Entropy, Enthalpy, and Gibbs Free Energy – Step-by-Step Chemistry Guidance

Q1. What can you say about the change of entropy when solid iodine sublimes to vapor?

Background

Topic: Entropy and Phase Changes

This question tests your understanding of how entropy changes during phase transitions, specifically sublimation (solid to gas).

Key Terms:

• Entropy ($S$): A measure of disorder or randomness in a system.

• Sublimation: The process where a solid turns directly into a gas.

Step-by-Step Guidance

1. Recall that entropy increases when a substance moves to a more disordered state (solid → liquid → gas).

2. Consider the molecular arrangement: solids have tightly packed molecules, gases have molecules spread far apart.

3. Think about the energy and freedom of movement in each phase.

Try solving on your own before revealing the answer!

Final Answer:

The entropy of iodine increases during sublimation because the molecules become more dispersed and disordered in the gas phase.

Q2. What is the entropy change when 1.00 mol liquid acetone vaporizes at 25°C?

Background

Topic: Entropy Change in Phase Transitions

This question asks you to calculate the entropy change ($\Delta S$) for the vaporization of acetone using enthalpy of formation data.

Key Terms and Formulas:

• Standard enthalpy of formation ($\Delta H_f^\circ$): The enthalpy change when 1 mol of a substance is formed from its elements in their standard states.

• Entropy change ($\Delta S$): $\Delta S = \frac{\Delta H_{vap}}{T}$

Step-by-Step Guidance

1. Write the balanced phase change: $\mathrm{CH_3COCH_3(l) \rightarrow CH_3COCH_3(g)}$

2. Calculate $\Delta H_{vap}$ using the enthalpy of formation values:

   $\Delta H_{vap} = \Delta H_f^\circ(\text{vapor}) - \Delta H_f^\circ(\text{liquid})$

3. Plug in the values: $\Delta H_{vap} = (-216.6\ \text{kJ/mol}) - (-247.6\ \text{kJ/mol})$

4. Convert $\Delta H_{vap}$ to J/mol if needed (1 kJ = 1000 J).

5. Use the formula $\Delta S = \frac{\Delta H_{vap}}{T}$, with $T = 298\ \text{K}$.

Try solving on your own before revealing the answer!

Final Answer: $\Delta S = 104\ \text{J/K}$

We calculated $\Delta H_{vap}$ and divided by the temperature in Kelvin to find the entropy change for vaporization.

Q3. What is the normal boiling point of o-xylene given $\Delta H^\circ_{vap} = 36.2$ kJ/mol and $\Delta S^\circ_{vap} = 86.7$ J/(mol·K)?

Background

Topic: Phase Change Thermodynamics

This question tests your ability to relate enthalpy and entropy changes to the boiling point using thermodynamic equations.

Key Formula:

• $\Delta S^\circ_{phase\ change} = \frac{\Delta H^\circ_{phase\ change}}{T_{phase\ change}}$

• Rearranged: $T_{phase\ change} = \frac{\Delta H^\circ_{phase\ change}}{\Delta S^\circ_{phase\ change}}$

Step-by-Step Guidance

1. Identify the given values: $\Delta H^\circ_{vap} = 36.2$ kJ/mol, $\Delta S^\circ_{vap} = 86.7$ J/(mol·K).

2. Convert $\Delta H^\circ_{vap}$ to J/mol: $36.2\ \text{kJ/mol} \times 1000 = 36200\ \text{J/mol}$

3. Set up the equation for boiling point: $T_{bp} = \frac{\Delta H^\circ_{vap}}{\Delta S^\circ_{vap}}$

4. Plug in the values: $T_{bp} = \frac{36200}{86.7}$

Try solving on your own before revealing the answer!

Final Answer: $T_{bp} \approx 418\ \text{K}$

By dividing the enthalpy by the entropy, you find the temperature at which o-xylene boils under normal conditions.

Q4. For each pair, which substance has the higher molar entropy?

Background

Topic: Molar Entropy Comparisons

This question asks you to compare the entropy of different substances based on their physical state, molecular complexity, and conditions.

Key Concepts:

• Entropy increases with molecular complexity, phase (gas > liquid > solid), and temperature.

• For gases, lower pressure means higher entropy.

