Question

The following data was collected for the reaction of ammonium ion and nitrite ion:
Reaction data for the reaction of ammonium and nitrite ions in water at 25 ^oC
NH₄⁺(aq) + NO₂^-(aq) --> N₂(g) + 2H₂O(l)

Experiment Number	Initial Concentrations (M)	Initial Concentrations (M)	Observed Initial Rate of Reaction (M s-1​)
	NH4​+	NO2-
1	0.0100	0.200	5.4 x 10-7
2	0.0200	0.200	10.8 x 10-7
3	0.0400	0.200	21.5 x 10-7
4	0.0600	0.200	32.3 x 10-7
5	0.200	0.0202	10.8 x 10-7
6	0.200	0.0404	21.6 x 10​-7
7	0.200	0.0606	32.4 x 10-7
8	0.200	0.0808	43.3 x 10-7

a) What is the rate law for this reaction? i) Rate = k[NH₄⁺][NO₂^-], ii) Rate = k[NH₄⁺]²[NO₂^-], iii) Rate = k[NH₄⁺][NO₂^-]²​, iv) Rate = k[NH₄⁺][NO₂^-]⁰
​b) What is the rate constant?​ i) 5.40 x 10^-7 s M^-1, ii) 3.23 x 10^-6 s M^-1, iii) 2.68 x 10^-4 s M^-1, iv) 7.63 x 10^-3 s M^-1
​c) If the initial concentrations of NH₄⁺ and NO₂^- are 0.300 and 0.500 M respectively at the beginning of an experiment, what is the concentration of NH₄⁺ after 2 hours? i) 0.137 M​, ii) 0.169 M, iii) 0.363 M, iv) 0.467 M

Answer 1

order w.r.t NH4+

r1/r2 = (a1/a2)^x

((5.4*10-7)/(10.8*10^-7)) = (0.01/0.02)^x

x = 1

order w.r.t NO2-

r5/r6 = (a5/a6)^y

((10.8*10^-7)/(21.6*10^-7)) = (0.0202/0.0404)^y

y = 1

over all order = 1+1 = 2

a) rate law , rate = k[NH4+[NO2-]

Answer: i

b) rate constant(k) = rate / [NH4+[NO2-]

                   = 43.3*10^-7/(0.2*0.808)

                   = 2.68*10^-5 M-1.S-1

answer: iii