The following data were collected for the decomposition of acetaldehyde, CH3CHO (used in the manufacture of a variety of chemicals including perfumes, dyes, and plastics), into methane and carbon monoxide. The data were collected at a temperature of 530 °C.

CH3CHO CH4 +CO

[CH3CHO]
(mol L-1) Time
(s)
0.200 0
0.153 20
0.124 40
0.104 60
0.090 80
0.079 100
0.070 120
0.063 140
0.058 160
0.053 180
0.049 200

1.Rate after 60 seconds?
Rate60 = _____________ mol L–1 s–1

2.Rate after 120 seconds?
Rate120 = ___________ mol L–1 s–1

I thought rate60 is 0.0016 and rate120 is 0.00108
but it was totally not correct.

Please help me. Thanks

Quote:

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Originally Posted by hslove142331

The following data were collected for the decomposition of acetaldehyde, CH3CHO (used in the manufacture of a variety of chemicals including perfumes, dyes, and plastics), into methane and carbon monoxide. The data were collected at a temperature of 530 °C.

CH3CHO CH4 +CO

Code:

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[CH3CHO]
(mol L-1)    Time
              (s)
0.200          0
0.153          20
0.124          40
0.104          60
0.090          80
0.079        100
0.070        120
0.063        140
0.058        160
0.053        180
0.049        200

---

1.Rate after 60 seconds?
Rate60 = _____________ mol L–1 s–1

2.Rate after 120 seconds?
Rate120 = ___________ mol L–1 s–1

I thought rate60 is 0.0016 and rate120 is 0.00108
but it was totally not correct.

Please help me. Thanks

---

How did you get those values of rate :confused:

[I am myself getting started with rate, and have not started with the calculations yet, but only the introductory theory.]

The problem just given us that data.
Even, I tried to solve but it wasn't correct...

I think what you need to do is to plot a graph, and the gradient at the different points will give you the rate.

Finding the rate and other things like that are often said to be better than just using two numbers. I guess the purpose of giving all the values are for you to make a graph.