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Byju's Answer
Standard XII
Chemistry
Introduction
Given, Cdiam...
Question
Given,
C
(
d
i
a
m
o
n
d
)
+
O
2
→
C
O
2
;
△
H
=
−
395
k
J
C
(
g
r
a
p
h
i
t
e
)
+
O
2
→
C
O
2
;
△
H
=
−
393
k
J
The heat of formation of the diamond from graphite is
A
+
2.0
k
J
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B
−
1.5
k
J
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C
−
788
k
J
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D
788
k
J
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Solution
The correct option is
A
+
2.0
k
J
C
(
d
i
a
m
o
n
d
)
+
O
2
→
C
O
2
;
△
H
=
−
395
k
J
.
.
.
.
(
i
)
C
(
g
r
a
p
h
i
t
e
)
+
O
2
→
C
O
2
;
△
H
=
−
393
k
J
.
.
.
.
(
i
i
)
on subtracting Eq. (i) by Eq. (ii), we get-
C
(
d
i
a
m
o
n
d
)
→
C
(
g
r
a
p
h
i
t
e
)
△
H
=
[
−
395
−
(
−
393
)
]
k
J
=
−
2.0
k
J
and
C
(
g
r
a
p
h
i
t
e
)
→
C
(
d
i
a
m
o
n
d
)
;
△
H
=
+
2.0
k
J
.
Hence, option A is correct.
Suggest Corrections
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Similar questions
Q.
Determine magnitude of the heat of transformation of
C
d
i
a
m
o
n
d
→
C
(
g
r
a
p
h
i
t
e
)
(
i
n
k
c
a
l
)
from the following data:
C
(
d
i
o
m
o
n
d
)
+
O
2
(
g
)
→
C
O
2
(
g
)
.
.
.
(
i
)
(
Δ
H
=
−
94.5
k
c
a
l
)
C
(
g
r
a
p
h
i
t
e
)
+
O
2
(
g
)
→
C
O
2
(
g
)
.
.
.
(
i
i
)
(
Δ
H
=
−
94.0
k
c
a
l
)
Q.
For the change,
C
d
i
a
m
o
n
d
⟶
+
C
g
r
a
p
h
i
t
e
;
Δ
H
=
−
1.89
k
J
, if 6 g of diamond and 6 g of graphite are separately burnt to yield
C
O
2
the heat liberated in first case is:
Q.
From the thermochemical reactions,
C
(
g
r
a
p
h
i
t
e
)
+
1
2
O
2
→
C
O
;
Δ
H
=
−
110.5
k
J
C
O
+
1
2
O
2
→
C
O
2
;
Δ
H
=
−
283.2
k
J
the heat of reaction of
C
(
g
r
a
p
h
i
t
e
)
+
O
2
→
C
O
2
is:
Q.
Calculate the
Δ
H
for conversion of C (diamond) to C (graphite) when the following reactions are given:
C
(
d
i
a
m
o
n
d
)
+
O
2
→
C
O
2
(
g
)
;
Δ
H
=
−
94.5
K
c
a
l
C
(
g
r
a
p
h
i
t
e
)
+
O
2
→
C
O
2
(
g
)
;
Δ
H
=
−
94
K
c
a
l
Q.
For the change,
C
d
i
a
m
o
n
d
⟶
C
g
r
a
p
h
i
t
e
;
Δ
H
=
−
1.89
k
J
, if
6
g
of diamond
and
6
g
of graphite are separately burnt to yield
C
O
2
the heat liberated in first case is:
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