In the reaction C(s) + CO_(2)(g)hArr2CO(g) the following amounts of sbstance were formed in 0.2 litre flask CO_(2) = 0.06 mole. The equilibrium constant is

In the reaction C(s) + CO_(2)(g)hArr2CO(g) the following amounts of sb

1.	In the reaction C(s) + CO_(2)(g)hArr2CO(g) the following amounts of sbstance were formed in 0.2 litre flask CO_(2) = 0.06 mole. The equilibrium constant is
Answer» `0.208` `4.10` `0.30` `0.416` SOLUTION :`K = ([CO]^(2))/([CO_(2)]) = (((0.05)/(0.2))^(2))/(((0.06)/(0.2))` `K = 0.208`