For CaCO3, = CaO(s) + CO2,, at
6400
equilibrium log Kp = 8-T then,
temperature at which Kp = 1?
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Answer:
In the preparation of CaO from CaCO
3
using the equilibrium
CaCO
3
(s)⇌CaO(s)+CO
2
(g)
K
p
is expressed as
logK
p
=7.282−
T
8500
For complete decomposition K
p
=1
logK
p
=0
logK
p
=7.282−
T
8500
0=7.282−
T
8500
7.282=
T
8500
T=1167.26K
T=1167.26−273.15
o
C
T=894.11
o
C
Thus, for the complete decomposition of CaCO
3
, the temperature in celsius to be used is T=894
o
C
Explanation:
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