PC The oxidation of nitrogen monoxide by oxygen at 25 c 2N0 +02-2 No. is second
ID: 591484 • Letter: P
Question
PC The oxidation of nitrogen monoxide by oxygen at 25 c 2N0 +02-2 No. is second order in NO and first order in O Complete the rate law for this reaction in the box below. 02 Use the form kIA]" B]".... where'1' is understood for m, n(don't enter 1) and concentrations taken to the zero power do not appear. Rate- In an experiment to determine the rate law, the rate of the reaction was determined to be 5.32 104 Ms 1 when [NO] 9.68 10-3 M and [O-] - 5.2610-4 M From this experiment, the rate constant isExplanation / Answer
Given second order in NO
third order overall
given reaction is
2N0 + 02 ---> 2N02
let the rate law be
rate = K [ N0]^a [02]^b
a and b are the orders
given
a = 2
overall order = a + b
given overall order is 3
so
3 = a + b
b = 3 - a
b = 3-2
b = 1
so the rate law is given by
rate = K [N0]^2 [02]
From the data, the reaction is second order in NO and first order in O2.
Thus, the rate law is,
rate = k[NO]2[O2]
rate = (5.32x10^-4 M-2 s-1)(9.68x10^-3 M)^2(5.26x10^-4 M)
rate =4.77*10^-4 M s-1
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