Initial rate data are listed in the table for the reaction: NH 4 + (aq) + NO 2 -
ID: 854886 • Letter: I
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 227 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
rate = k[NH4+]^x [NO2-]^y
[1] / [2]:
7.2e-4 / 3.6e-4 = {k(0.24)^x (0.10)^y} / {k(0.12)^x (0.10)^y}
2 = 2^x
x = 1
[3] / [2]:
5.4e-4 / 3.6e-4 = {k(0.12) (0.15)^y} / {k(0.12) (0.10)^y}
3/2 = (3/2)^y
y = 1
rate = k[NH4+][NO2-]
Calculating k with exp 1,
7.2e-4 M/s = k(0.24M)(0.10M)
k = 3.0e-2 (M-s)^(-1)
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If we assume the concentrations of NH4+ and NO2- remain the same as the reaction occurs (though decreasing), we can treat this as a second-order reaction (first-order in NH4+, first order in NO2-):
rate = k(x)(x)
rate = kx^2
1/[NH4+] - 1/(0.12M) = (3.0e-2 (M-s)^(-1))(477 s)
[NH4+] = 4.42e-2 M