Chapter 22, Problem 22.5P

(a)

Interpretation Introduction

Interpretation:

Number of plates and plate height and resolution has to be calculated.

Concept Introduction:

Expression to compute number of theoretical plate is as follows:

  $N=\frac{5.55t_{\text{r}}{}^2}{w_{1/2}^2}$

Here,

$t_{\text{r}}$ denotes analyte’s  retention time.

$w_{1/2}$ denotes peak width at half-height.

$N$ denotes number of theoretical plate.

It has been found that greater the equilibration steps narrower will be band so this suggests $N$ should be high for more resolution or narrower separation.

Expression to compute plate height is as follows:

  $H=\frac{L}{N}$

Here,

$N$ denotes number of theoretical plate.

$L$ denotes length of chromatographic column.

$H$ denotes plate height.

Explanation of Solution

Expression to compute number of theoretical plate is as follows:

  $N=\frac{5.55t_{\text{r}}{}^2}{w_{1/2}^2}$        (1)

Substitute $\text{50 min}$ for $t_{\text{r}}$ and $\text{0}\text{.263 mm }$ for $w_{1/2}$ in equation (1).

  $\begin{array}{c}N=\frac{5.55{\left(\text{50 min }\right)}^2}{{\left(\text{0}\text{.263 mm }\right)}^2}\\ =2\times {10}^5\end{array}$

Thus, number of theoretical plate is $2\times {10}^5$.

Expression to compute $H$ from number of plates from is as follows:

  $H=\frac{L}{N}$        (2)

Substitute $\text{100 m}$ for $L$ and $2\times {10}^5$ for $N$ in equation (2).

`

  $\begin{array}{c}H=\left(\frac{\text{100 m}}{2\times {10}^5}\right)\left(\frac{1000\text{ mm}}{1\text{ m}}\right)\\ =0.50\text{ mm}\end{array}$

Thus, plate height is $0.50\text{ mm}$.

Value of $\Delta t_{\text{r}}$ is calculated as follows:

  $\begin{array}{c}\Delta t_{\text{r}}=63.5\text{ min}-63\text{ min}\\ =\text{0}\text{.5 min}\end{array}$

Expression to compute resolution is as follows:

  $\text{Resolution}=\frac{\Delta t_{\text{r}}}{w_{\text{avg}}}$        (3)

Substitute $\text{0}\text{.5 min}$ for $\Delta t_{\text{r}}$, $\text{0}\text{.384 min}$ for $w_{\text{avg}}$ in equation (3).

  $\begin{array}{c}\text{Resolution}=\frac{\text{0}\text{.5 min}}{0.384\min }\\ =1.3\end{array}$

Thus, value of resolution is 1.3.

Ratio of plate height is calculated as follows:

  $\begin{array}{c}\frac{\text{Plate height in liquid chromatography}}{\text{Plate height in gas chromatography}}=\left(\frac{10\text{ μm}}{0.50\text{ mm}}\right)\left(\frac{1\text{ mm}}{1000\text{ μm}}\right)\\ =\frac{1}{50}\end{array}$

Reason behind such massive plate height difference is that rate of longitudinal diffusion is much more rapid in gas than in liquid chromatography.

(a)

Interpretation Introduction

Interpretation:

Number of plates and plate height for peak 11 and resolution for peaks 16 and 17 has to be calculated.

Concept Introduction:

Expression to compute number of theoretical plate is as follows:

  $N=\frac{5.55t_{\text{r}}{}^2}{w_{1/2}^2}$

Here,

$t_{\text{r}}$ denotes analyte’s  retention time.

$w_{1/2}$ denotes peak width at half-height.

$N$ denotes number of theoretical plate.

It has been found that greater the equilibration steps narrower will be band so this suggests $N$ should be high for more resolution or narrower separation.

Expression to compute plate height is as follows:

  $H=\frac{L}{N}$

Here,

$N$ denotes number of theoretical plate.

$L$ denotes length of chromatographic column.

$H$ denotes plate height.

Explanation of Solution

Expression to compute number of theoretical plate is as follows:

  $N=\frac{5.55t_{\text{r}}{}^2}{w_{1/2}^2}$        (1)

Substitute $\text{50 min}$ for $t_{\text{r}}$ and $\text{0}\text{.263 mm }$ for $w_{1/2}$ in equation (1).

  $\begin{array}{c}N=\frac{5.55{\left(\text{50 min }\right)}^2}{{\left(\text{0}\text{.263 mm }\right)}^2}\\ =2\times {10}^5\end{array}$

Thus, number of theoretical plate is $2\times {10}^5$.

Expression to compute $H$ from number of plates from is as follows:

  $H=\frac{L}{N}$        (2)

Substitute $\text{100 m}$ for $L$ and $2\times {10}^5$ for $N$ in equation (2).

`

  $\begin{array}{c}H=\left(\frac{\text{100 m}}{2\times {10}^5}\right)\left(\frac{1000\text{ mm}}{1\text{ m}}\right)\\ =0.50\text{ mm}\end{array}$

Thus, plate height is $0.50\text{ mm}$.

Value of $\Delta t_{\text{r}}$ is calculated as follows:

  $\begin{array}{c}\Delta t_{\text{r}}=63.5\text{ min}-63\text{ min}\\ =\text{0}\text{.5 min}\end{array}$

Expression to compute resolution is as follows:

  $\text{Resolution}=\frac{\Delta t_{\text{r}}}{w_{\text{avg}}}$        (3)

Substitute $\text{0}\text{.5 min}$ for $\Delta t_{\text{r}}$, $\text{0}\text{.384 min}$ for $w_{\text{avg}}$ in equation (3).

  $\begin{array}{c}\text{Resolution}=\frac{\text{0}\text{.5 min}}{0.384\min }\\ =1.3\end{array}$

Thus, value of resolution is 1.3.

(b)

Interpretation Introduction

Interpretation:

Extent of larger plate height has to be calculated and reason behind such massive plate height difference has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Ratio of plate height is calculated as follows:

  $\begin{array}{c}\frac{\text{Plate height in liquid chromatography}}{\text{Plate height in gas chromatography}}=\left(\frac{10\text{ μm}}{0.50\text{ mm}}\right)\left(\frac{1\text{ mm}}{1000\text{ μm}}\right)\\ =\frac{1}{50}\end{array}$

Reason behind such massive plate height difference is that rate of longitudinal diffusion is much more rapid in gas than in liquid chromatography.