Find out the value of `K_c` for each of the following equilibria from the value of `K_p` :
(i)`2NOCl_((g)) hArr 2NO_((g)) + Cl_(2(g)) , K_p=1.8xx10^(-2)` , 500 K
(ii)`CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) , K_p`=167 , 1073 K

Explanation

(i)`{:("Reaction equilibrium:",2NOCl_((g)) hArr, 2NO_((g))+, Cl_(2(g))),("Stoichiometric multiply:", 2,2,1):}`
All are gaseous , So,
`Deltan_((g))` = (Difference of coefficient of products and reactant gas compount)
`therefore Deltan_((g))`=(2+1)-(2)=+1
T=500 K, `K_p=1.8xx10^(-2)`
R=0.0831 L bar `"mol"^(-1) K^(-1)`
`K_p=K_c(RT)^(Deltan)_((g))`
`therefore K_c=K_p/(RT)^(Deltan_((g)))=(1.8xx10^(-2))/{{(0.0831)(500)}}^1`
`=(1.8xx10^(-2))/(0.0831xx5xx10^2)`
`=4.3321xx10^(-4) "mol L"^(-1)`
(If R=0.0821 L atm `"mol"^(-1) K^(-1)` So, answer `4.384xx10^(-4) "mol L"^(-1)` )
(ii)`CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g))`
In equilibrium solid is ignore,
`Deltan_((g))`=(Coefficient of gaseous product )-(Coefficient of gaseous reactants )
=+1-0=+1
`K_p`=167 T=1073 K
R=0.0831 L bar `"mol"^(-1) K^(-1)`