Question

For the formation of ammonia from N₂ and H₂, if rate of disappearance of N₂ is 2 M/s, then the rate of appearance of NH₃ and the rate of reaction respectively would be:

• A. 4 M/s and 2 M/s
• B. 2 M/s and 4 M/s
• C. 1 M/s and 2 M/s
• D. 4 M/s and 4 M/s

Answer 1

Answer: (A)

4 M/s and 2 M/s

Answer 2

The balanced chemical equation for the formation of ammonia from N₂ and H₂ is:

N₂(g) + 3H₂(g) → 2NH₃(g)

The rate of reaction in terms of the disappearance of reactants and appearance of products is given by:

Rate of reaction = $-\frac{1}{1}\frac{d[N₂]}{dt} = -\frac{1}{3}\frac{d[H₂]}{dt} = +\frac{1}{2}\frac{d[NH₃]}{dt}$

Given: Rate of disappearance of N₂ = $-\frac{d[N₂]}{dt}$ = 2 M/s

1. Calculate the Rate of Reaction:

From the definition, the rate of reaction is equal to the rate of disappearance of N₂ (since its stoichiometric coefficient is 1).

Rate of reaction = $-\frac{d[N₂]}{dt}$

Rate of reaction = 2 M/s

2. Calculate the Rate of Appearance of NH₃:

We relate the rate of reaction to the rate of appearance of NH₃:

Rate of reaction = $+\frac{1}{2}\frac{d[NH₃]}{dt}$

Substitute the calculated rate of reaction:

2 M/s = $+\frac{1}{2}\frac{d[NH₃]}{dt}$

Multiply both sides by 2 to find the rate of appearance of NH₃:

$\frac{d[NH₃]}{dt}$ = 2 × 2 M/s

$\frac{d[NH₃]}{dt}$ = 4 M/s

Therefore, the rate of appearance of NH₃ is 4 M/s, and the rate of reaction is 2 M/s.