What is the equilibrium concentrations of each of the substances in the equilibrium when the initial concentration of I Cl was 0.78 M ? 2I Cl(g) hArr I_2(g) + Cl_2(g) , K_c=0.14

What is the equilibrium concentrations of each of the substances in th

1.	What is the equilibrium concentrations of each of the substances in the equilibrium when the initial concentration of I Cl was 0.78 M ? 2I Cl(g) hArr I_2(g) + Cl_2(g) , K_c=0.14
Answer» Solution :Let at EQUILIBRIUM , `[I_2]=[Cl_2]="x mol L"^(-1)` `{:(,2ICl(g) hArr, I_2(g) +,Cl_2(g)),("Initial conc.",0.78 M, 0,0),("At EQU.", 0.78-2x, x ,x):}` `K_c=([I_2][Cl_2])/[IC l]^2` `0.14=(x xx x)/(0.78-2x)^2` or `x^2=0.14(0.78-2x)^2` or `x/(0.78-2x)=sqrt0.14=0.374` or x=0.292-0.748 x 1.748 x = 0.292 `therefore` x=0.167 Thus, at equilibrium , `[I_2]` =0.167 M , `[Cl_2]` =0.167 M , [I Cl]=0.78 - 2 x 0.167 = 0.446 M