Question

46.At 100°C, 0.1mole of N2O4 is heated in a one dm flask. At
equilibrium concentration of NO, was found to be 0.12 moles.
Calculate K, for the reaction.​

Answer 1

Answer:

At 100°C on heating,

N₂O₄ (g) ↔ 2 NO₂ (g)

Now,

No. of moles of N₂O₄ = 0.1 mole

No.of moles of NO₂ = 0.12 mole

Volume of flask = 1 dm = 1 litre

We know,  

Molarity (M) or Molar Concentration = (Moles of solute) / (Volume in litres of solution)  

Therefore,

Molarity of N₂O₄, [N₂O₄] = 0.1 mole / 1 L = 0.1 M

And,

Molarity of NO₂, [NO₂] = 0.12 mole / 1 L = 0.12 M

Thus, at equilibrium, the value of K for the reaction can be calculated as below:

K = [NO₂]² / [N₂O₄] = [0.1]² / [0.12] = 0.01/0.12 = 0.0833

Answer 2

Answer:

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