Degradation Kinetics Lab Report

How do we characterize the rate of chemical change?

The goal of Lab Assignments 5 and 6 is to characterize the degradation kinetics of a new drug in the human stomach. You will accomplish this by determining the drug’s rate of decomposition at room temperature in an aqueous (water-based) acidic solution similar to the environment of the human stomach. In particular, your objective is to determine the reaction order and rate constant for this degradation. With the order and rate constant known you can then infer the half-life of the drug (at room temperature) in an acidic solution similar to that encountered inside the stomach. You are further expected to analyze experiments that investigate how the degradation depends on temperature, and from the data, infer the activation energy for the decomposition reaction. With the activation energy known, you should be able to predict the half-life at physiologic (body) temperatures and estimate the amount drug remaining after 8 hours in the stomach.

Degradation Kinetics

Assume we are developing a new drug “Y” that must be taken orally, and we wish to explore its degradation in the presence hydrochloric acid (HCl) at the levels normally encountered in the human stomach. From lecture, you should understand that HCl in aqueous solutions exists as H+ and Cl. That is, HCl completely dissociates into H+ and Cl ions. The hydrogen ion, H+, is the reactive species here, whereas Cl remains a “spectator ion.” The rate of drug “Y” degradation by H+ follows the rate law:

𝑅𝑎𝑡𝑒 = −𝑘[𝑌]𝑛[𝐻+]𝑚 (1)

Where k is the rate constant for the overall reaction, n the rate order with respect to the molar concentration of drug Y, [Y], and m the rate order with respect to the molar concentration of the hydrogen ion [H+].

Drug Y has a characteristic λmax at 254 nm. At this wavelength, HCl does not absorb. Hence, measuring drug Y at low concentrations in the presence of high H+ concentrations via spectrophotometry is possible, and we can follow the degradation of drug Y directly.

Now, if the hydrogen ion concentration remains in vast excess over the drug Y concentration, which we can symbolize as [H+] >> [Y], the proportional decline in [Y] will be far greater than the corresponding decline in [H+]. Stated another way, the relative percent change in [H+] will be so small compared to the percent change of [Y] that for all practical purposes [H+] may be considered constant and thus combined with k to give a new constant kobs:

𝑘𝑜𝑏𝑠 = 𝑘[𝐻+]𝑚 (2)

In this definition, kobs is the observed rate constant given the assumption [H+] remains fixed. Substitution of (2) into the rate law for the hydrogen ion mediated degradation of Y (1) results in:

𝑅𝑎𝑡𝑒 = −𝑘𝑜𝑏𝑠[𝑌]𝑛 (3)

Expression (3) suggests that by maintaining [H+] constant, a condition met when [H+] >> [Y], the rate order n and observed rate constant kobs can be determined by measuring the drug Y molar concentration as a function of time and graphically analyzing the data.

 

Graphical Analysis of Kinetic Data

Let us now consider the graphical determination of n and kobs more closely. From lecture you should understand that different rate orders correspond to different trends in the change of reactant and/or product concentrations as a function of time. In particular, there are three possible rate orders for this system: zeroth order (n = 0), first order (n = 1) and second order (n = 2), each of which corresponds to a specific relationship between [Y] and t that gives a linear plot. For example, if the reaction is zeroth order in drug Y (n = 0), then a graph of the drug Y molar concentration versus time ([Y] vs. t) should be a straight line with kobs = – slope, while ln[Y] vs. t (for n = 1), and 1/[Y] vs. t (for n = 2) would return non-linear plots. (In practice, the R2 value from EXCEL can assist in determining the most “linear” plot; the plot featuring a trend-line with a R2 closest to 1.000 can be considered the most linear.) Hence, plotting the drug Y molar concentration as a function of time data as [Y] vs. t, ln[Y] vs. t and 1/[Y] vs. t can reveal the rate order n of the reaction and permit determination of the observed rate constant kobs.

