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Standard XII
Chemistry
Liquid-Vapour Equillibrium
Question
At what temperature liquid water will be in equilibrium with water vapour?
Δ
H
v
a
p
=
40.73
k
J
m
o
l
−
1
,
Δ
S
v
a
p
=
0.109
k
J
K
−
1
m
o
l
−
1
,
373.6 K
282.4 K
100 K
400 K
A
100 K
B
282.4 K
C
373.6 K
D
400 K
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Solution
Verified by Toppr
The given reaction is :-
H
2
O
(
l
)
⇌
H
2
O
(
g
)
Now, at equilibrium,
△
G
=
0
Now, from fibb's free energy change equation
△
G
=
△
H
−
T
△
S
⇒
△
H
=
T
△
S
⇒
T
=
△
H
△
S
=
40.73
0.1079
=
373.6
K
(
∵
△
H
r
a
p
=
40.73
K
J
/
m
o
l
,
△
S
r
a
p
=
0.109
K
J
K
−
1
m
o
l
−
1
)
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Similar Questions
Q1
At what temperature liquid water will be in equilibrium with water vapour?
Δ
H
v
a
p
=
40.73
k
J
m
o
l
−
1
,
Δ
S
v
a
p
=
0.109
k
J
K
−
1
m
o
l
−
1
,
View Solution
Q2
Calculate the temperature at which liquid water will be in equilibrium with water vapour.
Δ
H
v
a
p
=
40.73
k
J
m
o
l
−
1
and
Δ
S
v
a
p
=
0.109
k
J
m
o
l
−
1
K
−
1
View Solution
Q3
Δ
H
v
a
p
=
30
kJ/mol
and
Δ
S
v
a
p
=
75
J mol
−
1
K
−
1
. Find the temperature of vapour, at one atmosphere:
View Solution
Q4
Δ
H
v
a
p
=
30
k
J
/
m
o
l
and
Δ
S
v
a
p
=
75
J
m
o
l
−
1
K
−
1
. Find the temperature of vapour, at one atmosphere.
View Solution
Q5
Δ
H
v
a
p
=
30
k
J
/
m
o
l
e
and
Δ
S
v
a
p
=
75
Jmol
−
1
K
−
1
. Find temperature of vapour, at one atmosphere.
View Solution