Question

One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If 20.5% of nitrogen is converted into ammonia by the following reaction

N₂(g)+3H₂⇔2NH₃(g)
calculate the equilibrium constant of the reaction in concentration units.

Answer 1

To determine the equilibrium constant for the reaction of nitrogen and hydrogen forming ammonia.

Step-by-step explanation:

Initially, one mole of N₂ is mixed with three moles of H₂ in a 4-litre container. If 20.5% of N₂ is converted to NH₃, that would be 0.205 moles of N₂ (20.5% of 1 mole). Since the mole ratio from N₂ to NH₃ is 1:2, this means 0.205 moles of N₂ will produce 0.41 moles of NH₃.

According to the stoichiometry of the reaction, 3 times the moles of N₂ in H₂ are required; thus, 0.615 moles of H₂ (0.205 moles of N₂ multiplied by 3) will be consumed. At equilibrium, the concentration of N₂ will be: (1 - 0.205) / 4 = 0.19875 M, the concentration of H₂ will be: (3 - 0.615) / 4 = 0.59625 M, and the concentration of NH₃ will be: 0.41 / 4 = 0.1025 M.

Now, applying the equilibrium expression for this reaction: Kc = [NH₃]² / ([N₂][H₂]³), we substitute the concentration values to find Kc. Kc = (0.1025²) / (0.19875 x 0.59625³) = (0.01050625) / (0.19875 x 0.21167653125) = 0.01050625 / 0.04207923297 = 0.2496