Question

In: Chemistry

A certain compound, A, reacts to form products according to the reaction A → P. The...

A certain compound, A, reacts to form products according to the reaction A → P. The amount of A is measured as a function of time under a variety of different conditions and the tabulated results are shown here:

25.0 °C 35.0 °C 45.0 °C Time (s) [A] (M) [A] (M) [A] (M)

  

0

1.000

1.000

1.000

10

0.779

0.662

0.561

20

0.591

0.461

0.312

30

0.453

0.306

0.177

40

0.338

0.208

0.100

50

0.259

0.136

0.057

60

0.200

0.093

0.032

Make a graph of [A] vs. t, a graph of ln[A] vs. t, and a graph of 1/[A] vs. t using the data for 25 °C. Then make similar graphs for the other temperatures. What is the order of the reaction with respect to A? Explain your answer.

a. Use the data to determine the rate constant at each temperature. b. What is the activation energy for this reaction?

c. The same reaction is conducted in the presence of a catalyst, and the following data are obtained:

25.0 °C 35.0 °C 45.0 °C Time (s) [A] (M) [A] (M) [A] (M)

  

0

1.000

1.000

1.000

0.1

0.724

0.668

0.598

0.2

0.511

0.433

0.341

0.3

0.375

0.291

0.202

0.4

0.275

0.190

0.119

0.5

0.198

0.122

0.071

0.6

0.141

0.080

0.043

What effect does a catalyst have on the rate of the reaction? What is the activation energy for this reaction in the presence of the catalyst? How does it compare with the activation energy for the reaction when the catalyst isn’t present?

Solutions

Expert Solution

The data for plotting the graphs is as follows:

Time 1/[A] [A] ln[A]
0 1 1 0
10 1.283697 0.779 -0.24974
20 1.692047 0.591 -0.52594
30 2.207506 0.453 -0.79186
40 2.95858 0.338 -1.08471
50 3.861004 0.259 -1.35093
60 5 0.2 -1.60944

Since the graph fits best in ln [A] Vs time graph therefore the reaction follows 1st order kinetics.

The integrated equation of first order is

ln [A] = -Kt + ln [Ao]

For zero order the equation is [A] = [A0]- Kt

For second order reaction 1/[A] = 1/[A0] + Kt

Where K is the rate constant, [A] is the concentration of reactant and [A0] is the initial concentration.

On comparing the graph equation with the equation of first order reaction we will get slope equal to -0.0271, which is equal to -K from the integrate rate equation of first order

Hence the rate constant is 0.0271 mol L-1 s-1

Please note, for activation energy, the information regarding preexponential factor is not given.


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