Question

Nitrosyl chloride, NOCI, dissociates on heating as shown below. When a 1.50 gram
sample of pure NOCI is heated at 350°C in a volume of 1.00 liter, the percent
dissociation is found to be 57.2%. Calculate Kc for the reaction as written.
NOCI(9) = NO(g) + 12 Cl2(9)

Answer 1

From the equation of the reaction given, the equilibrium constant, Kc, of the reaction is 0.105 M^1/2

What is the valueof Kc of the reaction?

The equilibrium constant, Kc, of the reaction is determined from the equation of the reaction given below:

• NOCI(g) <===> NO(g) + 1/2 Cl2(g)
• Kc = [NO][Cl]^0.5/[NOCl]

number of moles of NOCl = mass/molar mass

molar mass of NOCl = 65.5

number of moles of NOCl = 1.5/65.5 = 0.0229 moles

Moles of NOCl reacted = 57.2/100 × 0.0229 = 0.013 moles

From equation of the reaction:

Number of moles of NO at equilibrium = 0.013 moles

Number of moles of Cl2 at equilibrium = 0.013/2 = 0.0065 moles

Number of moles of NOCl at equilibrium = 0.0229 - 0.013 = 0.0099 moles

Kc = [NO][Cl]^0.5/[NOCl]

Kc = 0.013 x 0.0065^0.5/0.0099

Kc = 0.105 M^1/2

Therefore, the equilibrium constant, Kc, of the reaction is 0.105 M^1/2

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