Textbook Problem

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Use values of Δ_fG° for solid and gaseous iodine at 25 °C (Appendix L) to calculate the equilibrium vapor pressure of iodine at this temperature.

Interpretation Introduction

Interpretation:

The equilibrium vapor pressure of iodine at $\text{25 }°\text{C}$ should be calculated using ${\text{Δ}}_{\text{f}}{\text{G}}^{\text{o}}$.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$. It can be calculated in a similar manner as entropy and enthalpy. The expression for the free energy change is:

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{°}}\text{= }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{products}\right)\text{- }\sum {\text{nΔ}}_{\text{f}}{\text{G}}^{\text{°}}\left(\text{reactants}\right)$

${\text{ΔG}}^{\text{o}}$ is related to the equilibrium constant $\text{K}$ by the equation, ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= - RT lnK}}_{\text{p}}$

Explanation of Solution

The equilibrium vapor pressure of iodine at equilibrium is calculated as follows,

The Appendix L was referred for the values of standard free energy values.

The given reaction is, ${\text{I}}_{\text{2}}\left(\text{s}\right)\rightleftarrows {\text{I}}_{\text{2}}\left(\text{g}\right)$

The standard entropy change for ${\text{I}}_{\text{2}}\left(\text{s}\right)$ is zero, thus ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ for the reaction is equal to the ${\text{Δ}}_{\text{f}}{\text{G}}^{\text{o}}$ of iodine gas.

Thus, the ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}$ value is $19.327\text{ kJ/mol-rxn}$.

${\text{ΔG}}^{\text{o}}$ is related to the equilibrium constant $\text{K}$ by the equation,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{= -RT lnK}}_{\text{p}}$

Here, ${\text{lnK}}_{\text{p}}\text{=}\frac{{\text{p}}_{{\text{I}}_{\text{2}}\left(\text{g}\right)}}{{\text{p}}_{{\text{I}}_{\text{2}}\left(\text{s}\right)}}$

Substituting the respective values,

  ${\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}\text{= - }\left(\text{0}\text{.008314 kJ/K×mol}\right)\left(\text{298 K}\right)\text{ln}\left(\frac{{\text{p}}_{{\text{I}}_{\text{2}}\left(\text{g}\right)}}{{\text{p}}_{{\text{I}}_{\text{2}}\left(\text{s}\right)}}\right)$

The value of ${\text{p}}_{{\text{I}}_{\text{2}}\left(\text{s}\right)}$ is zero