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 λ_{max} 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 k_{obs}:
𝑘_{𝑜𝑏𝑠 }= 𝑘[𝐻^{+}]^{𝑚} (2)
In this definition, k_{obs} 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 = −𝑘_{𝑜𝑏𝑠}[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 k_{obs} 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 k_{obs} 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 k_{obs} = – 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 R^{2} value from EXCEL can assist in determining the most “linear” plot; the plot featuring a trend-line with a R^{2} 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 k_{obs}.
Thus, the rate order n and observed rate constant k_{obs} 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 k_{obs}.
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 k_{obs} could serve the purpose. But the problem is how do you go from k_{obs} to determining the half-life of a drug?
Integrated Rate Laws: For n = 0: [𝑌]_{𝑡 }= −𝑘_{𝑜𝑏𝑠}𝑡 + [𝑌]_{0} For n = 1: [𝑌]_{𝑡} 𝑙𝑛 = −𝑘_{𝑜𝑏𝑠}𝑡 [𝑌]_{0 }For n = 2: 1 1 = 𝑘 [𝑌]𝑡 𝑜𝑏𝑠𝑡 + [𝑌]0 |
First, let’s define half-life. Half-life, t_{1/2}, 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 = t_{1/2} 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 = t_{1/2}, and solve for t_{1/2} to derive an equation for half-life (t_{1/2}) in terms of k_{obs}.
At this point we should have the means to determine half-life from k_{obs}. But we wish to know the t_{1/2} of a drug Y at various temperatures. For various reasons, the initial degradation experiments will be carried out at 23.5 ^{o}C (room temperature), but we wish to know t_{1/2} at physiologic temperatures (37 ^{o}C). Hence, we seek a relationship between k_{obs }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, E_{a} 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 k_{obs }and temperature.
Arrhenius Plot, ln(k_{obs}) 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:
𝐸_{𝑎 }1 𝑅 𝑇
𝑙𝑛(𝑘_{𝑜𝑏𝑠}) = −+𝑙𝑛(𝐴) (5)
By now you should recognize that equation (5) is linear with the activation energy E_{a} 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 k_{obs} 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 k_{obs}, 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 k_{obs} for each temperature block, you then transform the A(t) data for that whole degree temperature block into [Y(t)] and plot to find k_{obs}. 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 k_{obs}. Hence, if there are five whole degree temperature blocks, five graphs need be prepared (not 15!). Subsequently, the resulting k_{obs} vs. T data can be worked up into a ln(k_{obs}) vs. 1/T plot that should permit determination of E_{a}.
Two-Temperature Arrhenius Equation
Let us continue our development of an equation that relates k_{obs} to T. With E_{a} known, and given k_{obs} at one temperature, you can obtain k_{obs} corresponding to another temperature. To see how, write (5) for two different temperatures, T_{1} and T_{2}:
𝐸_{𝑎 }1
𝑙𝑛(𝑘_{𝑜𝑏𝑠,1}) = − +𝑙𝑛(𝐴)
𝑅 𝑇_{1 }𝐸_{𝑎 }1
𝑙𝑛(𝑘_{𝑜𝑏𝑠,2}) = − +𝑙𝑛(𝐴) 𝑅 𝑇_{2}
Where k_{obs,1} is the observed rate constant at T_{1}, and k_{obs,2} the observed rate constant at T_{2}. Note that E_{a} 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 [−^{𝐸}_{𝑅}^{𝑎}(_{𝑇}^{1}_{2}−_{𝑇}^{1}_{1})] (6)
=𝑒
𝑘𝑜𝑏𝑠,1
(see Unit 5 Module 3, Temperature Effects of your Chemical Thinking text). Thus, with E_{a} for the system and k_{obs,1} at T_{1} known, you can find k_{obs,2} at a desired T_{2} by (6) and from the t_{1/2} equation use k_{obs,2} to predict drug Y’s halflife at temperature T_{2}. Further, knowing k_{obs,2} at temperature T_{2}, the percent of drug Y remaining after a specific time period t, can be predicted for temperature T_{2} 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 λ_{max} = 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 ^{o}C, 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 10^{3} 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 k_{obs} (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 “k_{obs}” for the reaction. Paste-in ALL THE GRAPHS for the analysis, giving the linear trend-line equation with R^{2} value (from EXCEL) for each. Title the plots and label the axes correctly. Clearly explain how you derived the values of n and k_{obs}.
2. Half-Life (7p). Use the results from Question 1 to calculate the half-life (t_{1/2}) of Y at 23.5 ^{o}C. First show the derivation of a t_{1/2} equation based on the order you have determined for the reaction (hint: start by substituting [Y]_{t1/2} = 0.5[Y]_{0} when t = t_{1/2} 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, E_{a}_{,}. Recall Y has a unique _{max}at 254 nm, with a molar absorptivity ε of 4.55 x 10^{3} 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 k_{obs} 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 k_{obs} 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 R^{2} values. The summary table requires a title and properly labeled columns.
2. Activation Energy E_{a} (5p). Using EXCEL prepare an ln(k_{obs}) vs. 1/T plot from the Question 1 k_{obs} vs. T summary table that can be used to determine the activation energy E_{a} for the system. Paste-in this plot below and give the linear trend-line equation and R^{2} value (from EXCEL). State the E_{a} thus obtained. Show all work, explaining clearly how you derived the value of E_{a} from this plot.
3. Half-Life at 37.0 °C (5p). For the experimental conditions of the degradation study, calculate the t_{1/2} (half-life) for Y at 37.0 ^{o}C. Use equation (6) from the accompanying guide and the k_{obs} at 23.5 ^{o}C 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 ^{o}C. Show all work. Symbolically state the equations you are using.