Consider the following reaction: 2NO(g) + O2(g) 2NO2(g) The following rate data
ID: 679925 • Letter: C
Question
Consider the following reaction:
2NO(g) + O2(g) 2NO2(g)
The following rate data were obtained:
Experiment
[NO]0 (M)
[O2]0 (M)
Initial Rate (M/s)
1
0.0005
0.0005
0.89
2
0.0010
0.0005
3.56
3
0.0015
0.0005
7.99
4
0.0010
0.0010
7.12
(a) Determine the rate law.
(b) Calculate the rate constant for the reaction.
(c) Calculate the initial rate when [NO] = 0.0025 M and [O2] =0.0025 M.
Experiment
[NO]0 (M)
[O2]0 (M)
Initial Rate (M/s)
1
0.0005
0.0005
0.89
2
0.0010
0.0005
3.56
3
0.0015
0.0005
7.99
4
0.0010
0.0010
7.12
Explanation / Answer
To find the rate law, the rate law equation is: Rate = k[NO]x[O2]y You first have to find the exponents x and y, which are theorder of the reactions. To find the exponents, use thechart as your guide. First choose two experiments in which oneconcentration did no change. For example, in experiment 1,2, and 3,the O2 concentrations remain at 0.0005 M while the NOconcentration doubles from 0.0005 ---> 0.0010 ---->0.0015. Also notices as NO concentration doubles, the initial rateincreased by a factor of 4 or 22. This means that the NOconcentration is proportional to the rate since the NOconcentration double, the rate also doubles. Here is an example: Take any two [NO] concentration from experiment 1,2, and 3. I'mgoing to take experiment 2 and 3 as an example. [NO] M from experiment 3 is 0.0015 and 0.0010 from experiment 2.Divide 0.0015/0.0010 and you'll get 1.5. Take the initial rate fromexperiment 2 and 3 and divide it as well: 7.99/3.56 = 2.244. The concentration of [NO] of experiment 2 and 3 is proportional tothe initial rate because: (1.5)(2) ˜2.244 Pick another experiment such as 1 and 2: (0.0010/0.0005) = 2.Initial rate: (3.56/0.89) = 4. 2(2) = 4. Experiment 1and 3 will also give the same result. The [NO] is doubled, therefore [NO]x =[NO]2 To find [O2]y, look at the chart and see whichexperiment the concentration did not change. This time, [NO]remains constant at .0010 M at experiment 2 and 4. Therefore tofind [O2], you only need to look at experiment 2 and 4, where [NO]is constant and [O2] changes. Apply the same idea as above:(.0010/.0005) = 2. (7.12/3.56) = 2. Therefore: 2(1) = 2. [O2]y is [O2]1. The Rate Law is = k[NO]2[O2]1. If the explanation above is confusing, just know that to find theexponents choose two experiments in which one concentration doesnot change. If that's confusing as well, well...just use the ratelaw for (a) to answer question (b) and (c). For (b) solve for k by picking any one experiment as you want andplug numbers in. Rearrange the rate law to get k on one side. k =(Initial Rate)/[NO]2[O2]. Choose numbers for experiment1: (0.89 M/s)/((0.0005 M)2(0.0005 M)). The answer isabout 7.12 * 109 mol L-1 s-1. Picka different experiment and you'll see that they should also equal7.12 * 109 For (c) use the calculated constant k from (b) and solve: Rate =(7.12 * 109)(.0025)2(.0025). I hope this helps.
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