Initial rate data are listed in the table for the reaction: NH4+ (aq) + NO2- (aq

Question

Initial rate data are listed in the table for the reaction: NH4+ (aq) + NO2- (aq) N2 (g) + H2O (l) Experiment [NH4+]i [NO2-]i Initial rate (M/s)

1 0.24 0.10 7.2 x 10-4

2 0.12 0.10 3.6 x 10-4

3 0.12 0.15 5.4 x 10-4

4 0.12 0.12 4.3 x 10-4

First determine the rate law and rate constant. Under the same initial conditions as in Experiment 4, calculate [NH4+] at 207 seconds after the start of the reaction. In this experiment, both reactants are present at the same initial concentration. The units should be M, and should be calculated to three significant figures.

Explanation / Answer

consider the rate law as

rate = k [NH4+]^a [N02-]^b

now

for exp1 and exp2 , [N02-] = constant

so

rate2 / rate1 = ( [NH4+]2 / [NH4+]1)^a

3.6 x 10-4 / 7.2 x 10-4 = ( 0.12 / 0.24)^a

1/2 = (1/2)^a

a = 1

so the reaction is 1st order with respect to NH4+

now

for exp2 and exp3 , [NH4+] = constant

rate 3 / rate 2 = ( [NO2-]3 / [N02-]2)^b

5.4 x 10-4 / 3.6 x 10-4 = ( 0.15 / .10)^b

1.5 = (1.5)^b

b = 1

so the reaction is 1st order with respect to N02-

now

the rate law is

rate = k [NH4+] [N02-]

now

for exp1

we get

7.2 x 10-4 = k x [0.24] [0.1]

k = 0.03

so

the value of rate constant is 0.03

since it is 1+1 = 2 nd order , units of rate constant are M-1 s-1

now

for experiment 4 , the initial concentration of NH4+ and N02- are same

and the reaction is 1st order for both NH4+ and N02-

so

it can be considered as a 2nd order reaction

now

for a 2nd order reaction

1/[A] = kt + 1/[Ao]

so

1/[NH4+] = ( 0.03 x 207) + ( 1/0.12)

[NH4+] = 0.06876

so

the concentration of NH4+ after 207 seconds is 0.06876 M

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