Q62P - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

1. The given reaction standard free energy value is -228.6 kJ.
2. This reaction is not dependent on temperature T and enthalpy and entropy values are negative.
3. Given the reaction the temperature value is 7281.45 K and this reaction is non-spontaneous.

Step 1: Definition of Concept

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.

Step 2: Calculate △ H ° , △ S ° , and △ G ° per mole

$$\begin{gathered} ∆S_{rxn}^{°}=⅀m{S}_{product}^{°}-⅀m{S}_{reaction}^{°} \\ ∆S_{rxn}^{°}=\left[\left(1 mol H_{2}O\right)\left(S^{°}of H_{2}O\right)\right] \\ =-\left[\left(1 mol H_{2}O\right)\left(S^{°}of N_2\right)+\left(1/2 mol O_2\right)\left(S^{°}of O_2\right)\right] \\ ∆S_{rxn}^{°}=\left[\left(1 mol H_{2}O\right)\left(188.72 J/mol\times K\right)\right] \\ -\left[\left(1 mol H_{2}O\right)\left(130.6 J/mol\times K\right)+\left(1/2 mol O_2\right)\left(205.0 J/mol\times K\right)\right] \\ ∆S_{rxn}^{°}=-44.38 \\ =-44.4 J/K \\ ∆S_{rxn}^{°}=0.0444 KJ/K \end{gathered}$$(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,

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

The enthalpy change for the reaction is calculated as follows,

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

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,

 $$\begin{gathered} ∆S_{rxn}^{°}=⅀m{S}_{product}^{°}-⅀m{S}_{reaction}^{°} \\ ∆S_{rxn}^{°}=\left[\left(1 mol H_{2}O\right)\left(S^{°}of H_{2}O\right)\right] \\ =-\left[\left(1 mol H_{2}O\right)\left(S^{°}of N_2\right)+\left(1/2 mol O_2\right)\left(S^{°}of O_2\right)\right] \\ ∆S_{rxn}^{°}=\left[\left(1 mol H_{2}O\right)\left(188.72 J/mol\times K\right)\right] \\ -\left[\left(1 mol H_{2}O\right)\left(130.6 J/mol\times K\right)+\left(1/2 mol O_2\right)\left(205.0 J/mol\times K\right)\right] \\ ∆S_{rxn}^{°}=-44.38 \\ =-44.4 J/K \\ ∆S_{rxn}^{°}=0.0444 KJ/K \end{gathered}$$

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

Enthalpy and entropy values are

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

These values are plugging above standard free energy equation,

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

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

Step 3: Explain the spontaneity of this reaction is dependent on T

(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.

Step 4: Find at what temperature does the reaction become spontaneous

(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.