Chapter 7: Collection and Analysis of Rate Data

What Two Things are Wrong with This Solution?

Problem

The reaction

is carried out in a constant volume batch reactor. Determine the reaction order and specific reaction rate from the following data.

t (min)	0	10	20	30
CA(mol/dm3)	1	0.6	0.4	0.3

Solution

t	CA
0	1
		0.4	-(0.6-1.0)/(10-0)=0.04
10	0.6
		0.2	-(0.4-0.6)/(20-10)=0.02
20	0.4
		0.1	-(0.2-0.4)/(30-20)=0.01
30	0.3

First find

t	CA
0	1		0.05
		0.04
10	0.6		0.03
		0.02
20	0.4		0.015
		0.01
30	0.3		0.005

Now plotversus t and it should be a straight line.

The plot is essentially linear, therefore the reaction is zero order. From the slope of the line we find
k=0.00167 mol/dm³ min.

What two things are wrong with this solution?

Answer

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer

1) The graphical differentiation of the data is incorrect asis plotted as function of. First, even the data ofversusis not plotted correctly. According to this incorrect analysis, it should rise linearly. However, this fact is irrelevant because the x-axis should be time, not. The correct plot is as follows.

t	CA
0	1	0.053
10	0.6	0.028
20	0.4	0.017
30	0.3	0.014

2) The plot to determine the reaction order and reate constant is incorrect. Combined mole balance and postulated rate law is

taking the natural log of both sides

Plotversus ln C_A or plotversus C_A on log-log paper to determine the reaction order. The correct plot on log-log is as follows.

The reaction order is

therefore,

When C_A=1.0 mol/dm³ then

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