Question

Predict the sign of $\mathrm{\Delta }{S}^{\circ }$ and then calculate $\mathrm{\Delta }{S}^{\circ }$ for each of the following reactions. a. $2{\mathrm{H}}_2\mathrm{S}(g)+{\mathrm{SO}}_2(g)⟶3{\mathrm{S}}_{\text{ rhombic}}(s)+2{\mathrm{H}}_2\mathrm{O}(g)$ b. $2{\mathrm{SO}}_3(g)⟶2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)$ c. ${\mathrm{Fe}}_2{\mathrm{O}}_3(s)+3{\mathrm{H}}_2(g)⟶2\mathrm{Fe}(s)+3{\mathrm{H}}_2\mathrm{O}(g)$

Short answer

a. For the reaction $2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}_{\text { rhombic}}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$, we predict a negative sign for $\mathrm{\Delta }{S}^{\circ }$ because the number of moles of gas decreases. The calculated value of $\mathrm{\Delta }{S}^{\circ }$ is -194 J K^{-1} mol^{-1}. b. For the reaction $2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$, we predict a positive sign for $\mathrm{\Delta }{S}^{\circ }$ due to an increase in the number of moles of gas. c. For the reaction $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$, we predict a positive sign for $\mathrm{\Delta }{S}^{\circ }$ due to an increase in the number of moles of gas.

Step-by-step solution

Look up the standard molar entropies

From a standard molar entropy table, we find the following values: $S^{\circ }({\mathrm{H}}_2\mathrm{S})=205.8\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$ $S^{\circ }({\mathrm{SO}}_2)=248.1\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$ $S^{\circ }({\mathrm{S}}_{\text{ rhombic}})=31.7\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Multiply standard molar entropies by respective coefficients

We multiply the standard molar entropy values by their respective stoichiometric coefficients in the balanced reaction: $2\times 205.8\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}=411.6\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$ $3\times 31.7\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}=95.1\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Sum the entropy values of the products

We add the entropy values of the products: $S_{\text{products}}^{\circ }=95.1+370.6=465.7\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Sum the entropy values of the reactants

We add the entropy values of the reactants: $S_{\text{reactants}}^{\circ }=411.6+248.1=659.7\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Calculate $\mathrm{\Delta }{S}^{\circ }$

We subtract the total entropy of the reactants from the total entropy of the products: $\mathrm{\Delta }{S}^{\circ }=S_{\text{products}}^{\circ }-S_{\text{reactants}}^{\circ }=465.7-659.7=-194\mathrm{J}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$ Since the number of moles of gas is decreasing, the prediction of the negative sign for $\mathrm{\Delta }{S}^{\circ }$ is correct. Repeat the steps for the other two reactions: b. $2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$ c. $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$

Key concepts

Standard Molar Entropy

Standard molar entropy (denoted as $S^{\circ }$) is an essential concept in thermodynamics, important for calculating the entropy changes in chemical reactions. It's a measure of the amount of molecular disorder or randomness in a substance at a given standard state. This standard state is defined at one atmosphere of pressure and a specified temperature, usually 298 K (25°C). Each substance—be it a solid, liquid, or gas—possesses a unique standard molar entropy value, expressed in units of ${\text{J K}}^{-1}{\text{mol}}^{-1}$.

• Solids generally have lower $S^{\circ }$ values because their particles are arranged in an ordered lattice.
• Gases typically have higher $S^{\circ }$ because their particles move freely with more randomness.

When provided with these values for reactants and products, you can begin analyzing the entropy change in a chemical reaction. Standard molar entropies are foundational values in the calculations for predicting changes in entropy.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, and this transformation can affect entropy. Understanding the behavior of substances in these reactions is key to predicting the direction and spontaneous nature of reactions. When looking at a reaction, you should carefully observe:

• The states of matter (gas, liquid, solid) involved.
• The number of moles of gases produced or consumed.
• How the complexity of molecules changes from reactants to products.

For instance, when a reaction results in the formation of more moles of gas (from fewer moles of gas or solids), it usually indicates an increase in entropy. Conversely, forming a solid from gaseous reactants tends to decrease entropy. Chemists leverage these insights along with entropy calculations to predict whether a reaction is favorable or spontaneous based on the second law of thermodynamics.

Entropy Calculations

Entropy calculations help quantify the change in disorder from reactants to products within a chemical reaction. This process involves several steps, each building on the last to ensure accurate results. Calculating entropy change $\mathrm{\Delta }{S}^{\circ }$ requires these steps:

1. **Look up and Ananlyze:** Retrieve the standard molar entropy values $S^{\circ }$ for each substance involved from reference tables. - These tables list $S^{\circ }$ for various substances at standard conditions.
2. **Scale by Coefficients:** Multiply each substance's $S^{\circ }$ value by its stoichiometric coefficient from the balanced chemical equation.
3. **Sum Entropies:** Calculate the total $S^{\circ }$ for the products and the reactants separately.
4. **Determine $\mathrm{\Delta }{S}^{\circ }$:** Subtract the sum of the reactants' $S^{\circ }$ from the sum of the products' $S^{\circ }$: $$\mathrm{\Delta }{S}^{\circ }=S_{\text{products}}^{\circ }-S_{\text{reactants}}^{\circ }$$
These calculations not only predict the sign and magnitude of entropy change but also help decide whether a reaction proceeds spontaneously under constant temperature and pressure. A positive $\mathrm{\Delta }{S}^{\circ }$ typically favors spontaneity, aligning with the second law of thermodynamics.