What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl_((g)) hArr I_(2(g)) + Cl_(2(g)) , K_c=0.14

What is the equilibrium concentration of each of the substances in the

1.	What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl_((g)) hArr I_(2(g)) + Cl_(2(g)) , K_c=0.14
Answer» SOLUTION :`{:("Equilibrium reaction :", 2ICl_((g)) hArr , I_(2(g)) + , Cl_(2(g)), K_c=0.14),("Initial concentration:",0.78 M, 0,0),("Change in reaction :", -2x M , +X M, x M),("Concen. at equilibrium [M] :", (0.78-2x) , x, x):}` This EXPRESSION of equilibrium constant `K_c` for this reaction `K_c=([I_2][Cl_2])/([ICl]^2)=((x)(x))/((0.78-2x))` `therefore 0.14=x^2/(0.78-2x)^2` `therefore sqrt0.14 =x/(0.78-2x)` `therefore` 0.3742(0.78-2x)=x `therefore` 0.2918-0.7484 x= x `therefore` 1.7484 x = 0.2918 `therefore x=0.2918/1.7484`=0.1669 M `approx` 1.67 M So, at equilibrium [ICl] = 0.78-2x =0.78-2(0.1669) =0.78-0.3338=0.4462 M At equilibrium `[I_2]=[Cl_2]` = x= 0.1669 M