For the reaction n₂(g) + 3h₂(g) ⇌ 2nh₃(g) what is the value of kc at 500 ∘c if the equilibrium concentrations are as follows:
[h₂] = 0.50 m , [n₂] = 0.50 m , and [nh₃] = 2.1 m ?
express the equilibrium constant to two significant figures.

Kc = 70

Explanation:

This is the reaction:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Let's make the expression for K, with concentration at equilibrium and in molarity.

Kc = [NH₃]²  /  [H₂]³ . [N₂]

Kc = (2.1)² / (0.5)³ . 0.5

Kc = 4.41 / 0.0625 → 70

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the answer is na (sodium)