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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in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm