Question

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICI was 0.78 M? 2ICl (g) ⇆ I₂ (g) + Cl₂ (g); Kc = 0.14

Answer 1

Here equilibrium constant Kc= [I2] [Cl2] / [Icl]2.

From the given chemical reaction, as there are no reactants which are in solid form, there is no need to ignore any of the reactants.

By using required procedures and solving the equation by giving the values, we will get the [H2] = [I2] =0.167M and the [HI] = 0.446 M.  

Answer 2

Answer:

2ICL ⇆ I2 + CL2

Initially 0.78M 0 0

At eqm., 0.78-2x x x

Kc =([I2][CL2])/[ICL]

=>(X*X)/(0.78-2X)²=0.14

=> Computing this we get X=0.167

[ICL]=0.78-2×0.167=0.446M

[I2]=0.167M

[CL2]=0.167M.