As a fuel, H₂(g) produces only nonpolluting when it burns. Moreover, it combines with O₂(g) in a fuel cell (Chapter 21) to provide electrical energy.

(a) Calculate $\mathbf{△}{\mathbf{H}}^{\mathbf{°}}\mathbf{,}\mathbf{△}{\mathbf{S}}^{\mathbf{°}}$and $\mathbf{△}{\mathbf{G}}^{\mathbf{°}}$per mole of at 298 K.

(b) Is the spontaneity of this reaction dependent on T? Explain.

(c) At what temperature does the reaction become spontaneous?

Solubility: The maximum amount of solute that can be dissolved in the solvent at equilibrium is defined as solubility.

Constant of soluble product: For equilibrium between solids and their respective ions in a solution, the solubility product constant is defined. In general, the term "solubility product" refers to water-equilibrium insoluble or slightly soluble ionic substances.

$$\begin{gathered} ∆S_{rxn}^{°}=⅀m{S}_{product}^{°}-⅀m{S}_{reaction}^{°} \\ ∆S_{rxn}^{°}=\left[\left(1mol{H}_{2}O\right)\left(S^{°}of{H}_{2}O\right)\right] \\ =-\left[\left(1mol{H}_{2}O\right)\left(S^{°}of{N}_2\right)+\left(1/2mol{O}_2\right)\left(S^{°}of{O}_2\right)\right] \\ ∆S_{rxn}^{°}=\left[\left(1mol{H}_{2}O\right)\left(188.72J/mol\times K\right)\right] \\ -\left[\left(1mol{H}_{2}O\right)\left(130.6J/mol\times K\right)+\left(1/2mol{O}_2\right)\left(205.0J/mol\times K\right)\right] \\ ∆S_{rxn}^{°}=-44.38 \\ =-44.4J/K \\ ∆S_{rxn}^{°}=0.0444KJ/K \end{gathered}$$(a)

The reaction of H₂ with O₂ to give H₂O formation is,

${\text{H}}_{\text{2}}\left(\mathrm{g}\right)\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\left(\mathrm{g}\right){\text{→H}}_{\text{2}}\text{O(g)}$

The coefficient are written this way instead of

$2{\mathrm{H}}_2\left(\mathrm{g}\right)+{\text{O}}_{\text{2}}\left(\mathrm{g}\right){\text{→2H}}_2\text{O(g)}$

The specific thermodynamic values (per 1 mol H₂) not per 2 mol H₂ .

Standard enthalpy change is

$△{\mathrm{H}}^{°}$Formation of values,

role="math" localid="1663563404863" $$\begin{gathered} {\text{H}}_{\text{2}}\text{(g)= 0kJ/mol} \\ {\text{O}}_{\text{2}}\text{(g) = 0kJ/mol} \\ {\text{H}}_{\text{2}}\text{O(g)=-241kJ/mol} \end{gathered}$$

$$\begin{gathered} -\left[\left(1\mathrm{mol}{\mathrm{H}}_2\right)\left(△{\mathrm{H}}_{\mathrm{ff}}^{°}\mathrm{of}{\mathrm{H}}_2\right)\left(△{\mathrm{H}}_{\mathrm{f}}^{°}\mathrm{of}{\mathrm{O}}_2\right)\right] \\ △{\mathrm{H}}_{\mathrm{rxn}}^{°}=\left[\left(1\mathrm{mol}{\mathrm{H}}_2\mathrm{O}\right)\left(-241.826\mathrm{kJ}/\mathrm{mol}\right)\right] \\ =\left[\left(1\mathrm{mol}{\mathrm{H}}_2\right)\left(0\mathrm{kJ}/\mathrm{mol}\right)+\left(1/2\mathrm{mol}{\mathrm{O}}_2\right)\left(0\mathrm{kJ}/\mathrm{mol}\right)\right] \\ △{\mathrm{H}}_{\mathrm{rxn}}^{°}=-241.826\mathrm{kJ} \end{gathered}$$

The enthalpy change is negative.

Hence, the enthalpy $\left(△{\mathrm{H}}_{\mathrm{rxn}}^{°}\right)\text{changes is - 241.826kJ}$.

