Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

6. Thermodynamics and Kinetics

Entropy

Organic Chemistry

6. Thermodynamics and Kinetics

Entropy

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3:38 minutes

Problem 53

Textbook Question

When ethene is treated in a calorimeter with H₂ and a Pt catalyst, the heat of reaction is found to be –137 kJ/mol (–32.7 kcal/mol), and the reaction goes to completion. When the reaction takes place at 1400 K, the equilibrium is found to be evenly balanced, with K_eq =1. Compute the value of ΔS for this reaction.

Verified step by step guidance

Step 1: Understand the problem. The reaction involves ethene (CH2=CH2) reacting with hydrogen (H2) in the presence of a platinum catalyst to form ethane (CH3—CH3). The heat of reaction (ΔH) is given as -137 kJ/mol, and at 1400 K, the equilibrium constant (Keq) is 1. We need to compute the entropy change (ΔS) for this reaction.

Step 2: Recall the relationship between Gibbs free energy (ΔG), enthalpy change (ΔH), and entropy change (ΔS). The equation is:

Δ G = Δ H - T Δ S

, where T is the temperature in Kelvin.

Step 3: At equilibrium, the Gibbs free energy change (ΔG) is zero because Keq = 1. This simplifies the equation to:

Δ H = T Δ S

. Rearrange this equation to solve for ΔS:

Δ S = \frac{Δ}{H}

Step 4: Substitute the given values into the equation. ΔH is -137 kJ/mol, and T is 1400 K. Ensure that the units are consistent; ΔH should be converted to joules (1 kJ = 1000 J) if necessary.

Step 5: Perform the division to calculate ΔS. The result will be in units of J/(mol·K). This value represents the entropy change for the reaction under the given conditions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Enthalpy (ΔH)

Enthalpy is a thermodynamic property that represents the total heat content of a system. In this reaction, the negative value of ΔH (-137 kJ/mol) indicates that the reaction is exothermic, meaning it releases heat to the surroundings. Understanding enthalpy is crucial for analyzing energy changes during chemical reactions and helps predict the spontaneity of the reaction.

Entropy (ΔS)

Entropy is a measure of the disorder or randomness in a system. It plays a key role in determining the spontaneity of a reaction alongside enthalpy. In this context, calculating ΔS will help assess how the reaction's disorder changes as ethene and hydrogen gas convert into ethane, particularly under the given conditions of temperature and equilibrium.

Gibbs Free Energy (ΔG)

Gibbs Free Energy combines enthalpy and entropy to determine the spontaneity of a reaction at constant temperature and pressure. The relationship is given by the equation ΔG = ΔH - TΔS. Since the equilibrium constant (Keq) is 1 at 1400°K, ΔG is zero, indicating that the system is at equilibrium. This relationship is essential for calculating ΔS using the provided ΔH and temperature.

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Related Practice

Textbook Question

For the following values of ∆H° , ∆S°, and T, tell whether the process would be favored.

(b) ∆H° = +7.34 kcal/mol ; ∆S° = +43 cal/mol•K ; T = 325 K

Textbook Question

Considering the process described in Assessment 5.13, will it be favored or disfavored at a temperature higher than the one you calculated? How about at a temperature below what you calculated?

Textbook Question

For the following values of ∆H° , ∆S°, and T, tell whether the process would be favored.

(d) ∆H° = -8.3 kcal/mol ; ∆S° = -12 cal/mol•K ; T = 298 K

Textbook Question

Calculate ∆G°, ∆H°, and ∆S° for the following acid–base reactions. Rationalize the value of ∆H° based on the structure of the conjugate bases. [Assume T = 298 K.]

(b)

Textbook Question

For the following values of ∆H° , ∆S°, and T, tell whether the process would be favored.

(a) ∆H° = -15 kcal/mol ; ∆S° = +37 cal/mol•K ; T = 273 K

Textbook Question

For each reaction, estimate whether ΔS° for the reaction is positive, negative, or impossible to predict.

(a)

Textbook Question

When ethene is mixed with hydrogen in the presence of a platinum catalyst, hydrogen adds across the double bond to form ethane. At room temperature, the reaction goes to completion. Predict the signs of ΔH° and ΔS° for this reaction. Explain these signs in terms of bonding and freedom of motion.

Textbook Question

For each reaction, estimate whether ΔS° for the reaction is positive, negative, or impossible to predict.

(b)

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