Which are heterogenous equilibrium ?state the unit of K_c. (i)H_2O_((l)) hArr H_2O_((g)) (ii)Ca(OH)_(2(s)) + aq hArr Ca_((aq))^(2+) + 2OH_((aq))^(-) (iii)CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) (iv)CH_3COOC_2H_(5(l)) + H_2O_((l)) hArr CH_3COOH_((l)) + C_2H_5OH_((l)) (v)Fe_((aq))^(3+) + SCN_((aq))^(-) hArr [Fe(SCN)]_((aq))^(2+)

Which are heterogenous equilibrium ?state the unit of K_c. (i)H_2O_((

1.	Which are heterogenous equilibrium ?state the unit of K_c. (i)H_2O_((l)) hArr H_2O_((g)) (ii)Ca(OH)_(2(s)) + aq hArr Ca_((aq))^(2+) + 2OH_((aq))^(-) (iii)CaCO_(3(s)) hArr CaO_((s)) + CO_(2(g)) (iv)CH_3COOC_2H_(5(l)) + H_2O_((l)) hArr CH_3COOH_((l)) + C_2H_5OH_((l)) (v)Fe_((aq))^(3+) + SCN_((aq))^(-) hArr [Fe(SCN)]_((aq))^(2+)
Answer» Solution :(i)`K_c=[H_2O]="MOL L"^(-1)` (ii)`K_c=[Ca^(2+)][OH^-]^2 ("mol L"^(-1))^3` (iii) `K_c=[CO_2] "mol L"^(-1)` (iv)`K_c=([(CH_3COOH)][(C_2H_5OH)])/([CH_3COOC_2H_5][H_2O])` no UNIT (V)`K_c=([Fe(SCN)^(2+)])/([Fe^(3+)][SCN^(2+)])` Thus unit of `K_c` is `("mol" L^(-1))^(-1)` or L `mol^(-1)`