Step-by-Step Guidance

1. Compare the number of atoms in each molecule (more atoms = higher entropy).

2. Compare phase: aqueous > solid; gas > liquid > solid.

3. Compare pressure for gases: lower pressure = higher entropy.

4. Compare temperature: higher temperature = higher entropy.

Try solving on your own before revealing the answer!

Final Answer:

For each pair, the substance with more atoms, higher phase (gas/aqueous), lower pressure, or higher temperature has higher molar entropy.

Q5. Predict the sign of $\Delta S^\circ$ for the reaction: Ba(OH)$_2$·8H$_2$O(s) + 2NH$_4$NO$_3$(s) → 2NH$_3$(g) + 10H$_2$O(l) + Ba(NO$_3$)$_2$(aq)

Background

Topic: Entropy Change in Chemical Reactions

This question asks you to predict whether the entropy increases or decreases during a reaction based on the states and number of moles of reactants and products.

Key Concepts:

• Entropy increases when solids produce gases, liquids, or aqueous solutions.

• Compare the number of moles and phases before and after the reaction.

Step-by-Step Guidance

1. Count the total number of moles of reactants and products.

2. Identify the phases: solids on the reactant side, gases/liquids/aqueous on the product side.

3. Recall that producing more moles and more disordered phases increases entropy.

Try solving on your own before revealing the answer!

Final Answer:

$\Delta S^\circ$ is positive because the reaction produces more moles and more disordered phases (gas, liquid, aqueous) from solid reactants.

Q6. Predict the sign of $\Delta S$ for each reaction:

Background

Topic: Qualitative Entropy Changes

This question asks you to predict whether entropy increases or decreases for various reactions based on the change in phase and number of moles.

Key Concepts:

• Entropy increases when a reaction produces more moles of gas or moves to a more disordered phase.

• Entropy decreases when a reaction produces fewer moles of gas or moves to a more ordered phase.

Step-by-Step Guidance

1. For each reaction, count the moles of gas on both sides.

2. Identify any phase changes (solid to gas, liquid to gas, etc.).

3. Determine if the system becomes more or less disordered.

Try solving on your own before revealing the answer!

Final Answer:

Reactions that produce more gas or move to a more disordered phase have positive $\Delta S$; those that produce less gas or move to a more ordered phase have negative $\Delta S$.

Q7. Calculate $\Delta S^\circ_{rxn}$ for N$_2$H$_4$(l) + 2H$_2$O$_2$(l) → N$_2$(g) + 4H$_2$O(g)

Background

Topic: Calculating Entropy Change for a Reaction

This question asks you to use standard molar entropy values to calculate the entropy change for a chemical reaction.

Key Formula:

• $\Delta S^\circ_{rxn} = \sum n S^\circ_m(\text{products}) - \sum n S^\circ_m(\text{reactants})$

Step-by-Step Guidance

1. List the standard molar entropy ($S^\circ_m$) values for each substance involved.

2. Multiply each $S^\circ_m$ by the number of moles in the balanced equation.

3. Add up the total entropy for products and reactants.

4. Subtract the sum for reactants from the sum for products.

Try solving on your own before revealing the answer!

Final Answer: $\Delta S^\circ_{rxn} = +605.8$ J/mol·K

The reaction produces more moles of gas, leading to a large positive entropy change.

Q8. Calculate $\Delta S^\circ_{rxn}$ for 2H$_2$O(g) → 2H$_2$(g) + O$_2$(g)

Background

Topic: Entropy Change for a Reaction

This question asks you to calculate the entropy change using standard molar entropy values for each species.

Key Formula:

• $\Delta S^\circ_{rxn} = \sum n S^\circ_m(\text{products}) - \sum n S^\circ_m(\text{reactants})$

Step-by-Step Guidance

1. Write down the $S^\circ_m$ values for H$_2$O(g), H$_2$(g), and O$_2$(g).

2. Multiply each value by the number of moles in the balanced equation.

3. Add up the total entropy for products and reactants.

4. Subtract the sum for reactants from the sum for products.

Try solving on your own before revealing the answer!

Final Answer: $\Delta S^\circ_{rxn} = +88.8$ J/K

The reaction increases entropy because it produces more moles of gas from fewer moles.