Thus, the rate order n and observed rate constant kobs can be experimentally determined by measuring the molar concentration of Y as a function of time. However, we cannot measure concentration directly. Instead, we take advantage of the fact that Y uniquely absorbs at a particular wavelength (that is, no other chemical species in the reaction mixture has an appreciable absorbance at this wavelength). Hence, one collects absorbance measurements as a function of time A(t).

As you have learned in the past, at low concentrations the light absorbed by a substance (absorbance A) tends to be proportional to its molar concentration C:

A = ε b C

Using this relationship and the molar absorptivity, you can transform the absorbance collected as a function of time A(t) into the molar concentration as a function of time [Y(t)]. Graphical analysis of the [Y(t)] data set (e.g., [Y] vs. t, ln[Y] vs. t and 1/[Y] vs. t) and NOT the absorbance versus time (e.g., A vs. t, ln(A) vs. t and 1/A vs. t) should be done to properly determine the rate order n and the observed rate constant kobs.

Half-life

For all kinds of practical reasons, it is desirable to know the half-life of a drug. So, we need some way of quantitating half-life. From equation (3) you might suspect kobs could serve the purpose. But the problem is how do you go from kobs to determining the half-life of a drug?

Integrated Rate Laws: For n = 0:

[𝑌]𝑡 = −𝑘𝑜𝑏𝑠𝑡 + [𝑌]0

For n = 1:

[𝑌]𝑡

𝑙𝑛 = −𝑘𝑜𝑏𝑠𝑡

[𝑌]0 For n = 2:

1 1

= 𝑘

[𝑌]𝑡 𝑜𝑏𝑠𝑡 + [𝑌]0

First, let’s define half-life. Half-life, t1/2, is the length of time for a substance’s concentration to decrease from its original value to half of this original value. For drugs, this half-life is the time at which the product retains 50 % of its original potency (or the time at which 50 % of the drug has decomposed and 50 % of the active drug still remains). In other words, at t = t1/2 the Y molar concentration must be half that of the initial molar concentration, or [Y]1/2 = 0.5[Y]0. This suggests that if we know the rate order n of the degradation reaction, we can use the integrated form of the rate law, substitute in [Y]1/2 = 0.5[Y] 0 when t = t1/2, and solve for t1/2 to derive an equation for half-life (t1/2) in terms of kobs.

At this point we should have the means to determine half-life from kobs. But we wish to know the t1/2 of a drug Y at various temperatures. For various reasons, the initial degradation experiments will be carried out at 23.5 oC (room temperature), but we wish to know t1/2 at physiologic temperatures (37 oC). Hence, we seek a relationship between kobs and temperature T. Enter the Arrhenius equation:

𝐸𝑎

𝑘𝑜𝑏𝑠 = 𝐴𝑒− 𝑅𝑇 (4)

In which A is the frequency factor, or frequency of collisions with the proper configuration, Ea the activation energy, R the ideal gas constant (8.314 J/K·mol) and T the temperature in K (kelvins). So, now we have a relationship between kobs and temperature.

Arrhenius Plot, ln(kobs) vs. 1/T

Equation (4) can be written in a logarithmic form which is more convenient to apply and graphically interpret.

Taking the natural logarithm of both sides and rearranging yields:

𝐸𝑎 1 𝑅 𝑇

𝑙𝑛(𝑘𝑜𝑏𝑠) = −+𝑙𝑛(𝐴) (5)

By now you should recognize that equation (5) is linear with the activation energy Ea for hydronium ion mediated drug Y degradation determined from the slope (you may find Unit 5 Module 4, Activation Energy of your Chemical Thinking text helpful).

