Question

Consider the reaction of ammonium ions (NH₄⁺) and nitrite ions (NO₂^-) shown in equation below:

                  NH₄⁺ (aq) + NO₂^-(aq) → N₂(g) + 2 H₂O(l)

Determination Number	Initial [NH4+], M	Initial [NO2-], M	Initial rate, Ms-1
1	0.15	0.0075	3.0 x 10-7
2	0.15	0.015	6.0 x 10-7
3	0.30	0.015	1.2 x 10-6

(a) Use method of initial rates to find the order of the reaction with respect to NH₄⁺.

(b) Use the method of initial rates to find the order of the reaction with respect to NO₂^-.

(c) Calculate the rate constant, k for the reaction of NH₄⁺ and NO₂^-.

Answer 1

Divide the rate 2 / rate 1 = 0.15 x 0.15 / 0.15 x 0.0075 = 6 x 10^-7 / 3 x 10^-7

2^[NO2-] = 2¹

According to [NO2-] the order of the reaction is one

Divide the rate 3 / rate 2 = 0.3 x 0.015 / 0.15 x 0.015 = 1.2 x 10^-6 / 6 x 10^-7

2^[NH4+] = 2¹

According to [NH4+] the order of the reaction is one

hence we can write the rate law

R = k [NO₂^-] [NH₄⁺]

To calculate rate constant substitute the values in the equation

Rate constant k = 0.15 x 0.0075 / 3.0 x 10^-7

Rate constant k = 3.75 x 10³ M/S