Q9. For the reverse reaction 2H$_2$(g) + O$_2$(g) → 2H$_2$O(g), what is $\Delta S^\circ_{rxn}$?

Background

Topic: Entropy Change for Reverse Reactions

This question asks you to consider how the entropy change for a reaction relates to its reverse.

Key Concept:

• The entropy change for the reverse reaction is the negative of the forward reaction.

Step-by-Step Guidance

1. Recall the calculated $\Delta S^\circ_{rxn}$ for the forward reaction.

2. Reverse the sign for the reverse reaction.

Try solving on your own before revealing the answer!

Final Answer: $\Delta S^\circ_{rxn} = -88.8$ J/K

The reverse reaction has the opposite entropy change.

Q10. How do you calculate $\Delta S_{surroundings}$ for a reaction?

Background

Topic: Entropy Change of Surroundings

This question asks you to relate the heat exchanged with the surroundings to the entropy change of the surroundings.

Key Formula:

• $\Delta S_{surroundings} = \frac{q_{surroundings}}{T}$

• For a reaction at constant pressure, $q_{surroundings} = -\Delta H_{system}$

Step-by-Step Guidance

1. Calculate $\Delta H_{rxn}$ for the system using standard enthalpies of formation.

2. Recognize that $\Delta H_{surroundings} = -\Delta H_{system}$

3. Plug $q_{surroundings}$ and $T$ into the formula for $\Delta S_{surroundings}$.

Try solving on your own before revealing the answer!

Final Answer:

Calculate $\Delta S_{surroundings}$ by dividing the heat absorbed by the surroundings by the temperature in Kelvin.

Q11. What is $\Delta S_{universe}$ for a reaction?

Background

Topic: Second Law of Thermodynamics

This question asks you to combine the entropy changes of the system and surroundings to find the total change for the universe.

Key Formula:

• $\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}$

Step-by-Step Guidance

1. Calculate $\Delta S_{system}$ for the reaction.

2. Calculate $\Delta S_{surroundings}$ using the heat released and temperature.

3. Add the two values to get $\Delta S_{universe}$.

Try solving on your own before revealing the answer!

Final Answer:

$\Delta S_{universe}$ is positive for a spontaneous process, even if $\Delta S_{system}$ is negative.

Q12. For each reaction, state when, if ever, the reaction will be spontaneous given $\Delta H$ and $\Delta S$.

Background

Topic: Spontaneity and Gibbs Free Energy

This question asks you to use $\Delta H$ and $\Delta S$ to predict spontaneity at different temperatures.

Key Formula:

• $\Delta G = \Delta H - T\Delta S$

Step-by-Step Guidance

1. Identify the sign of $\Delta H$ (enthalpy change) and $\Delta S$ (entropy change).

2. Recall that a reaction is spontaneous when $\Delta G < 0$.

3. Analyze how temperature affects $\Delta G$ based on the signs of $\Delta H$ and $\Delta S$.

Try solving on your own before revealing the answer!

Final Answer:

Reactions with negative $\Delta H$ and positive $\Delta S$ are always spontaneous; those with both negative are spontaneous at low T; both positive at high T; both unfavorable are never spontaneous.

Q13. For the vaporization of a liquid at a given pressure, how does $\Delta G$ depend on temperature?

Background

Topic: Gibbs Free Energy and Phase Changes

This question asks you to analyze how $\Delta G$ changes with temperature for a phase change where both $\Delta H$ and $\Delta S$ are positive.

Key Formula:

• $\Delta G = \Delta H - T\Delta S$

Step-by-Step Guidance

1. Recognize that for vaporization, $\Delta H > 0$ and $\Delta S > 0$.

2. As temperature increases, the $-T\Delta S$ term becomes more negative.

3. At low T, $\Delta G$ is positive; at high T, $\Delta G$ can become negative.

Try solving on your own before revealing the answer!

Final Answer:

$\Delta G$ is positive at low temperatures and negative at high temperatures for vaporization.

Q14. If a reaction is endothermic and non-spontaneous at 25°C, can it ever be spontaneous?

Background

Topic: Temperature Dependence of Spontaneity

This question asks you to consider how temperature affects the spontaneity of an endothermic reaction.