Considering (5), apparently you need to know kobs for many T. But, notice that temperature does not appear in equation (3). This is because (3) assumes T is constant. Put another way, when finding kobs, you must use A vs. t data at the same temperature. This means the resulting data should be parsed (segregated) into whole degree temperature blocks. To obtain kobs for each temperature block, you then transform the A(t) data for that whole degree temperature block into [Y(t)] and plot to find kobs. Since you know the order n from previous work, you should know whether to plot the [Y(t)] data as [Y] vs. t, or ln[Y] vs. t, or 1/[Y] vs. t, to find kobs. Hence, if there are five whole degree temperature blocks, five graphs need be prepared (not 15!). Subsequently, the resulting kobs vs. T data can be worked up into a ln(kobs) vs. 1/T plot that should permit determination of Ea.

Two-Temperature Arrhenius Equation

Let us continue our development of an equation that relates kobs to T. With Ea known, and given kobs at one temperature, you can obtain kobs corresponding to another temperature. To see how, write (5) for two different temperatures, T1 and T2:

𝐸𝑎 1

𝑙𝑛(𝑘𝑜𝑏𝑠,1) = − +𝑙𝑛(𝐴)

𝑅 𝑇1 𝐸𝑎 1

𝑙𝑛(𝑘𝑜𝑏𝑠,2) = − +𝑙𝑛(𝐴) 𝑅 𝑇2

Where kobs,1 is the observed rate constant at T1, and kobs,2 the observed rate constant at T2. Note that Ea and ln(A) are assumed to be constants for a given chemical process, hence we have a system of two equations with two unknowns that yields with some algebra:

𝑘𝑜𝑏𝑠,2 [−𝐸𝑅𝑎(𝑇12𝑇11)] (6)

=𝑒

𝑘𝑜𝑏𝑠,1

(see Unit 5 Module 3, Temperature Effects of your Chemical Thinking text). Thus, with Ea for the system and kobs,1 at T1 known, you can find kobs,2 at a desired T2 by (6) and from the t1/2 equation use kobs,2 to predict drug Y’s halflife at temperature T2. Further, knowing kobs,2 at temperature T2, the percent of drug Y remaining after a specific time period t, can be predicted for temperature T2 via the appropriate integrated rate law. In this characterization, solving the integrated rate law for the ratio [Y]t/[Y]o and then multiplying by 100% would give the percent of drug Y remaining at time t.

Lab Assignment 5

Your name: ___________________________________ Your section: ________ GRADE ____ /25 p

All work must be very neat and organized. Significant figures must be reasonable, and correct units (where applicable) must be present.

Table 1 Absorbance at λmax = 254 nm for Y at Various Time Points

Time (hours)

0.00

24.0

48.0

72.0

96.0

120

144

Absorbance

0.678

0.549

0.446

0.366

0.299

0.245

0.201

Studying the degradation kinetics of Y in the presence of excess H+ so that [H+] >> [Y] at 23.5 oC, you collect the following absorbance data at a max = 254 nm as a function of time from a sample of Y with a starting (initial) concentration of 1.49 x 10-4 M. The molar absorptivity ε of Y is 4.55 x 103 cm-1 M-1 at 254 nm and the optical path length can be taken as b = 1.00 cm.

1. Graphical Analysis to Determine n and kobs (16p). Using EXCEL transform (convert) the above absorbance values into concentration (mol/L or M). Perform a graphical analysis with EXCEL to determine the rate order “n and the observed rate constant “kobs for the reaction. Paste-in ALL THE GRAPHS for the analysis, giving the linear trend-line equation with R2 value (from EXCEL) for each. Title the plots and label the axes correctly. Clearly explain how you derived the values of n and kobs.

2. Half-Life (7p). Use the results from Question 1 to calculate the half-life (t1/2) of Y at 23.5 oC. First show the derivation of a t1/2 equation based on the order you have determined for the reaction (hint: start by substituting [Y]t1/2 = 0.5[Y]0 when t = t1/2 into the appropriate integrated rate law). To receive credit, you must show all work in a way that justifies the mathematical relationships you are using.

Lab Assignment 6

Your name: ___________________________________ Your section: ________ GRADE ____ /25 p

All work must be very neat and organized. Significant figures must be reasonable, and correct units (where applicable) must be present.

