Question

At 35°C, K = 1.6 × 10^-5 for the reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

Calculate the concentrations of all species at equilibrium for each of the following original mixtures.
a. 2.0 mol pure NOCl in a 2.0 L flask
b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Answer 1

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2] = 1.004 M

Step-by-step explanation:

Step 1: Data given

Temperature = 35°C = 308K

K = 1.6 × 10^-5

Step 2: The reaction

2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)

For 2 moles NOCl we'll have 2 moles NO and 1 mol Cl2

Step 3

a. 2.0 mol pure NOCl in a 2.0 L flask

Concentration at the start:

Concentration = mol / volume

[NOCl] = mol / volume

[NOCl] = 2.0 / 2.0 L

[NOCl] = 1.0 M

[NO] = 0 M

[Cl] = 0M

Concentration at the equillibrium

[NOCl] = 1.0M - 2x

[NO] = 2x

[Cl2]= x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6*10^-5 = ((2x)² * x) / (1.0-2x)²

x = 0.016

[NOCl] = 1.0 - 2*0.016 = 0.968 M

[NO] = 2*0.016 = 0.032M

[Cl²] = 0.016M

b. 2.0 mol NOCl and 1.0 mol Cl2 in a 1.0 L flask

Concentration at the equillibrium

[NOCl] = 2.0 mol / 1.0 L = 2.0 M

[NO] = 0 M

[Cl2]= 1.0 mol / 1.0 L = 1.0 M

Concentration at the equillibrium

[NOCl] = 2.0M - 2x

[NO] = 2x

[Cl2]= 1.0 + x

K = [Cl2][NO]² / [NOCl]² = 1.6*10^-5

1.6 *10^-5 = (2x)²*(1.0+x) / ((2.0-2x)²)

1.6 *10^-5= (2x)² * 1 )/2.0²

1.6 *10^-5= 4x² / 4 = x²

x =
`√(1.6 *10^-5)` = 4.0*10^-3

[NOCl] = 2.0 - 2*0.004 = 1.992M

[NO] = 2*0.004 = 0.008 M

[Cl2] = 1+ 0.004M = 1.004 M