The following data were collected for the rate of disappearance of NO in the rea
ID: 862879 • Letter: T
Question
The following data were collected for the rate of disappearance of NO in the reaction 2 NO(g) + O2(g) ? 2 NO2(g).
(a) What is the rate law for the reaction? (Rate equations take the general form 'rate = k . [A] . [B]'.)
(b) What is the average value of the rate constant calculated from the three data sets?
(c) What is the rate of disappearance of NO when [NO] = 0.0750 M and [O2] = 0.0100 M?
(d) What is the rate of disappearance of O2 at the concentrations given in part (d)?
Experiment [NO] (M) [O2] (M) Initial Rate(M/s) 1 0.0126 0.0125 1.41 10-2 2 0.0252 0.0125 5.64 10-2 3 0.0252 0.0250 1.13 10-1
Explanation / Answer
(a)
Let the rate law be, rate = k [NO]^x[O2]^y
From the trial 1 and 2, it is clear that [O2] is constant.
So, trial1 / trial 2 = (1.41 x 10^-2 / 5.64 x 10^-2) = (0.0126 / 0.0252)^x
0.25 = (0.5)^x
x = 2.
Thus, the order of reaction with respect to [NO] is 2.
From the trial 2 and 3, it is clear that [NO] is constant.
So, trial 2/trial 3 = (5.64 x 10^-2 / 1.13 x 10^-1) = (0.0125 / 0.0250)^y
0.499 = (0.5)^y
y = 1
Thus, the order of the reaction with respect to [O2] is 1.
Hence, the rate law is rate = k[NO]^2[O2]^1
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(b)
Rate = k [ NO]^2[O2]
Plugging in the values of trial 1 in the equation,
1.41 x 10^-2 = k [0.0126]^2[0.0125]
k = 7105.064 M^-2 s^-1
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(c)
Rate = k [ NO]^2[O2]
Rate = 7105.064 [0.0750]^2[0.01]
Rate = 0.399 M/s
rate = -1/2 d[NO]/dt
The rate of disspareance of NO is 2 times the rate. so, 2 x 0.399 = 0.798 M/s.
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(d)
Rate = k [ NO]^2[O2]
Rate = 7105.064 [0.0750]^2[0.01]
Rate = 0.399 M/s
rate = -1/2 d[NO]/dt
rate = -d[O2]/dt
The rate of disspareance of O2 is equal to the rate. so, 0.399 M/s.
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