Continuing with your study of the degradation kinetics of Y in the presence of excess H+ so that [H+] >> [Y], you are interested in finding the activation energy, Ea,. Recall Y has a unique maxat 254 nm, with a molar absorptivity ε of 4.55 x 103 cm-1 M-1 at 254 nm and the optical path length of b = 1.00 cm. The following data is collected from samples of Y with a starting (initial) concentration of 1.49 x 10-4 M:

Time

(hours)

Absorbance

(at λ = 254 nm)

Temperature (°C)

 

Time

(hours)

Absorbance

(at λ = 254 nm)

Temperature (°C)

       

1.0

0.668

16

 

1.0

0.669

18

2.0

0.664

16

 

2.0

0.662

18

3.0

0.660

16

 

3.0

0.654

18

4.0

0.656

16

 

4.0

0.647

18

5.0

0.652

16

 

5.0

0.640

18

6.0

0.648

16

 

6.0

0.633

18

7.0

0.643

16

 

7.0

0.626

18

8.0

0.639

16

 

8.0

0.619

18

9.0

0.635

16

 

9.0

0.612

18

10.0

0.630

16

 

10.0

0.605

18

11.0

0.626

16

 

11.0

0.598

18

12.0

0.622

16

 

12.0

0.592

18

13.0

0.618

16

 

13.0

0.586

18

14.0

0.614

16

    

15.0

0.610

16

 

1.0

0.669

19

16.0

0.606

16

 

2.0

0.659

19

17.0

0.602

16

 

3.0

0.649

19

18.0

0.598

16

 

4.0

0.639

19

19.0

0.594

16

 

5.0

0.630

19

    

6.0

0.621

19

1.0

0.665

17

 

7.0

0.612

19

2.0

0.659

17

 

8.0

0.603

19

3.0

0.653

17

 

9.0

0.594

19

4.0

0.647

17

 

10.0

0.586

19

5.0

0.642

17

 

11.0

0.577

19

6.0

0.636

17

 

12.0

0.568

19

7.0

0.631

17

    

8.0

0.625

17

 

1.0

0.664

20

9.0

0.620

17

 

2.0

0.652

20

10.0

0.614

17

 

3.0

0.639

20

11.0

0.609

17

 

4.0

0.627

20

12.0

0.604

17

 

5.0

0.615

20

13.0

0.599

17

 

6.0

0.603

20

14.0

0.594

17

 

7.0

0.591

20

15.0

0.589

17

 

8.0

0.581

20

16.0

0.584

17

 

9.0

0.569

20

    

10.0

0.559

20

1. Rate Constant kobs vs. Temperature T (12p). Build a summary table based only on the data from the previous page. In the table present the observed rate constant kobs values at the five different temperatures, T. Show all the graphical work, by pasting in all five EXCEL graphs (one for each temperature) below. Remember to properly title the plots, label the axes correctly and include the linear trend-line equations and R2 values. The summary table requires a title and properly labeled columns.

2. Activation Energy Ea (5p). Using EXCEL prepare an ln(kobs) vs. 1/T plot from the Question 1 kobs vs. T summary table that can be used to determine the activation energy Ea for the system. Paste-in this plot below and give the linear trend-line equation and R2 value (from EXCEL). State the Ea thus obtained. Show all work, explaining clearly how you derived the value of Ea from this plot.

3. Half-Life at 37.0 °C (5p). For the experimental conditions of the degradation study, calculate the t1/2 (half-life) for Y at 37.0 oC. Use equation (6) from the accompanying guide and the kobs at 23.5 oC from Lab Assignment 5. Show all work. Symbolically state the equations you are using. (Hint, the material in Unit 5 Module 3 under “Temperature Effects” may prove helpful.)

4. Percent of Drug Y Remaining (3p). For the experimental conditions of the degradation study, calculate the percent of drug Y remaining after 8.00 hours at 37.0 oC. Show all work. Symbolically state the equations you are using.

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