Question

Consider the reaction for the synthesis of ammonia:

N₂(g) + 3H₂(g) → 2NH₃(9)

Experiment [N₂] (M) [H₂] (M) Initial Rate (M/s)
1 0.231 0.077 0.03759
2 0.462 0.077 0.07519
3 0.462 0.154 0.07519
4 0.692 0.077 0.11278

Calculate the rate of the reaction when [N₂] = 0.0154M:

Answer 1

The rate of the ammonia synthesis reaction when [N₂] = 0.0154 M is approximately 0.0025 M/s.

We notice that doubling the [H₂] concentration (from experiment 2 to 3) while keeping [N₂] constant does not affect the reaction rate and a suggestion that the reaction is first-order in N₂ and zero-order in H₂.

The reaction order and rate constant remain constant at different concentrations.

Rate = k * [N₂]
`^1`

0.03759 M/s = k * (0.231 M)
`^1`

k = 0.1628
`M^-^1 s^-^1`

We find the Rate at [N₂] = 0.0154 M:

Rate = k * [N₂]
`^1`

Rate = 0.1628
`M^-^1 s^-^1` * (0.0154 M)
`^1`

Rate = 0.0025 M/s