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JEE Main Chemistry Class 11 Chemical Thermodynamics Part 2 Questions
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© examsnet.com
Question : 5
Total: 88
Combustion of 1 mole of benzene is expressed at
C
6
H
6
(
1
)
+
15
2
O
2
(
g
)
⟶
CO
2
(
g
)
+
3
H
2
O
(
1
)
.
The standard enthalpy of combustion of
2
mol
of benzene is - '
x
'
kJ
.
x
=
_______.
(1) standard Enthalpy of formation of
1
mol
of
C
6
H
6
(
1
)
, for the reaction
6
C
(graphite)
+
3
H
2
(
g
)
⟶
C
6
H
6
(
1
)
is
48.5
kJ
mol
−
1
.
(2) Standard Enthalpy of formation of
1
mol
of
CO
2
(
g
)
, for the reaction
C
(graphite)
+
O
2
(
g
)
⟶
CO
2
(
g
)
is
−
393.5
kJ
mol
−
1
.
(3) Standard and Enthalpy of formation of
1
mol
of
H
2
O
(
1
)
, for the reaction
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
1
)
is
−
286
kJ
mol
−
1
.
[5 Apr 2024 Shift 2]
Your Answer:
Validate
Solution:
6
C
(
graphite
)
+
3
H
2
(
g
)
⟶
C
6
H
6
(
ℓ
)
;
∆
H
=
48.5
kJ
∕
mol
C
(
graphite
)
+
O
2
(
g
)
⟶
CO
2
(
g
)
;
∆
H
=
−
393.5
kJ
∕
mol
H
2
(
g
)
+
1
2
(
g
)
→
H
2
O
(
ℓ
)
;
∆
H
=
−
286
kJ
∕
mol
equation
−
(
1
)
×
1
+
(
2
)
×
6
+
(
3
)
×
3
−
48.5
−
6
×
393.5
−
3
×
286
=
−
3267.5
kJ
for
1
mol
=
−
6535
kJ
for
2
mol
Ans.
6535
kJ
© examsnet.com
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