Question

In an analysis of interhalogen reactivity, 0.350 mol ICl was placed in a 5.00 L flask and allowed to decompose at a high temperature.
                2 ICl(g) I2(g) + Cl2(g)
                
                Calculate the equilibrium concentrations of I2, Cl2, and ICl. (Kc = 0.110 at this temperature.)
                I2 M
                Cl2 M
                ICl M

Answer 1

Given:

Mol of ICl = 0.350 mol , volume = 5.00 L

Lets calculate [ICl]

[ ICl]= 0.350 mol / 5.00 L = 0.07 M

Lets set up ICE chart

                           2 ICl(g)----- > I2(g) +    Cl2(g)

I                        0.07 M                           0                   0

C                        -2x                   +x               +x

E                         (0.07-2x)              x                     x

Equilibrium expression for this reaction:

Kc = [I2] [Cl2]/ [ICl]²

Lets plug all the value.

0.110 = x² / (0.07 – 2x)²

Lets take a square root of both side

0.332 = x/ (0.07 – 2x)

0.332 (0.07 -2x) = x

0.02322 – 0.663 x = x

x + 0.663 x = 0.02322

1.663 x = 0.02322

x = 0.014

Now equilibrium concentration

[ICl ]= (0.07- 2x ) = 0.07 – 2 x 0.014 = 0.042 M

[I2]= [Cl2]= x = 0.014 M