Question

Consider the chromatography of $\mathbf{n}\mathbf{-}{\mathbf{c}}_{\mathbf{12}}{\mathbf{H}}_{\mathbf{26}}$on a 25 m × 0.53 mm open tubular column of 5% phenyl–95% methyl polysiloxane with a stationary phase thickness of 3.0 $\mathrm{\mu m}$and He carrier gas at 125ºC. The observed retention factor for $\mathbf{n}\mathbf{-}{\mathbf{c}}_{\mathbf{12}}{\mathbf{H}}_{\mathbf{26}}$is 8.0. Measurements were made of plate height, H, at various values of linear velocity,${\mathrm{\mu }}_{\mathrm{x}}\left(\mathrm{m}/\mathrm{s}\right)$. A least-squares curve through the data points is given by

localid="1654864230381" $$\begin{gathered} \mathbf{H}\mathbf{\left(}\mathbf{m}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{}\mathbf{/}{\mathbf{\mu }}_{\mathbf{x}}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{09}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{s}\mathbf{)}{\mathbf{\mu }}_{\mathbf{x}} \\ \mathbf{}\mathbf{the}\mathbf{}\mathbf{coefficients}\mathbf{}\mathbf{of}\mathbf{}\mathbf{the}\mathbf{}\mathbf{van}\mathbf{}\mathbf{Deemter}\mathbf{}\mathbf{equation}\mathbf{,}\mathbf{}\mathbf{find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{diffusion}\mathbf{}\mathbf{coefficient}\mathbf{} \\ \mathbf{of}\mathbf{}\mathbf{n}\mathbf{-}{\mathbf{c}}_{\mathbf{12}}{\mathbf{H}}_{\mathbf{26}}\mathbf{}\mathbf{}\mathbf{in}\mathbf{}\mathbf{the}\mathbf{}\mathbf{mobile}\mathbf{}\mathbf{and}\mathbf{}\mathbf{stationary}\mathbf{}\mathbf{phases}\mathbf{.}\mathbf{}\mathbf{Why}\mathbf{}\mathbf{is}\mathbf{}\mathbf{one}\mathbf{}\mathbf{of}\mathbf{}\mathbf{these}\mathbf{}\mathbf{diffusion}\mathbf{}\mathbf{coefficients}\mathbf{}\mathbf{so}\mathbf{}\mathbf{much}\mathbf{}\mathbf{greater}\mathbf{}\mathbf{than}\mathbf{}\mathbf{the}\mathbf{}\mathbf{other}\mathbf{?} \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{From}\mathrm{substituting}\mathrm{the}\mathrm{values}\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{n}-C_{\mathit{12}}{H}_{\mathit{26}}\mathit{}in\mathit{}mobile\mathit{}phase \\ is3.0\times {10}^{-5\frac{{\mathrm{m}}^2}{\mathrm{s}}}\mathrm{and}\mathrm{the}\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{n}\mathit{-}{C}_{\mathit{12}}{H}_{\mathit{16}}\mathit{}\mathrm{in}\mathrm{stationary}\mathrm{phase}\mathrm{is}5.0\times {10}^{-10}{\mathrm{m}}^2/\mathrm{s}.\mathrm{The}\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{mobile}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{the}\mathrm{stationary} \\ \mathrm{phase}\mathrm{because}\mathrm{it}\mathrm{is}\mathrm{easier}\mathrm{for}\mathrm{the}\mathrm{diffusion}\mathrm{of}\mathrm{solute}\mathrm{through}\mathrm{Helium}\mathrm{gas}\mathrm{than}\mathrm{through}\mathrm{a}\mathrm{viscous}\mathrm{liquid}\mathrm{phase}. \end{gathered}$$

Step-by-step solution

The deemter equation is taken in the form

$$\begin{gathered} H=\frac{B}{{\mu }_x}+C_{{\mu }_x}=(C_s+C_m){\mu }_x \\ {C}_s=\frac{2k}{3{\left(k+1\right)}^2}\frac{d^2}{d_s} \\ {C}_m=\frac{1+6k11k^2}{24{\left(k+1\right)}^2}\frac{r^2}{D_m} \\ B=2D_m \end{gathered}$$

$$\begin{gathered} \mathrm{Where}{\mathrm{D}}_{\mathrm{m}}=\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{solute}\mathrm{in}\mathrm{mobile}\mathrm{phase} \\ k=\mathrm{retention}\mathrm{factor} \\ d=\mathrm{thickness}\mathrm{of}\mathrm{stationary}\mathrm{phase} \\ {D}_s=\mathrm{diffusion}\mathrm{coefficient}\mathrm{of}\mathrm{solute}\mathrm{in}\mathrm{stationary}\mathrm{phase} \end{gathered}$$

Diffusion coefficient of solute in mobile phase

$$\begin{gathered} H=\left(6.0\times {10}^{-5}{m}^2/s\right){\mu }_x+(2.09\times {10}^{-3}s){\mu }_x \\ B=2D_M=\left(6.0\times {10}^{-5}{m}^2/s\right) \\ {D}_m=3.0\times {10}^5{m}^2/s \\ Forthesecondterm, \\ {C}_s+C_m=2.09\times {10}^{-3}s=\frac{2k}{3{(k+1)}^2}\frac{d^2}{D_s}+\frac{1+6k+11k^2}{24{(k+1)}^2}\frac{r^2}{d_m} \\ consumptionofcitrateandactivationofglycolysis. \end{gathered}$$

Diffusion coefficient of solute in stationary phase

Substituting the values of all parameters to solve $D_s$

$$\begin{gathered} 2.09\times {10}^{-3}\mathrm{s}=\frac{2\left(8.0\right)}{3{\left(8.0+1\right)}^2}\frac{{(3.0\times {10}^{-6}\mathrm{m})}^2}{{\mathrm{D}}_{\mathrm{s}}} \\ +\frac{1+6(8.0)+11{\left(80\right)}^2}{24({(8.0)+1)}^2}\frac{{(2.65\times {10}^{-4}\mathrm{m})}^2}{(3.0\times {10}^{-5}{\mathrm{m}}^2/\mathrm{s})} \\ {D}_{\mathit{s}}\mathit{=}5.0\times {10}^{-10}{m}^{\mathit{2}}\mathit{/}s \end{gathered}$$

Diffusion coefficient in mobile phase

Diffusion coefficient in mobile phase=$\frac{3.0\times {10}^{-5}{m}^2/s}{5.0\times {10}^{-10}{m}^2/s}=6.0\times {10}^4$

Reason why one of these diffusion coefficient is greater than the other

Diffusion coefficient in mobile phase is $6.0\times {10}^{-4}$times higher than the diffusion coefficient in stationary phase. The diffusion coefficient of mobile is greater than the diffusion coefficient in the stationary phase because it is easier for the diffusion of solute through Helium gas than through liquid phase.