Chemical Kinetics and Rate Constants Essay

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Experiment 7 Worksheet

THE KINETICS OF HYDROLYSIS OF tert–BUTYL
CHLORIDE

Write a brief aim for the experiment:
1. To determine the first order rate constants for the HYDROLYSIS OF tert–BUTYL
CHLORIDE (TBC) reaction at the three difference temperatures: 25°C,35°C , 47°C. We need to make out of date of experiments in 25°C,35°C by using conductimetric methods and the date in 47°C already has given. Then we need to make graphs by Excels to get the rate constants.
2. To use these rate constants to determine the activation energy for the process by Arrhenius plot (ln(K) vs 1/T). The formula is that:

Explain briefly how you are following the progress of the reaction:
Q1. Why are the two methods used in this experiment only necessary for when a quantity proportional to the product is being monitored rather than the loss of a reactant in a first order reaction? There are a quantity proportional relationship between product (Ion) concentration and conductivity. The product concentration is higher, the electrical conductivity is stronger. So just monitoring products concentration is enough for this experiment.

Results.
Q2. Attach a plot (drawn with Excel or equivalent) of conductance, , against time, t, and draw a smooth curve through the data for the two temperatures you measured and the
47 +/1 0.5 oC data provided on LMS.

The plot of conductance against time t in 25°C

The plot of conductance against time t in 35°C

The plot of conductance against time t in 47°C

Q3. From the above smooth curve, (or using every 4 th point or so) read off values for t at regular intervals so you can construct a ln (t+ – t) versus time plot of equation (13).
Choose values of t so that you can use  ~400 s for 35 °C and ~800 s for 25 °C case.
The values of t should increase by about 120 s , typical (x,y) pairs would be (0, t),
(120, t), (240, t) etc

For the 47 oC data what value of would appropriate?
120s is good for making graph.
Q4

Attach plots of the data from Q3 done in Linplot and extract the rate constant k with errors. the rate constant k =0.001 in 35°C

the rate constant k =0.0007 in 25°C

the rate constant k =0.0037 in 47°C

Q5. Use your results for the rate constants at 25 °C , 35 °C and 47 oC and draw an Arrhenius plot (ln(K) vs 1/T) using Linplot and calculate Ea the activation energy of the reaction and A

the pre-exponential factor.

Ea/R is equal to the slop of the line. So Ea/R is 2623.3 in our experiment. R is gas constant which is equal to 8.314472(15) J·K−1·mol−1.
Ea=
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