For `CaCO_(3)(s) hArr CaO(s)+CO_(2)(g), K_(c)` is equal to …………..

For the reaction CaCO_(3(s)) hArr CaO_((s))+CO_(2(g))k_p is equal to

For the reaction CaCO_(3(s)) hArr CaO_((s))+CO_(2(g))k_p is equal to

Find out the value of K_(c) for each of the following equilibrium from the value of K_(p) : CaCO_(3)(s) hArr CaO(s) + CO_(2)(g), K_(p) = 167 atm at 1073 K.

In the preparation of CaO from CaCO_(3) using the equilibrium, CaCO_(3)(s) hArr CaO(s)+CO_(2)(g) K_(p) is expressed as log K_(p)=7.282-8500/T For complete decomposition of CaCO_(3) , the temperature in celsius to be used is:

In the preparation of CaO from CaCO_(3) using the equilibrium, CaCO_(3)(s) hArr CaO(s)+CO_(2)(g) K_(p) is expressed as log K_(p)=7.282-8500/T For complete decomposition of CaCO_(3) , the temperature in celsius to be used is:

At 27^(@)C, K_(p) value for the reaction CaCO_(3)(s) hArr CaO (s) + CO_(2)(g) , is 2 atm. K_(C) value for this reaction is

For the reaction CaCO_(3)(s) hArr CaO(s) + CO_(2)(g) K_(p)=1.16 atm at 800^(@)C . If 20.0g of CaCO_(3) was put into a 10.0 L flask and heated to 800^(@)C, what percentage of CaCO_(3) would remain unreacted at equilibrium ?

At 27^(@)C, K_(p) value for the reaction CaCO_(3)(s) hArr CaO (s) + CO_(2)(g) , is 0.2 atm. K_(C) value for this reaction is