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Byju's Answer
Standard XII
Chemistry
Third Law of Thermodynamics
What is the e...
Question
What is the entropy change for the conversion of one gram of ice to water at
273
K
and one atmospheric pressure?
(
Δ
H
f
u
s
i
o
n
=
6.025
k
J
m
o
l
−
1
)
Open in App
Solution
Δ
H
f
u
s
i
o
n
=
6.025
×
1000
J
m
o
l
−
1
=
6025
18
J
g
−
1
=
334.72
J
g
−
1
Δ
S
f
u
s
i
o
n
=
Δ
H
f
u
s
i
o
n
T
=
334.72
273
=
1.226
J
K
−
1
m
o
l
−
1
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1
Similar questions
Q.
The entropy change for the conversion of the one gram of ice to water at 273 K and one atmospheric pressure
[
Δ
H
f
u
s
i
o
n
=
6.025
k
J
m
o
l
−
1
]
is:
Q.
One mole of ice is converted into water at
273
K
. The entropies of
H
2
O
(
s
)
and
H
2
O
(
l
)
are
38.20
J
m
o
l
−
1
and
60.01
J
m
o
l
−
1
K
−
1
respectively. The enthalpy of change for the conversion is:
Q.
The entropy change associated with the conversion of
1
k
g
of ice at
273
K
to water vapours at
383
K
is:
[Given that : Specific heat of water liquid and water vapour are
4.2
k
J
K
−
1
k
g
−
1
and
2.0
k
J
K
−
1
k
g
−
1
; latent heat of fusion and vapourisation of water are
344
k
j
k
g
−
1
and
2491
k
J
k
g
−
1
, respectively].
[Take :
l
o
g
273
=
2.436
,
l
o
g
373
=
2.572
,
l
o
g
383
=
2.583
]
Q.
At
0
o
C
, ice and water are in equilibrium and enthalpy change for the process
H
2
O
(
s
)
⟶
H
2
O
(
l
)
is
6
k
J
m
o
l
−
1
. Calculate the entropy change for the conversion of ice into liquid water.
Q.
One mole of ice is converted into water at
273
K. The entropies of
H
2
O
(
s
)
and
H
2
O
(
l
)
are
38.20
and
60.01
J
.
m
o
l
−
1
K
−
1
respectively. The enthalpy change for the conversion is:
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