Tardigrade
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Tardigrade
Question
Chemistry
Given standard enthalpy of formation of CO and CO 2 are (-110 kJ mol -1) and (-394 kJ mol -1) respectively. The heat of combustion when one mole of graphite burns is
Q. Given standard enthalpy of formation of
CO
and
C
O
2
are
(
−
110
k
J
m
o
l
−
1
)
and
(
−
394
k
J
m
o
l
−
1
)
respectively. The heat of combustion when one mole of graphite burns is
3508
180
Thermodynamics
Report Error
A
−
110
k
J
0%
B
−
284
k
J
34%
C
−
394
k
J
26%
D
−
504
k
J
39%
Solution:
Combustion of graphite,
C
+
O
2
→
C
O
2
Δ
H
=
−
394
k
J
m
o
l
−
1
=
heat of formation of
C
O
2