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.

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.