The following data was collected for the reaction of ammonium ion and nitrite io

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)

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

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

Explanation / Answer

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

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