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KCET 2014 Chemistry Solved Paper
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© examsnet.com
Question : 59
Total: 60
The ratio of heats liberated at 298 K from the combustion of one kg of coke and by burning water gas obtained from 1 kg of coke is
(Assume coke to be 100% carbon.)
(Given enthalpies of combustion of
C
O
2
,
C
O
and
H
2
as 393.5 kJ, 285 kJ, 285 kJ respectively all at 298 K.)
0.69 : 1
0.96 : 1
0.79 : 1
0.86 : 1
Validate
Solution:
One kg of coke
=
1000
12
=
83.33
moles of carbon By burning of one kg of coke
C
Coke
+
O
2
→
C
O
2
;
Δ
c
H
=
83.33
×
393.5
kJ → (1)
By burning of water gas so obtain
C
+
H
2
O
→
C
O
+
H
2
C
O
+
H
2
+
O
2
→
C
O
2
+
H
2
O
;
∆
c
H
=
83.33
×
283.5
k
J
+
83.33
×
285.5
k
J
→ (2)
=
83.33
×
569
k
J
Divide Eq. (1) by Eq. (2), we get
=
83.33
×
393.5
k
J
83.33
×
569
k
J
=
0.69
:
1
Therefore, the ratio of heat liberated is 0.69:1
© examsnet.com
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