Answer:
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