Question

Using the data from appendix 4, calculate $\mathbf{∆}\mathbf{G}$for the reaction $\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{S}\left(g\right)\mathbf{+}{\mathbf{SO}}_{\mathbf{2}}\left(g\right)\mathbf{\iff }\mathbf{3}\mathbf{S}\left(s\right)\mathbf{+}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(g\right)$for the following conditions at $\mathbf{25}\mathbf{°}\mathbf{C}$:

$$\begin{gathered} {\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{s}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{atm} \\ {\mathbf{P}}_{{\mathbf{SO}}_{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{atm} \\ {\mathbf{P}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{atm} \end{gathered}$$

Short answer

Answer

The $∆G$determined for the given reaction at $\mathbf{25}\mathbf{°}\mathbf{C}$ and given pressureis-50.3kJ/mol

Step-by-step solution

Step 1: Introduction to the Concept

The difference in the standard enthalpy of formation of the products and that of the reactants is used to calculate the standard Gibbs free energy change for a reaction $\mathbf{∆}\mathbf{G}\mathbf{°}$measured at 25°C and 1 atm pressure.

$\Delta G_R^{o}298=\sum n_p\Delta G_f^o\left(Products\right)-\sum n_r\Delta G_f^o\left(Reactants\right)-----\left(1\right)$

The pressure dependency of a reaction's free energy change is written as

$\mathit{\Delta }\mathit{G}\mathbf{=}\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{+}\mathit{R}\mathit{T}\mathit{l}\mathit{n}\frac{\left[{\mathit{P}}_{\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{s}}\right]}{\left[{\mathit{P}}_{\mathit{r}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathit{s}}\right]}-----\left(2\right)$

Where, P_products, P_reactants= Partial pressure pf the products and reactants

$\left[\frac{{\mathrm{P}}_{\mathrm{products}}}{{\mathrm{P}}_{\mathrm{reactants}}}\right]=Q$

Where Q= reaction quotient

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}\mathbf{G}\mathbf{°}\mathbf{+}\mathbf{RT}\mathbf{}\mathbf{In}\mathbf{}\mathbf{Q}----\left(3\right)$

Determination of ∆GR° at 25°C and 1 atm

The given reaction is,

$2{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)+{\mathrm{SO}}_2\left(\mathrm{g}\right)\iff 3\mathrm{S}\left(\mathrm{s}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Let's compute $\mathbf{∆}{\mathbf{G}}_{\mathbf{R}}^{\mathbf{°}}$at $\mathbf{25}\mathbf{°}\mathbf{C}$and 1atm

From equation (1),

$\Delta G_R^o=\left[3\Delta G_f^o\left(S\left(s\right)\right)+2\Delta G_f^o\left(H_{2}O\left(g\right)\right)\right]-\left[2\Delta G_f^o\left(H_{2}S\left(s\right)\right)+\Delta G_f^o\left(S{O}_2\left(g\right)\right)\right]$

$∆{\mathrm{G}}_{\mathrm{R}}^{°}=\left[3\mathrm{mole}\times 0\mathrm{kJ}/\mathrm{mol}+2\mathrm{mole}\times 229\mathrm{kJ}/\mathrm{mol}\right]-\left[2\mathrm{mole}\times -34\mathrm{kJ}/\mathrm{mol}+1\mathrm{mole}\times -300\mathrm{kJ}/\mathrm{mol}\right]$

$$\begin{gathered} ∆{\mathrm{G}}_{\mathrm{R}}^{°}=0-458+68+300 \\ ∆{\mathrm{G}}_{\mathrm{R}}^{°}=-90\mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Determination of ∆G at 25°C

Let’s compute $∆\mathrm{G}$at $25°\mathrm{C}$and obtained pressure,

From equation (1),

$\mathit{\Delta }\mathit{G}\mathbf{=}\mathit{\Delta }{\mathit{G}}^{\mathbf{o}}\mathbf{+}\mathit{R}\mathit{T}\mathit{l}\mathit{n}\frac{\left[{\mathit{P}}_{\mathit{p}\mathit{r}\mathit{o}\mathit{d}\mathit{u}\mathit{c}\mathit{t}\mathit{s}}\right]}{\left[{\mathit{P}}_{\mathit{r}\mathit{e}\mathit{a}\mathit{c}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathit{s}}\right]}$

$\Delta G=\Delta G^o+RTln\frac{{\left[P_{H_{2}O}\right]}^2}{\left[P_{S{O}_2}\right]\times {\left[P_{H_{2}S}\right]}^2}$

$∆\mathrm{G}=-90\mathrm{kJ}/\mathrm{mol}+.008314\mathrm{kJ}/\mathrm{mol}-\mathrm{K}\times \mathrm{In}\times \frac{{\left[3\times {10}^{-2}\right]}^2}{\left[1.0\times {10}^{-2}\right]\times {\left[1.0\times {10}^{-4}\right]}^2}$

$∆\mathrm{G}=-50.3\mathrm{kJ}/\mathrm{mol}$

Therefore, the$\mathbf{∆}\mathbf{G}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{reaction}\mathrm{at}\mathbf{25}\mathbf{°}\mathbf{C}\mathbf{}\mathbf{is}\mathbf{}\mathbf{-}\mathbf{50}\mathbf{.}\mathbf{3}\mathbf{kJ}\mathbf{/}\mathbf{mol}$