Question

Plot the following data, and from the graph determine ∆H_vap for magnesium and lithium. In which metal is the bonding stronger?

$\begin{array}{l}\mathbf{T}\mathbf{e}\mathbf{m}\mathbf{p}\mathbf{e}\mathbf{r}\mathbf{a}\mathbf{t}\mathbf{u}\mathbf{r}\mathbf{e}\\ \mathbf{V}\mathbf{a}\mathbf{p}\mathbf{o}\mathbf{r}\mathbf{p}\mathbf{r}\mathbf{e}\mathbf{s}\mathbf{s}\mathbf{u}\mathbf{r}\mathbf{e}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{L}\mathbf{i}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{M}\mathbf{g}\\ \mathbf{\left(}\mathbf{m}\mathbf{m}\mathbf{H}\mathbf{g}\mathbf{\right)}\\ \mathbf{1}\mathbf{.}\mathbf{750}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{620}\mathbf{.}\\ \mathbf{10}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{890}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{740}\\ \mathbf{100}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1080}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{900}\\ \mathbf{400}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1240}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1040}\mathbf{.}\\ \mathbf{760}\mathbf{.}\mathbf{1310}\mathbf{.}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1110}\mathbf{.}\end{array}$

Short answer

We get the slope by taking two points (x, y) that are on the line we draw. For a line,

the slope is ∆y/∆x, or we can determine the straight-line equation using a calculator.

The general straight-line equation is y = mx + b where m=slope and by the intercept.

The graph is plotted by taking Temperature = 10⁴/T on the x-axis and ln P_vap on the y-axis.

Step-by-step solution

Enthalpy of vaporization of lithium

The equation of the Li line is:

$\begin{array}{l}\mathrm{In}{\mathrm{P}}_{\mathrm{vap}}=-1.90\mathrm{x}{10}^4(1/\mathrm{T})+18.6,\\ \mathrm{slope}=-1.90\mathrm{x}{10}^4\mathrm{K}\end{array}$

$\begin{array}{c}\mathrm{Slope}={\mathrm{\Delta H}}_{\mathrm{vap}}/\mathrm{R},\\ {\mathrm{\Delta H}}_{\mathrm{vap}}=-\mathrm{slope}\mathrm{x}\mathrm{R}\\ {\mathrm{\Delta H}}_{\mathrm{vap}}=1.90\mathrm{x}{10}^4\mathrm{K}\times 8.3145\mathrm{J}/\mathrm{K}\mathrm{mol}\\ {\mathrm{H}}_{\mathrm{vap}}=1.58\times {10}^5\mathrm{J}/\mathrm{mol}\\ =158\mathrm{kJ}/\mathrm{mol}\end{array}$

Enthalpy of vaporization of magnesium

The equation of the line is:

$\begin{array}{c}\mathrm{In}{\mathrm{P}}_{\mathrm{vap}}=-1.67\times {10}^4(1/\mathrm{T})+18.7,\\ \mathrm{slope}=-1.67{10}^4\mathrm{K}\\ {\mathrm{\Delta H}}_{\mathrm{vap}}=-\mathrm{slope}\times \mathrm{R}\\ =1.67\mathrm{x}{10}^4\mathrm{K}\mathrm{x}8.3145\mathrm{J}/\mathrm{K}\mathrm{mol}\\ {\mathrm{H}}_{\mathrm{vap}}=1.39\times {10}^5\mathrm{J}/\mathrm{mol}\\ =139\mathrm{kJ}/\mathrm{mol}\end{array}$

The bonding is stronger in Li because ∆ H_vapis larger for Li.

Explanation

We get the slope by taking two points (x, y) on the line we draw. For a line,

the slope is ∆y/∆x, or we can determine the straight-line equation using a calculator.

The general straight-line equation is y = mx + b where m=slope and by the intercept.

The graph is plotted by taking Temperature = 10⁴/T on the x-axis and ln P_vap on the y-axis.