Answer:
Kc = 4.86×10⁻⁶
Explanation:
We begin from the equation:
N₂ + 2H₂ ⇄ 2NH₃
We start from 3.47×10⁻² moles of N₂(g) and 6.38×10⁻² moles of H₂(g), so when we reach the equilibrium, we get 6.25×10⁻² moles of H₂.
This data means, that in the reaction we made react:
6.38×10⁻² - x = 6.25×10⁻²
x = 1.3×10⁻³ moles of H₂
As stoichiometry is 1:3, we will know that the moles of N₂ that have been reacted were:
1.3×10⁻³ moles / 3 = 4.33×10⁻⁴ moles of N₂
So, in the equilibrium we would have:
3.47×10⁻² moles of N₂ - 4.33×10⁻⁴ moles of N₂ = 0.0343 moles of N₂
How many ammonia, would we have in the equilibrium?
4.33×10⁻⁴ mol . 2 = 8.66×10⁻⁴ moles (from stoichiometry with N₂, 1:2)
(1.3×10⁻³ mol . 2) / 3 = 8.66×10⁻⁴ moles (from stoichiometry with H₂, 2:3)
Let's make the expression for Kc
Kc = [NH₃]³ / [N₂] . [H₂]²
(8.66×10⁻⁴ )³ / (0.0343 . (6.25×10⁻²)² = 4.86×10⁻⁶
