Question

The equilibrium constant Kc for the reaction below is 6.16 ✕ 10−6 at 185°C. COCl2(g) equilibrium reaction arrow CO(g) + Cl2(g) COCl2 at an initial concentration of 6.70 ✕ 10−2 M is placed into an empty reaction vessel at 185°C and allowed to equilibrate. The concentration of Cl2 measured after equilibration is 6.39 ✕ 10−4 M. Create an I.C.E. table showing the initial concentrations, change in concentrations, and equilibrium concentrations of the reactants and products of this reaction. PLEASE HELP ICE TABLES ARE WHAT I AM THE WORSE AT

Answer 1

The reaction is COCl2 (g)ß-> CO(g)+Cl2(g)

Kc= [CO][Cl2]/[COCl2]

1mole of COCl2 reacts to give 1 mole of Co and 1 mole of Cl2

Let x= change in concentration to reach equilibrium

Preparing iCE table

S.No	Compound	Initial concentration (M)	Change in concentration (M)	Equilibrium concentration (M)
1	CoCl2	6.7*10-2M	-x	6.7*10-2-x
2	CO	0	x	x
3	Cl2	0	x	x

Given at Equilibrium [Cl2]=x= 6.39*10^-4= [Cl2],

[CoCl2] =6.7*10^-2-6.39*10^-4=0.0663

Hence K= 6.39*10^-4*6.39*10^-4/0.0663= 6.16*10^-6