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Question
Chemistry
Calculate Δ H in kJ for the following reaction C (g)+ O 2(g) longrightarrow CO 2(g) Given that, H 2 O (g)+ C (g) longrightarrow CO (g)+ H 2(g);Δ H=+131 kJ CO (g)+(1/2) O 2(g) longrightarrow CO 2(g);Δ H=-282 kJ H 2(g)+(1/2) O 2(g) longrightarrow H 2 O (g);Δ H=-242 kJ
Q. Calculate
Δ
H
in
k
J
for the following reaction
C
(
g
)
+
O
2
(
g
)
⟶
C
O
2
(
g
)
Given that,
H
2
O
(
g
)
+
C
(
g
)
⟶
CO
(
g
)
+
H
2
(
g
)
;
Δ
H
=
+
131
k
J
CO
(
g
)
+
2
1
O
2
(
g
)
⟶
C
O
2
(
g
)
;
Δ
H
=
−
282
k
J
H
2
(
g
)
+
2
1
O
2
(
g
)
⟶
H
2
O
(
g
)
;
Δ
H
=
−
242
k
J
1790
211
EAMCET
EAMCET 2008
Report Error
A
-393
B
+393
C
+655
D
-655
Solution:
Given,
H
2
O
(
g
)
+
C
(
g
)
⟶
CO
(
g
)
+
H
2
(
g
)
;
Δ
H
=
131
k
J
…
(
i
)
CO
(
g
)
+
2
1
O
2
(
g
)
⟶
C
O
2
(
g
)
;
Δ
H
=
−
282
k
J
…
(
ii
)
H
2
(
g
)
+
2
1
O
2
(
g
)
⟶
H
2
O
(
g
)
;
Δ
H
=
−
242
k
J
…
(
iii
)
C
(
g
)
+
O
2
(
g
)
⟶
C
O
2
(
g
)
;
Δ
H
=
?
…
(
i
v
)
On adding Eqs (i), (ii) and (iii), we get Eq (iv)
H
2
O
(
g
)
+
C
(
g
)
⟶
CO
(
g
)
+
H
2
(
g
)
;
Δ
H
=
+
131
k
J
CO
(
g
)
+
2
1
O
2
(
g
)
⟶
C
O
2
(
g
)
;
Δ
H
=
−
282
k
J
H
2
(
g
)
+
2
1
O
2
(
g
)
⟶
H
2
O
(
g
)
;
Δ
H
=
−
242
k
J
__________________________________________
C
(
g
)
+
O
2
(
g
)
⟶
C
O
2
(
g
)
Δ
H
=
(
131
−
282
−
242
)
k
J
=
−
393
k
J