Question

Retention time depends on temperature, T, according to the equation log t’_r =(a/T) + b, where a and b are constants for a specific compound on a specific column. A compound is eluted from a gas chromatography column at an adjusted retention time t’_r =15.0 min when the column temperature is 373K. At 363 K, t9r 5 20.0 min. Find the parameters a and b and predict t’_r at 353K

Short answer

The relationship between temperature T and retention time$({t^{\text{'}}}_r)$is shown below:

$\log \left({t_r}^{\text{'}}\right)=\frac{a}{T}+b$

where: a and b are constants for a specific compound and specific column

List the given:

-$$\begin{gathered} {{\mathrm{t}}_{\mathrm{r}}}^{\text{'}}=15.0\min \mathrm{at}\mathrm{T}=373\mathrm{K} \\ {{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}=20.0\mathrm{minatT}=363\mathrm{K} \end{gathered}$$

Step-by-step solution

constant a and b

Find:

- a and b

- ${{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}\mathrm{at}353\mathrm{K}$

First, to find the parameters a and b, we can set-up two equations from the given ${{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}$and T.

Setting up Equation 1

$$\begin{gathered} \log \left(15\right)=\frac{a}{373}+b \\ \mathrm{multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}373: \\ 438.6820=\mathrm{a}+373\mathrm{b}(\mathrm{Equation}1) \\ \mathrm{Setting}\mathrm{up}\mathrm{Equation}2: \\ \log \left(20\right)=\frac{\mathrm{a}}{363}+\mathrm{b} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}363: \\ 472.2739=\mathrm{a}+363\mathrm{b}(\mathrm{Equation}2) \end{gathered}$$

Solve for parameters a and b using Equations 1 and 2:

$$\begin{gathered} \mathrm{a}=1691.658 \\ \mathrm{b}=-3.3592 \end{gathered}$$

Step 2:b) t'r at 353 K 

$$\begin{gathered} \log ({t_r}^{\text{'}})=\frac{a}{T}+b\log \left({t^{\text{'}}}_r\right)=\frac{1691.658}{353}+(-3.3592) \\ \log ({t_r}^{\text{'}})=1.4330 \\ {t^{\text{'}}}_r=27.10min \end{gathered}$$

final answer

${{\mathrm{t}}^{\text{'}}}_{\mathrm{r}}=27.10\min$