$2\mathrm{NOCl}\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{Cl}}_2( \mathrm{g})$
In an experiment, 2.0 moles of NOCl was placed in a one-litre flask and the concentration of NO after equilibrium established, was found to be 0.4 $\mathrm{mol}/\mathrm{L}$. The equilibrium constant at $30°\mathrm{C}$ is ________$\times {10}^{-4}$

The given reaction is: 2 NOCl(g) ⇌ 2 NO(g) + Cl₂(g)

In an experiment, 2.0 moles of NOCl were placed in a one-litre flask, and the concentration of NO at equilibrium was found to be 0.4 mol/L. We are tasked with calculating the equilibrium constant at 30°C.

Step-by-step Solution:

1. Initial Setup:

• We start with 2.0 moles of NOCl in a 1-litre flask.

• The initial concentration of NOCl is:

   [NOCl]₀ = 2.0 moles / 1 L = 2.0 mol/L 

• The initial concentrations of NO and Cl₂ are both 0.

2. Change in Concentration:

• Let x be the amount of NOCl that dissociates at equilibrium.
• The change in concentration will be:
  • For NOCl: -2x
  • For NO: +2x
  • For Cl₂: +x

3. Equilibrium Concentrations:

• At equilibrium, the concentrations will be:
  • [NOCl] = 2.0 - 2x
  • [NO] = 2x
  • [Cl₂] = x

4. Given Information:

• The concentration of NO at equilibrium is given as 0.4 mol/L.

• Therefore, we can set up the equation:

  2x = 0.4 ⟹ x = 0.2

5. Substituting x Back:

• Now, we can find the equilibrium concentrations:
  • [NOCl] = 2.0 - 2(0.2) = 2.0 - 0.4 = 1.6 mol/L
  • [NO] = 2(0.2) = 0.4 mol/L
  • [Cl₂] = 0.2 mol/L

6. Equilibrium Constant Expression:

• The equilibrium constant, K, is given by the expression:

  K = ([NO])² ([Cl₂]) / ([NOCl])²        

• Substitute the equilibrium concentrations into the expression:

  K = (0.4)² (0.2) / (1.6)²        

7. Calculating K:

• First, calculate the numerator:

  (0.4)² = 0.16 and 0.16 × 0.2 = 0.032        

• Next, calculate the denominator:

  (1.6)² = 2.56        

• Therefore:

  K = 0.032 / 2.56 = 0.0125        

8. Expressing in Scientific Notation:

• We can express K in scientific notation:

  K = 1.25 × 10^-2

• Since the problem asks for the answer in the form of x × 10^-4, we express:

  K = 125 × 10^-4

Final Answer:

The equilibrium constant K at 30°C is 125 × 10^-4.