Entropy change$△{\mathrm{S}}_{\mathrm{system}}^{°}$

Calculate the change in entropy for this reaction as follows,

$$\begin{gathered} ∆S_{rxn}^{°}=⅀m{S}_{product}^{°}-⅀m{S}_{reaction}^{°} \\ ∆S_{rxn}^{°}=\left[\left(1mol{H}_{2}O\right)\left(S^{°}of{H}_{2}O\right)\right] \\ =-\left[\left(1mol{H}_{2}O\right)\left(S^{°}of{N}_2\right)+\left(1/2mol{O}_2\right)\left(S^{°}of{O}_2\right)\right] \\ ∆S_{rxn}^{°}=\left[\left(1mol{H}_{2}O\right)\left(188.72J/mol\times K\right)\right] \\ -\left[\left(1mol{H}_{2}O\right)\left(130.6J/mol\times K\right)+\left(1/2mol{O}_2\right)\left(205.0J/mol\times K\right)\right] \\ ∆S_{rxn}^{°}=-44.38 \\ =-44.4J/K \\ ∆S_{rxn}^{°}=0.0444KJ/K \end{gathered}$$

Hence, the $\left(△{\mathrm{S}}_{\mathrm{rxn}}^{°}\right)$of the reaction is -44.4 J/K.

Enthalpy and entropy values are

$$\begin{gathered} △{\mathrm{H}}_{\mathrm{rxn}}^{°}=-241.826 \\ △{\mathrm{G}}_{\mathrm{rxn}}^{°}=-0.0444\mathrm{kJ}/\mathrm{K} \end{gathered}$$

$$\begin{gathered} △{\mathrm{G}}_{\mathrm{rxn}}^{°}=\left(-241.826\mathrm{kJ}\right)-\left[\left(298\mathrm{K}\right)-\left(0.04438\mathrm{Kj}/\mathrm{K}\right)\right] \\ =-228.6\mathrm{kJ} \end{gathered}$$

Therefore, the given reaction standard free energy value is -228.6 kJ.

(b)

Given reaction,

${\text{2H}}_{\text{2}}{\text{(g) +O}}_{\text{2}}{\text{(g)→2H}}_{\text{2}}\text{O(g)}$

This reaction is does not dependent on temperature T, because enthalpy $△\mathrm{H}<0$and $△\mathrm{H}<0$ values are negative.

This reaction will become non-spontaneous at higher temperature because the positive $\left(-\mathrm{T}△\mathrm{S}\right)$term becomes larger than negative term. Further in this reaction.

(c)

Given reaction,

${\text{2H}}_{\text{2}}{\text{(g) +O}}_{\text{2}}{\text{(g)→2H}}_{\text{2}}\text{O(g)}$

The reaction becomes spontaneous below the temperature where$△{\mathrm{G}}_{\mathrm{rxn}}^{°}=0$

Consider the following free energy equation,

$$\begin{gathered} △{\mathrm{G}}_{\mathrm{rxn}}^{°}=0=∆{\mathrm{H}}_{\mathrm{rxn}}^{°}-\mathrm{T}∆\mathrm{S}-----\left[1\right] \\ △{\mathrm{H}}_{\mathrm{rxn}}^{°}=\mathrm{T}∆{\mathrm{S}}_{\mathrm{rxn}}^{°}-----\left[2\right] \end{gathered}$$

Rearrange equation (2) to calculate temperature T,

Hence,

$$\begin{gathered} \mathrm{T}=\frac{-241.826\mathrm{kJ}}{-0.04438\mathrm{kJ}/\mathrm{K}} \\ \text{= 5448.986} \\ \mathrm{T}=5.45\times {10}^3\mathrm{K} \end{gathered}$$

At temperature become above 7281.45 K, this reaction is non-spontaneous. Because both enthalpy $△{\mathrm{H}}_{\mathrm{rxn}}^{°}$and entropy $△{\mathrm{S}}_{\mathrm{rxn}}^{°}$values are negative, the reaction becomes non-spontaneous calculated temperature T.