Question

The following data were collected for the reaction between hydrogen and nitric oxide at 700°C: 2H2(g) + 2NO(g) -+ 2H20(g) + N2(g) Experiment [H2]/M [NO]/M Initial rate/M. s-1 1 0.010 0.025 2.4 X 10-6 2 0.0050 0.025 1.2 X 10-6 3 0.010 0.0125 0.60 X 10-6 (a) What is the rate law for the reaction? (b) Calculate the rate constant for the reaction. (c) Suggest a plausible reaction mechanism that is consistent with the rate

Answer 1

Step-by-step explanation:

Given that:

2H₂(g) + 2NO(g) → 2H₂O(g) + N₂(g)

Experiment [H₂] (M) [NO] (M) Initial Rate (M/s)

1 0.010 0.025 2.4 × 10⁻⁶

2 0.0050 0.025 1.2 × 10⁻⁶

3 0.010 0.0125 0.60 × 10⁻⁶

The general formula for rate law is:

`rate = k[H_2]_x [NO]_y`

From (1) and (2), it is obvious that the concentration of NO looks constant unlike that of H₂ which has decreased by 1/2. Similarly, the initial rate also reduced by 1/2. Hence, the initial rate is proportional to the concentration of H₂. So, x = 1

Also;

From (1) and (3), it is obvious that the concentration of H₂ looks constant unlike that of NO which has decreased by 1/2. Similarly, the initial rate also reduced by 1/4. Hence, the initial rate is proportional to the concentration of NO. So, y = 2

∴

The overall rate law is:
`rate = k[H_2][NO]_2`

`Order \ of \ reaction = 1 + 2 = 3`

(b)

From (1)

The rate constant is:

`rate = k[H_2][NO]_2`

∴

`k = (rate)/( [H_2] [NO]^2)`

`k = (2.4 * 10^(-6) \ M/s)/((0.010 \ M)(0.025 \ M)^2 )`

k = 0.38 / M².s

(c)

From the rate law, it is pertinent to understand that the slow step in the reaction includes one molecule of H₂ and two molecules of NO, where O atoms serve as an intermediary.

SO;

H₂ + 2NO → N₂ + H₂O + O slow step

O + H₂ → H₂O fast step

2H₂ + 2NO → 2H₂O + N₂