The following initial rate data are for the oxidation of nitrogen monoxide by ox
ID: 702307 • Letter: T
Question
The following initial rate data are for the oxidation of nitrogen monoxide by oxygen at 25 oC: 2 NO + O22 NO2 Experiment [NO]o, M [O2]o, M Initial Rate, M s-1 1 7.46×10-3 1.29×10-3 6.18×10-4 2 1.49×10-2 1.29×10-3 2.47×10-3 3 7.46×10-3 2.58×10-3 1.24×10-3 4 1.49×10-2 2.58×10-3 4.93×10-3 Complete the rate law for this reaction in the box below. Use the form k[A]m[B]n , where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n. Rate = From these data, the rate constant is M-2s-1.
Explanation / Answer
Consider NO = A
O2 = B
Rate r = k[A]m [B]n
From trial 1 and 2
r1 /r2 = ([A1]m [B1]n) /( [A2]m [B2]n)
(6.18×10-4) / (2.47×10-3) = (7.46×10-3 / 1.49×10-2)m
0.25 = 0.5m
0.5^2 = 0.5^m
m = 2
Similarly from trial 1 and trial 3
r1 /r3 = ([A1]m [B1]n) /( [A3]m [B3]n)
(6.18×10-4) / (1.24×10-3) = (1.29×10-3 / 2.58×10-3)^n
0.5 = 0.5n
n = 1
Rate = k [A]2 [B]
Rate = k [NO]2 [O2]
From trial 1
6.18×10-4 = k (7.46×10-3)2 (1.29×10-3)
k = 8608.37 = 8.608*10^3 M-2s-1
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