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

Answer:

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:

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

∴

$k = \dfrac{rate}{ [H_2] [NO]^2}$

$k = \dfrac{2.4 \times 10^{-6} \ M/s}{(0.010 \ M)(0.025 \ M)^2 }$

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