Consider the reaction N2 (g) + 3H2 (g) → 2NH3 (g) Suppose that at a particular moment during the reaction, molecular hydrogen is reacting at the rate of - 0.091 M/s.

(a) At what rate is ammonia being formed? M/s

(b) At what rate is molecular nitrogen consumed? M/s

The solution to the question is as follows

(a) The rate of ammonia formation = 0.061 M/s

(b) the rate of N₂ consumption = 0.0303 M/s

Explanation:

(a) To solve the question we note that the reaction consists of one mole of N₂ combining with three moles of H₂ to form 2 moles of NH₃

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

The rate of reaction of molecular hydrogen = 0.091 M/s, hence we have

3 moles of H₂ reacts to form 2 moles of NH₃, therefore

0.091 M of H₂ will react to form 2/3 * 0.091 M or 0.061 M of NH₃

Hence the rate of ammonia formation is 0.061 M/s

(b) From the reaction equation we have 3 moles of H₂ and one mole of N₂ being consumed at the same time hence

0.091 M of H₂ is consumed simultaneously with 1/3 * 0.091 M or 0.0303 M of N₂

Therefore the rate of consumption of N₂ = 0.0303 M/s