Chapter 12, Problem 12.3IA

Interpretation Introduction

Interpretation:

The exchange rate constant at $280\text{K}$ and $300\text{K}$ has to be calculated and the activation energy of the interconversion has to be determined.

Concept introduction:

The NMR spectra appear differently if the magnetic nuclei jump rapidly between different environments.  When the rate is fast, the spectrum shows a single line at the mean of the two chemical shifts.  When the rate is intermediate, the lines broaden and coalesce into a single broad line.

Answer to Problem 12.3IA

The exchange rate constant at $280\text{K}$ is $\underset{\_}{0.452\times 10^3{s}^{-1}}$.  The exchange rate constant at $300\text{K}$ is $\underset{\_}{0.226\times 10^4{s}^{-1}}$.  The activation energy is $\underset{\_}{-56219.268Jmo{l}^{-1}}$.

Explanation of Solution

The two lines collapse into one when the lifetime $\left(\tau \right)$ is given by the expression below.

    $\tau =\frac{1}{\text{2π}\nu °\left(\delta \text{'}-\delta \right)\times {10}^{-6}}$        (1)

Where,

• $\nu °$ is the operating frequency of the spectrometer.
• $\delta \text{'}$ and $\delta$ are the chemical shifts.

The operating frequency is $\text{60}\text{MHz}$.

The conversion of $\text{MHz}$ to $\text{Hz}$ is shown below.

    $\text{1}\text{MHz}={\text{10}}^6\text{Hz}$

Therefore, the conversion of $\text{60}\text{MHz}$ to $\text{Hz}$ is shown below.

    $\text{60}\text{MHz}=6\text{0}\times {\text{10}}^6\text{Hz}$

The value of $\delta \text{'}$ is $5.2$.

The value of $\delta$ is $4.0$.

Substitute the value of $\nu °$, $\delta$, and $\delta \text{'}$ in equation (1).

    $\begin{array}{c}{\tau }_{60}=\frac{1}{2\times 3.14\times 6\text{0}\times {\text{10}}^6\text{Hz}\left(5.2-4.0\right)\times {10}^{-6}}\\ =\frac{1}{452.16\text{Hz}}\times \frac{1\text{Hz}}{1{\text{s}}^{-1}}\\ =\text{2}\text{.212}\times {10}^{-3}\text{s}\end{array}$

The rate constant ($k$) needed for the interconversion process is calculated by the expression shown below.

    $k=\frac{1}{\tau }$        (2)

Substitute the value of ${\tau }_{60}$ in equation (2).

    $\begin{array}{c}k=\frac{1}{\text{2}\text{.212}\times {10}^{-3}\text{s}}\\ =\underset{\_}{0.452\times 10^3{s}^{-1}}\end{array}$

Therefore, the rate constant is $\underset{\_}{0.452\times 10^3{s}^{-1}}$.

The operating frequency is $\text{300}\text{MHz}$

The conversion of $\text{MHz}$ to $\text{Hz}$ is shown below.

    $\text{1}\text{MHz}={\text{10}}^6\text{Hz}$

Therefore, the conversion of $\text{300}\text{MHz}$ to $\text{Hz}$ is shown below.

    $\text{300}\text{MHz}=30\text{0}\times {\text{10}}^6\text{Hz}$

The value of $\delta \text{'}$ is $5.2$.

The value of $\delta$ is $4.0$.

Substitute the value of $\nu °$, $\delta$, and $\delta \text{'}$ in equation (1).

    $\begin{array}{c}{\tau }_{300}=\frac{1}{2\times 3.14\times 300\times {\text{10}}^6\text{Hz}\left(5.2-4.0\right)\times {10}^{-6}}\\ =\frac{1}{2260.8\text{Hz}}\times \frac{1\text{Hz}}{1{\text{s}}^{-1}}\\ =\text{4}\text{.4232}\times {10}^{-4}\text{s}\end{array}$

Substitute the value of ${\tau }_{300}$ in equation (2).

    $\begin{array}{c}k=\frac{1}{\text{4}\text{.4232}\times {10}^{-4}\text{s}}\\ =\underset{\_}{0.226\times 10^4{s}^{-1}}\end{array}$

Therefore, the rate constant is $\underset{\_}{0.226\times 10^4{s}^{-1}}$.

The relation between the rate $\left(r\right)$ and the activation energy $\left(E_a\right)$ is given below.

    $\ln \left(\frac{r_{60}}{r_{300}}\right)=\frac{-E_a}{R}\left(\frac{1}{T_{60}-T_{300}}\right)$        (3)

The lifetimes are directly proportional to the rate.  Therefore equation (3) can be written as shown below.

    $\ln \left(\frac{{\tau }_{60}}{{\tau }_{300}}\right)=\frac{-E_a}{R}\left(\frac{1}{T_{60}}-\frac{1}{T_{300}}\right)$        (4)

Where,

• $E_a$ is the activation energy.
• $R$ is the gas constant. $\left(\text{8}\text{.314}\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-\text{1}}\right)$
• $T_{60}$ is the temperature when the operating frequency is $\text{60}\text{MHz}$.
• $T_{300}$ is the temperature when the operating frequency is $\text{300}\text{MHz}$.

The value of $T_{60}$ is $280\text{K}$.

The value of $T_{300}$ is $300\text{K}$.

The value of ${\tau }_{60}$ is $\text{2}\text{.212}\times {10}^{-3}\text{s}$.

The value of ${\tau }_{300}$ is $\text{4}\text{.4232}\times {10}^{-4}\text{s}$.

Substitute the value of $R$, $T_{60}$, $T_{300}$, ${\tau }_{60}$, and ${\tau }_{300}$ in equation (4).

    $\begin{array}{c}\ln \left(\frac{\text{2}\text{.212}\times {10}^{-3}\text{s}}{\text{4}\text{.4232}\times {10}^{-4}\text{s}}\right)=\frac{-E_a}{\text{8}\text{.314}\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-\text{1}}}\left(\frac{1}{280\text{K}}-\frac{1}{300\text{K}}\right)\\ \mathrm{l}.61=\frac{-E_a}{\text{8}\text{.314}\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-\text{1}}}\times \frac{20}{84000\text{K}}\\ E_a=-\frac{1.61\times \text{8}\text{.314}\text{J}{\text{K}}^{-\text{1}}{\text{mol}}^{-\text{1}}\times 84000\text{K}}{20}\\ =\underset{\_}{-56219.268Jmo{l}^{-1}}\end{array}$

Therefore, the activation energy is $\underset{\_}{-56219.268Jmo{l}^{-1}}$.