The liquid phase reaction A + B →C+D was carried out in a batch reactor in two different experiments, both at 25°C. In the first run, CAO = CB0 = 5 mol/L. In the second run, CA0 = 0.5 mol/L and CB0 = 5 mol/L. Data for both runs are given in the table below. Time (h) 0 1 2 4 6 8 10 Run 1 CA (mol/L) 5.0 1.31 0.931 0.672 0.551 0.478 0.429 Run 2 CA (mol/L) 0.5 0.302 0.217 0.140 0.104 0.083 0.069 a. Determine the rate law for this reaction. b. Calculate the conversion that could be achieved if the reaction were run at the same temperature in a 25-L CSTR if the feed were equal molar in A and B, with CA0 = 5 mol/L and vo = 2 L/h.

Introduction to Chemical Engineering Thermodynamics
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Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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The liquid phase reaction A + B →C+D was carried out in a batch reactor in two different
experiments, both at 25°C. In the first run, CAO = CB0 = 5 mol/L. In the second run, CA0 = 0.5
mol/L and CB0 = 5 mol/L. Data for both runs are given in the table below.
Time (h) 0
1
2
4
6
8
10
Run 1 CA (mol/L)
5.0
1.31
0.931
0.672
0.551
0.478
0.429
Run 2 CA (mol/L)
0.5
0.302
0.217
0.140
0.104
0.083
0.069
a. Determine the rate law for this reaction.
b. Calculate the conversion that could be achieved if the reaction were run at the same
temperature in a 25-L CSTR if the feed were equal molar in A and B, with CA0 = 5 mol/L
and vo = 2 L/h.
Transcribed Image Text:The liquid phase reaction A + B →C+D was carried out in a batch reactor in two different experiments, both at 25°C. In the first run, CAO = CB0 = 5 mol/L. In the second run, CA0 = 0.5 mol/L and CB0 = 5 mol/L. Data for both runs are given in the table below. Time (h) 0 1 2 4 6 8 10 Run 1 CA (mol/L) 5.0 1.31 0.931 0.672 0.551 0.478 0.429 Run 2 CA (mol/L) 0.5 0.302 0.217 0.140 0.104 0.083 0.069 a. Determine the rate law for this reaction. b. Calculate the conversion that could be achieved if the reaction were run at the same temperature in a 25-L CSTR if the feed were equal molar in A and B, with CA0 = 5 mol/L and vo = 2 L/h.
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