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

a) [NOCl] = 0.968 M

[NO] = 0.032M

[Cl²] = 0.016M

b) [NOCl] = 1.992M

[NO] = 0.008 M

[Cl2]  = 1.004 M

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 = = 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

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

if it's not useful, feel free to report for mistake in answer.