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Byju's Answer
Standard XII
Chemistry
Valence Bond Theory
Given that : ...
Question
Given that :
2
C
(
s
)
+
2
O
2
(
g
)
→
2
C
O
2
(
g
)
;
Δ
H
=
−
787
kJ
H
2
(
g
)
+
1
/
2
O
2
(
g
)
→
H
2
O
(
l
)
;
Δ
H
=
−
286
kJ
C
2
H
2
(
g
)
+
5
2
O
2
(
g
)
→
2
C
O
2
(
g
)
+
H
2
O
(
l
)
;
Δ
H
=
−
1310
kJ.
Heat of formation of acetylene is:
A
+
1802
KJ
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B
−
1802
KJ
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C
−
800
KJ
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D
+
237
KJ
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Solution
The correct option is
B
+
1802
KJ
Reversing first two equation and
adding upper and then subtract from third equation we get:
C
2
H
2
(
g
)
=
H
2
(
g
)
+
2
C
(
s
)
Δ
H
=
−
1310
K
J
−
(
+
787
K
J
+
286
K
J
)
Δ
H
=
−
1802
K
J
but for the formation of
C
2
H
2
we have to reverse the equation
So
Δ
H
of reaction also reverse
Therefore
Δ
H
=
+
1802
K
J
Suggest Corrections
0
Similar questions
Q.
Consider the following reactions.
(a)
H
+
(
a
q
)
+
O
H
−
(
a
q
)
=
H
2
O
(
l
)
,
Δ
H
=
−
X
1
kJ
m
o
l
−
1
(b)
H
2
(
g
)
+
1
2
O
2
(
g
)
=
H
2
O
(
l
)
,
Δ
H
=
X
2
kJ
m
o
l
−
1
(c)
C
O
2
(
g
)
+
H
2
(
g
)
=
C
O
(
g
)
+
H
2
O
(
l
)
−
X
3
kJ
m
o
l
−
1
(d)
C
2
H
2
(
g
)
+
5
2
O
2
(
g
)
=
2
C
O
2
(
g
)
+
H
2
O
(
l
)
+
X
4
kJ
m
o
l
−
1
Enthalpy of formation of
H
2
O
(
l
)
is:
Q.
2
C
(
s
)
+
2
O
2
(
g
)
→
2
C
O
2
(
g
)
,
Δ
H
=
−
787
k
J
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
,
Δ
H
=
−
286
k
J
C
2
H
2
(
g
)
+
5
2
O
2
(
g
)
→
2
C
O
2
(
g
)
+
H
2
O
(
l
)
,
Δ
H
=
−
1310
k
J
From the above data, heat of formation of acetylene is:
Q.
Given that:
2
C
(
s
)
+
2
O
2
(
g
)
⟶
2
C
O
2
(
g
)
;
Δ
H
=
−
787
k
J
.
.
.
(
i
)
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
l
)
;
Δ
H
=
−
286
k
J
.
.
.
(
i
i
)
C
2
H
2
(
g
)
+
5
2
O
2
(
g
)
⟶
2
C
O
2
(
g
)
+
3
H
2
O
(
l
)
.
.
.
.
(
i
i
i
)
Δ
H
=
−
1301
k
J
Heat formation of acetylene is:
Q.
The enthalpy of vapourisation of water from the following equations is:
H
2
(
g
)
+
1
2
O
2
(
g
)
→
H
2
O
(
l
)
,
△
H
=
−
286
kJ
H
2
(
g
)
+
1
2
O
2
(
g
)
,
→
H
2
O
(
g
)
,
△
H
=
−
245.5
kJ
Q.
For the two equations given below:
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
l
)
+
x
1
k
J
H
2
(
g
)
+
1
2
O
2
(
g
)
⟶
H
2
O
(
g
)
+
x
2
k
J
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