Question

A 2.320 mol quantity of NOCl was initially placed in a 1.750 L reaction chamber at 400°C. After equilibrium was established, it was found that 28.70 percent of the NOCl has dissociated: 2NOCl(g) ⇆ 2NO(g) + Cl2(g) Calculate the equilibrium constant Kc for the reaction.

Answer 1

Answer:

Kc = 0.0307

Explanation:

Step 1: Data given

Number of moles NOCl = 2.320 moles

Volume = 1.750 L

Temperature = 400 °C

NOCl was 28.70 % dissociated

Step 2: The balanced equation

2NOCl(g) ⇔ 2NO(g) + Cl2(g)

Step 3: Calculate molarity NOCl

[NOCl] = moles / volume

[NOCl] = 2.320 moles / 1.750 L

[NOCl] = 1.326 M

Step 4: Initial concentration

[NOCl] = 1.326 M

[NO] = 0M

[Cl2] = 0M

Step 5: Concentration at equilibrium

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

[NOCl] = (1.326 -2X)M

[NO] = 2X M

[Cl2] = XM

Since NOCl is 28.70 % dissociated

([NOCl]0 -[NOCl]equi) = 0.2870[NOCl]0

2x = 0.2870(1.326) ⇒ x  =0.190

Step 6: Calculate Kc

Kc = [NO]²[Cl2] / [NOCl]²

Kc = (2x)²*x /(1.326 - 2x)²

Kc = [2(0.190)]² * (0.190)  / [1.326 - 2(0.190)]²

Kc = 0.0307