Chapter 7: Collection and Analysis of Rate Data

What Two Things are Wrong with This Solution?

Problem

The reaction

\( A \rightarrow B \)

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

\( -r_A = k C_A^\alpha = - \frac{dC_A}{dt} \)

t CA \( -\Delta C_A \) \( -\frac{\Delta C_A}{\Delta t} \) \( -\frac{dC_A}{dt} \)
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 \(\left[ \frac{dC_A}{dt} \right]\)

t CA \( -\frac{\Delta C_A}{\Delta t} \) \( -\frac{dC_A}{dt} \)
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 plot \( -\frac{dC_A}{dt} \) versus t and it should be a straight line.

Graph showing a negative slope with the label 'Slope = 1.67 x 10⁻³ = k'. The x-axis represents time (t), and the y-axis represents the rate of change of concentration (-dC_A/dt). Data points are marked with black dots, and the line represents the slope.

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 as \( -\frac{\Delta C_A}{\Delta t} \) is plotted as function of \( \Delta C_A \). First, even the data of \( -\frac{\Delta C_A}{\Delta t} \) versus \( - \Delta C_A \) is 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 \( \Delta C_A \). The correct plot is as follows.

Graph showing a decay curve with several rectangular areas under the curve, representing the rate of change of concentration (-dC_A/dt) over time (t). The y-axis shows the rate of change, and the x-axis represents time. The data points are connected by a smooth curve.

t CA \( -\frac{dC_A}{dt} \)
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

\( -\frac{dC_A}{dt} = k C_A^\alpha \)

taking the natural log of both sides

\( \ln \left( - \frac{dC_A}{dt} \right) = \ln k + \alpha \ln C_A \)

Plot \( \ln \left( - \frac{dC_A}{dt} \right) \) versus ln CA or plot \( -\frac{dC_A}{dt} \) versus CA on log-log paper to determine the reaction order. The correct plot on log-log is as follows.

Graph showing a linear relationship between concentration (C_A) and the rate of change of concentration (-dC_A/dt). Data points are marked with circles, and the line represents the relationship. The slope of the line is labeled '2.0' and '2.1' at different sections of the graph.

The reaction order is

\( \alpha = \frac{2.1}{2.0} = 1.0 \)

therefore,

\( -r_A = k C_A \)

\( -r_A = - \frac{dC_A}{dt} \)

\( k = \frac{- \frac{dC_A}{dt}}{C_A} \)

When CA=1.0 mol/dm3 then

\( - \frac{d(C_A)}{dt} = 0.053 \, \frac{\text{mol}}{\text{dm}^3 \, \text{min}} \)

\( k = \frac{0.053 \, \frac{\text{mol}}{\text{dm}^3 \, \text{min}}}{1.0 \, \frac{\text{mol}}{\text{dm}^3}} \)

\( k = 0.053 \, \text{min}^{-1} \)


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