Question

a. Calculate the activation parameters $\left(\Delta H^{+}\right.$and $\left.\Delta S^{+}\right)$at $40^{\circ} \mathrm{C}$ for the acetolysis of 3-chlorobenzyl tosylate from the data given below: (DIAGRAM CAN'T COPY) $$ \begin{array}{cc} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k \times 10^5 \mathrm{~s}^{-1} \\ \hline 25.0 & 0.0136 \\ 40.0 & 0.085 \\ 50.1 & 0.272 \\ 58.8 & 0.726 \\ \hline \end{array} $$ (DIAGRAM CAN'T COPY) b. Calculate the activation parameters ( $E_\alpha, \Delta H^1$, and $\Delta S^2$ ) at $100^{\circ} \mathrm{C}$ from the data given for the reaction below. $$ \begin{array}{cr} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k \times 10^4 \mathrm{~s}^{-1} \\ \hline 60.0 & 0.30 \\ 70.0 & 0.97 \\ 75.0 & 1.79 \\ 80.0 & 3.09 \\ 90.0 & 8.92 \\ 95.0 & 15.90 \\ \hline \end{array} $$

   a. Calculate the activation parameters $\left(\Delta H^{+}\right.$and $\left.\Delta S^{+}\right)$at $40^{\circ} \mathrm{C}$ for the acetolysis of 3-chlorobenzyl tosylate from the data given below:
(DIAGRAM CAN'T COPY)
$$
\begin{array}{cc}
\hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k \times 10^5 \mathrm{~s}^{-1} \\
\hline 25.0 & 0.0136 \\
40.0 & 0.085 \\
50.1 & 0.272 \\
58.8 & 0.726 \\
\hline
\end{array}
$$
(DIAGRAM CAN'T COPY)
b. Calculate the activation parameters ( $E_\alpha, \Delta H^1$, and $\Delta S^2$ ) at $100^{\circ} \mathrm{C}$ from the data given for the reaction below.
$$
\begin{array}{cr}
\hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k \times 10^4 \mathrm{~s}^{-1} \\
\hline 60.0 & 0.30 \\
70.0 & 0.97 \\
75.0 & 1.79 \\
80.0 & 3.09 \\
90.0 & 8.92 \\
95.0 & 15.90 \\
\hline
\end{array}
$$
Show more…
 Advanced Organic Chemistry. Part A. Structure and Mechanisms
Advanced Organic Chemistry. Part A. Structure and Mechanisms
Francis A. Carey,… 5th Edition
Chapter 3, Problem 4 ↓
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a. Calculate the activation parameters $\left(\Delta H^{+}\right.$and $\left.\Delta S^{+}\right)$at $40^{\circ} \mathrm{C}$ for the acetolysis of 3-chlorobenzyl tosylate from the data given below: (DIAGRAM CAN'T COPY) $$ \begin{array}{cc} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k \times 10^5 \mathrm{~s}^{-1} \\ \hline 25.0 & 0.0136 \\ 40.0 & 0.085 \\ 50.1 & 0.272 \\ 58.8 & 0.726 \\ \hline \end{array} $$ (DIAGRAM CAN'T COPY) b. Calculate the activation parameters ( $E_\alpha, \Delta H^1$, and $\Delta S^2$ ) at $100^{\circ} \mathrm{C}$ from the data given for the reaction below. $$ \begin{array}{cr} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k \times 10^4 \mathrm{~s}^{-1} \\ \hline 60.0 & 0.30 \\ 70.0 & 0.97 \\ 75.0 & 1.79 \\ 80.0 & 3.09 \\ 90.0 & 8.92 \\ 95.0 & 15.90 \\ \hline \end{array} $$
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Transcript

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00:01 There are a couple ways in which you can determine the activation energy and frequency factor.
00:05 One is to graph the data, or two is to use a simple equation that relates one rate constant to another rate constant at two different temperatures.
00:17 Let me show you how to get that equation.
00:21 So if this is the uranus equation found in your textbook, if we take a second form of this equation at t2, a second temperature, and a second activation energy, and then subtract the two.
00:36 If we subtract the two, the frequency factors cancel, and this is what we're left with.
00:43 Natural log k1 minus natural log k2 equals ea over r multiplied by 1 over t2 minus 1 over t1.
00:54 Or we could rewrite it, rearranging this single equation to solve for ea, and ea is equal to r, multiplied by the natural log of the ratio of k1 over k2 divided by 1 over t2 minus 1 over t1 and then simply choose any rate constant at any temperature and then choose a second rate constant at a second temperature so any two of those four data sets you could use so i'm going to use dataset 1 and dataset 2 at 25 and 30 degrees celsius respectively make sure i convert the celsius temperature into kelvin temperature, and this gives me an activation energy of 163 ,000 joules per mole, or 163 kilojoules per mole.
01:51 Now that i know the activation energy, i can go back to this form of the arraneous equation and solve for a.
01:59 Rearrangement gives me a is equal to any rate constant with its corresponding temperature, rate constant divided by e to the negative ea over rt.
02:10 So i'll choose the first data set with a rate constant of 7 .95 times 10 of the negative 8, put in my activation energy in joules per mole, put in my r value in the corresponding kelvin temperature, 25 degrees celsius is 298 kelvin.
02:30 And that gives me a frequency, factor of 3 .5 times 10 to the 20.
02:37 Now it asks for you to determine the rate constant at 100 degrees celsius.
02:44 Well, to do this, we'll just use this equation, plug in the rate constant at one temperature, 25 degrees celsius, have the rate constant at an unknown rate constant at 100 degrees celsius, and then set that equal to ea over r, multiplied by 1 over the second temperature, 100 degrees celsius is 373 kelvin, minus 1 over the first temperature that corresponds to this rate constant, 25 degrees celsius or 298 kelvin.
03:18 This then gives us negative natural log of the rate constant we're solving for is 3 .08, or that rate constant is e to the negative 3 .08.
03:28 4 .58 times 10 to the negative 2, and at a higher temperature, it is expected to be a greater value than at a lower temperature.
03:38 If, however, we decide to plot the data, we should get similar values, so long as the data has a small amount of uncertainty, rather than a large amount of uncertainty.
03:49 So to plot the data, we're going to use the equation that was found in the book of natural log k as our y -py.
04:01 Value equals negative ea over r, and that's our slope, multiplied by 1 over the kelvin temperature, and that's our x value, plus the natural log of a, and that's our y intercept...
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