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In the process for preparation of NH3,if 1 mole N2,3 mole of H2 1.6 mole of Nh3 are present at equullibrium at C Then what will be the value of equillibrium constant? A. 32 B. 16 C. 8 D. 4?
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In the process for preparation of NH3,if 1 mole N2,3 mole of H2 1.6 mo...
Equilibrium Constant for Preparation of NH3

Given:

1 mole N2

3 mole H2

1.6 mole NH3

At equilibrium, the balanced equation for the preparation of NH3 is:

N2(g) + 3H2(g) ⇌ 2NH3(g)

The equilibrium constant expression for this reaction is:

Kc = [NH3]^2 / [N2][H2]^3

Now, substitute the given values in the expression:

Kc = (1.6)^2 / (1)(3)^3

Kc = 0.284

Therefore, the equilibrium constant for the preparation of NH3 is 0.284.

Answer: None of the given options (since none matches with the calculated value).

Note: The equilibrium constant is a dimensionless quantity, so it cannot have units. Also, the equilibrium constant does not depend on the initial concentrations of the reactants and products, but only on their concentrations at equilibrium.
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In the process for preparation of NH3,if 1 mole N2,3 mole of H2 1.6 mole of Nh3 are present at equullibrium at C Then what will be the value of equillibrium constant? A. 32 B. 16 C. 8 D. 4?
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