In the preparation of CaO from CaCO 3 using the equilibrium, CaCO _ 3 ( ~s ) CaO ( ~s )+…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
In the preparation of CaO from CaCO3 using the equilibrium, \(\mathrm{CaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{CaO}(\mathrm{~s})+\mathrm{CO}_{2}(\mathrm{~g}),\) (At 1 atm)
Kp is expressed as log Kp = 7.282\(-\frac{8500}{\mathrm{~T}}\).
For complete decomposition of CaCO3 the temperature in celsius to be used is :
Kp is expressed as log Kp = 7.282\(-\frac{8500}{\mathrm{~T}}\).
For complete decomposition of CaCO3 the temperature in celsius to be used is :
- 1167
- 894
- 8500
- 850
Answer
(B) 894
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