2NOCI (g)
→ 2NO(g) + Cl2 (g)
Problem:
Initially
the above reaction contains
0.50 M of NOCI,
0.00 M of NO,
0.00 M of C12.
Equilibrium Constant Kc = 1.6 (10^5)
What is the Equilibrium
Concentration of Cl2?

0.0097 mol·L⁻¹

Explanation:

The balanced equation is

2NOCl ⇌ 2NO₂ + Cl₂

Data:

Your value of Kc is incorrect. It should be

       Kc =1.6 × 10⁻⁵

[NOCl] = 0.50 mol·L⁻¹

   [NO] = 0.00 mol·L⁻¹

   [Cl₂] = 0.00 mol·L⁻¹

Calculations:

1. Set up an ICE table.

2. Calculate the equilibrium concentrations

This is a cubic equation. Some calculators can solve cubic equations, but we can solve it by the method of successive approximations.

We will make changes to the right-hand side until the calculated value of x no longer changes,

(a) 1st approximation

Assume that 2x is negligible compared to 0.50. Then

(b) 2nd approximation

Assume that x= 0.010. Then

(b) 3rd approximation

Assume that x= 0.0097. Then

No change, so x = 0.0097.

[Cl₂] = x mol·L⁻¹ = 0.0097 mol·L⁻¹

Check:

It checks.

can you list the answers so i can answer