustion of diamond and graphite are
−
395.4
k
J
m
o
l
−
1
and
−
393.6
k
J
m
o
l
−
1
,
The enthalpy change for the
C
(
g
r
a
p
h
i
t
e
)
⟶
C
(
d
i
a
m
o
n
d
)
is :
Q.
Calculate the enthalpy change accompaning the conversion of
10
g
of graphite into diamond if the heats of combustion of
C
(
g
r
a
p
h
i
t
e
)
and
C
(
d
i
a
m
o
n
d
)
are
−
94.05
k
c
a
l
and
−
94.50
k
c
a
l
respectively.
Q.
What will be the enthalpy change of conversion of graphite into diamond?
Given
C
g
r
a
p
h
i
t
e
,
△
c
o
m
b
H
= -391.25 kJ;
C
d
i
a
m
o
n
d
,
△
c
o
m
b
H
= -393.12 kJ
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