Key Formula:

• $\Delta G = \Delta H - T\Delta S$

Step-by-Step Guidance

1. Endothermic means $\Delta H > 0$; non-spontaneous means $\Delta G > 0$ at 25°C.

2. If $\Delta S > 0$, increasing T can make $\Delta G$ negative.

3. If $\Delta S < 0$, the reaction will never be spontaneous.

Try solving on your own before revealing the answer!

Final Answer:

If $\Delta S > 0$, the reaction may become spontaneous at higher temperatures.

Q15. What does a positive standard Gibbs energy of formation ($\Delta G_f^\circ$) for CS$_2$(g) mean?

Background

Topic: Gibbs Free Energy of Formation

This question asks you to interpret the meaning of a positive $\Delta G_f^\circ$ value for a compound.

Key Concept:

• A positive $\Delta G_f^\circ$ means the compound is less stable than its elements in their standard states.

Step-by-Step Guidance

1. Recall the definition of $\Delta G_f^\circ$.

2. Compare the stability of the compound to its constituent elements.

Try solving on your own before revealing the answer!

Final Answer:

CS$_2$(g) is less stable than its elements; it tends to decompose into C(s) and S$_8$(s).

Q16. Calculate the free-energy change, $\Delta G^\circ$, for the oxidation of ethyl alcohol to acetic acid using standard free energies of formation.

Background

Topic: Calculating Gibbs Free Energy Change

This question asks you to use tabulated $\Delta G_f^\circ$ values to calculate the free energy change for a reaction.

Key Formula:

• $\Delta G^\circ_{rxn} = \sum n \Delta G_f^\circ(\text{products}) - \sum n \Delta G_f^\circ(\text{reactants})$

Step-by-Step Guidance

1. Write the balanced chemical equation.

2. List the $\Delta G_f^\circ$ values for each reactant and product.

3. Multiply each value by the number of moles in the equation.

4. Add up the total for products and reactants.

5. Subtract the sum for reactants from the sum for products.

Try solving on your own before revealing the answer!

Final Answer: $\Delta G^\circ = -454.9$ kJ

The negative value indicates the reaction is spontaneous under standard conditions.

Q17. Calculate the standard Gibbs energy change for the formation of methane at 298 K given $\Delta H^\circ_f$ and $S^\circ_m$ for reactants and products.

Background

Topic: Calculating $\Delta G^\circ$ from $\Delta H^\circ$ and $\Delta S^\circ$

This question asks you to use enthalpy and entropy data to calculate the Gibbs free energy change for a reaction.

Key Formula:

• $\Delta G^\circ_{rxn} = \Delta H^\circ_{rxn} - T \Delta S^\circ_{rxn}$

Step-by-Step Guidance

1. Write the balanced equation: C(s) + 2H$_2$(g) → CH$_4$(g)

2. Find $\Delta H^\circ_{rxn}$ using standard enthalpies of formation.

3. Calculate $\Delta S^\circ_{rxn}$ using standard molar entropies.

4. Convert $\Delta S^\circ_{rxn}$ to kJ by dividing by 1000 if needed.

5. Plug values into $\Delta G^\circ_{rxn} = \Delta H^\circ_{rxn} - T \Delta S^\circ_{rxn}$ with $T = 298$ K.

Try solving on your own before revealing the answer!

Final Answer: $\Delta G^\circ_{rxn} = -50.7$ kJ/mol at 298 K

The reaction is spontaneous at 298 K, but not at higher temperatures.

Q18. Consider the reaction N$_2$(g) + O$_2$(g) → 2NO(g). What does the standard free energy of formation of NO say about this reaction?

Background

Topic: Gibbs Free Energy and Reaction Spontaneity

This question asks you to interpret the meaning of a positive $\Delta G^\circ$ for the formation of NO.

Key Concept:

• A positive $\Delta G^\circ$ means the reaction is nonspontaneous under standard conditions.

Step-by-Step Guidance

1. Note the value of $\Delta G^\circ$ for the reaction and for 1 mol NO.

2. Compare the stability of NO to its elements.

3. Predict the direction of spontaneity (decomposition vs. formation).

Try solving on your own before revealing the answer!

Final Answer:

The reaction is nonspontaneous; NO is less stable than N$_2$ and O$_2$ and tends to decompose.