Question

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

Short answer

For reaction a, the sign of ΔS° is negative with a value of -439.6 J/molK. For reaction b, the sign of ΔS° is positive with a value of 187.6 J/molK. For reaction c, the sign of ΔS° is slightly positive with a value of 11.3 J/molK.

Step-by-step solution

Predict the sign of ΔS° for reaction a

For reaction a, we have: \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \longrightarrow 3 \mathrm{S_{thombic}}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)\) We can see that there are 3 moles of gaseous reactants and 2 moles of gaseous products. As a result, the reaction leads to more order because we have fewer moles of gas products than the reactants. So, the sign of ΔS° will be negative.

Calculate ΔS° for reaction a

To calculate ΔS° for reaction a, we need the standard entropy values for each substance involved in the reaction: \(S^{\circ}_{H2S(g)} = 206.8 \, J/molK\) \(S^{\circ}_{SO2(g)} = 248.2 \, J/molK\) \(S^{\circ}_{S_{thombic}(s)} = 31.9 \, J/molK\) \(S^{\circ}_{H2O(g)} = 188.7 \, J/molK\) Now, we can use the equation to calculate ΔS°: \(\Delta S^{\circ}=\sum S^{\circ}_{\text {products }}-\sum S^{\circ}_{\text {reactants }}\) \(\Delta S^{\circ} = [3(31.9) + 2(188.7)] - [2(206.8) + 1(248.2)]\) \(\Delta S^{\circ} = -439.6 \, J/molK\) So, ΔS° for reaction a is -439.6 J/molK

Predict the sign of ΔS° for reaction b

For reaction b, we have: \(2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)\) We can notice that there are 3 moles of gaseous products and 2 moles of gaseous reactants. So, the reaction is leading to more disorder with more moles of gas products than the reactants. Thus, the sign of ΔS° will be positive.

Calculate ΔS° for reaction b

To calculate ΔS° for reaction b, we need the standard entropy values for each substance involved in the reaction: \(S^{\circ}_{SO3(g)} = 256.8 \, J/molK\) \(S^{\circ}_{SO2(g)} = 248.2 \, J/molK\) \(S^{\circ}_{O2(g)} = 205.0 \, J/molK\) Now, we can use the equation to calculate ΔS°: \(\Delta S^{\circ}=\sum S^{\circ}_{\text {products }}-\sum S^{\circ}_{\text {reactants }}\) \(\Delta S^{\circ} = [2(248.2) + 1(205)] - 2(256.8)\) \(\Delta S^{\circ} = 187.6 \, J/molK\) So, ΔS° for reaction b is 187.6 J/molK

Predict the sign of ΔS° for reaction c

For reaction c, we have: \(\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)\) Comparing the moles of gas in reactants and products, we have 3 moles of gaseous reactants and 3 moles of gaseous products. So, there is no significant change in the number of moles of gas. As a result, the sign of ΔS° will be close to zero or slightly positive due to the formation of water molecules.

Calculate ΔS° for reaction c

To calculate ΔS° for reaction c, we need the standard entropy values for each substance involved in the reaction: \(S^{\circ}_{Fe2O3(s)} = 87.4 \, J/molK\) \(S^{\circ}_{H2(g)} = 130.6 \, J/molK\) \(S^{\circ}_{Fe(s)} = 27.2 \, J/molK\) \(S^{\circ}_{H2O(g)} = 188.7 \, J/molK\) Now, we can use the equation to calculate ΔS°: \(\Delta S^{\circ}=\sum S^{\circ}_{\text {products }}-\sum S^{\circ}_{\text {reactants }}\) \(\Delta S^{\circ} = [2(27.2) + 3(188.7)] - [1(87.4) + 3(130.6)]\) \(\Delta S^{\circ} = 11.3 \, J/molK\) So, ΔS° for reaction c is 11.3 J/molK

Key concepts

Predicting Entropy Sign

Entropy, \(\Delta S^{\circ}\), is a measure of the disorder or randomness in a system. When it comes to chemical reactions, predicting the sign of the entropy change involves comparing the moles of gaseous reactants to the moles of gaseous products. Here's how you can do it:

• Increase in gas moles: When a reaction results in more moles of gas products than reactants, the system becomes more disordered, leading to a positive \(\Delta S^{\circ}\).
• Decrease in gas moles: Conversely, if there are fewer moles of gas products than reactants, the system becomes more ordered, and \(\Delta S^{\circ}\) is negative.
• No change in gas moles: If the number of gas moles remains constant, the entropy change might be zero or close to zero, but other factors, like the nature of substances formed, can still influence it.

Understanding these basic principles can significantly aid you in predicting whether the entropy change in a reaction will be positive, negative, or negligible!

Calculating Standard Entropy Change

To calculate the standard entropy change (\(\Delta S^{\circ}\)) for a chemical reaction, you need to apply the formula:\[\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\]The key steps involved are:

• Determine the standard molar entropy (\(S^{\circ}\)) of each reactant and product involved.
• Multiply each \(S^{\circ}\) by its stoichiometric coefficient from the balanced chemical equation.
• Subtract the total entropy of reactants from the total entropy of products to find \(\Delta S^{\circ}\).

Using the correct units and values ensures accuracy. Remember, all \(S^{\circ}\) values should be in \(J/molK\) for consistency in calculations. This method reflects the overall increase or decrease in disorder from reactants to products.

Chemical Thermodynamics

Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or physical changes of state. It involves concepts like free energy, enthalpy, and entropy. Entropy (\(S\)) is a central player in thermodynamics and is often considered alongside other thermodynamic quantities:

• Gibbs Free Energy (\(G\)): It indicates the spontaneity of a reaction. The relation is given by \(\Delta G = \Delta H - T\Delta S\), where \(\Delta H\) is the enthalpy change, and \(T\) is the temperature in Kelvin.
• Enthalpy (\(H\)): This is the total heat content of a system. Reactions can be exothermic (release heat) or endothermic (absorb heat).
• Temperature and enthalpy strongly influence entropy changes because they contribute to the disorder during a reaction.

Understanding how entropy fits into the larger framework of thermodynamics is crucial for predicting reaction behavior and the feasibility of chemical processes.

Standard Entropy Values

Standard entropy values (\(S^{\circ}\)) are tabulated entropy data for various substances measured under standard conditions. These values give insight into the randomness or disorder associated with a mole of substance in its standard state.

• Standard Conditions: These are usually defined as 1 atm pressure and a specified temperature, typically 25 °C (298 K).
• Each substance has a unique \(S^{\circ}\) value, reflecting its atomic and molecular complexity and phase.
• Gases tend to have higher standard entropy values compared to liquids and solids due to their greater degrees of freedom and mobility.

When calculating entropy changes in reactions, these standardized data help ensure consistency and accuracy, making them an invaluable resource in chemical thermodynamics.

Gas Phase Reactions

In gas phase reactions, moles react in the gaseous state, significantly influencing the entropy changes due to their naturally high disorder and mobility.

• Impact on Entropy: The decrease or increase in moles of gases during a reaction has the greatest impact on \(\Delta S^{\circ}\). More moles of gas typically mean more disorder, influencing entropy positively.
• Reactions where gases condense into solids or liquids result in negative entropy changes because of decreased randomness.
• Le Chatelier's Principle also applies, where a system at equilibrium responds to changes in temperature or pressure by favoring either the forward or reverse reaction, impacting gas phase reactions significantly.

Overall, understanding how gas phase reactions contribute to entropy is key in predicting and calculating the thermodynamic behaviors of chemical reactions.