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Byju's Answer
Standard XII
Chemistry
Difference between Internal Energy and Enthalpy
What is the e...
Question
What is the enthalpy of vaporisation of liquid water in
k
J
m
o
l
−
1
if:
( i )
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
l
)
+
285.77
k
J
m
o
l
−
1
( ii )
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
v
)
+
241.84
k
J
m
o
l
−
1
A
+
43.93
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B
−
43.93
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C
+
527.61
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D
−
527.61
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Solution
The correct option is
A
+
43.93
( i )
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
l
)
+
285.77
k
J
m
o
l
−
1
( ii )
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
v
)
+
241.84
k
J
m
o
l
−
1
Subtracting equation (ii) from equation (i), we get-
H
2
O
(
l
)
⟶
H
2
O
(
g
)
−
43.93
∴
Δ
H
=
+
43.93
k
J
m
o
l
−
1
Hence, option A is correct.
Suggest Corrections
0
Similar questions
Q.
The enthalpy of vapourisation of liquid water using the data
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
;
Δ
H
=
−
285.77
K
J
m
o
l
−
1
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
g
)
;
Δ
H
=
−
241.84
K
J
m
o
l
−
1
Q.
The enthalpy of vaporisation of liquid water using data:
H
2
(
g
)
+
1
/
2
O
2
(
g
)
⟶
H
2
O
(
l
)
;
Δ
H
=
−
285.77
k
J
/
m
o
l
e
H
2
(
g
)
+
1
/
2
O
2
(
g
)
⟶
H
2
O
(
l
)
;
Δ
H
=
−
241.84
k
J
/
m
o
l
e
Q.
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
g
)
;
Δ
H
=
x
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
;
Δ
H
=
y
Heat of vaporisation of water is:
Q.
If for
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
g
)
;
△
H
1
is the enthalpy of reaction and for
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
;
△
H
2
is enthalpy of reaction then:
Q.
If X is the lattice enthalpy of
C
a
C
l
2
, find X. Given that the enthalpy of
(i) Sublimation of
C
a
(
s
)
is
121
k
J
m
o
l
−
1
(ii) Dissociation of
C
l
2
(
g
)
to
2
C
l
(
g
)
is
242.8
k
J
m
o
l
−
1
(iii) Ionization of
C
a
(
g
)
to
C
a
2
+
(
g
)
is
2422
k
J
m
o
l
−
1
(iv) Electron gain enthalpy for
C
l
(
g
)
to
C
l
−
1
(
g
)
is
−
355
k
J
m
o
l
−
1
(v)
△
f
H
overall is
−
795
k
J
m
o
l
−
1
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