Textbook Question
For the following equilibrium processes and the corresponding ∆G°, calculate (i) Keq and (ii) the % composition of the equilibrium mixture (% reactants, % products) at 298 K.
(b)
1
views
Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 2 - General Chemistry Translated: Finding the Electrons114
- Ch. 3 - Alkanes and Cycloalkanes: Properties and Conformational Analysis109
- Ch. 4 - Acids and Bases: Electron Flow144
- Ch. 5 - Chemical Reaction Analysis: Thermodynamics and Kinetics118
- Ch. 6 - Stereoisomerism: Arrangement of Atoms in Space112
- Ch. 7 - Principles of Green (Organic) Chemistry3
- Ch. 8 - Alkenes I: Properties and Electrophilic Additions158
- Ch. 9 - Alkenes II: Oxidation and Reduction150
- Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions101
- Ch. 11 - Properties and Synthesis of Alkyl Halides: Radical Reactions108
- Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes172
- Ch. 13 - Alcohols, Ethers and Related Compounds: Substitution and Elimination239
- Ch. 14 - Structural Identification I: Infrared Spectroscopy and Mass Spectrometry54
- Ch. 15 - Structural Identification II: Nuclear Magnetic Resonance Spectroscopy93
- Ch. 16 - Metals in Organic Chemistry48
- Ch. 17 - Carbonyl Addition Reactions: Aldehydes and Ketones45
- Ch. 18 - Nucleophilic Acyl Substitution I: Carboxylic Acids18
- Ch. 19 - Nucleophilic Acyl Substitution II: Carboxylic Acid Derivatives39
- Ch. 20 - Enolates: Carbonyl Addition and Substitution13
- Ch. 21 - Conjugated Systems I: Stability and Addition Reactions56
- Ch. 22 - Conjugated Systems II: Pericyclic Reactions54
- Ch. 23 - Benzene I: Aromatic Stability and Substitution Reactions64
- Ch. 24 - Benzene II: Reactions Influenced by the Aromatic Ring62
- Ch. 25 - Amines: Structure, Reactions, and Synthesis27
- Ch. 26 - Amino Acids, Proteins, and Peptide Synthesis9
- Ch. 27 - Carbohydrates, Nucleic Acids, and Lipids54
Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Not the one you use?Change textbook
All textbooks
Mullins 1st EditionCh. 5 - Chemical Reaction Analysis: Thermodynamics and KineticsProblem 11b
Mullins 1st EditionCh. 5 - Chemical Reaction Analysis: Thermodynamics and KineticsProblem 11bChapter 4, Problem 11b
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
Verified step by step guidance1
Step 1: Recall the Gibbs free energy equation: G = ΔH - TΔS. This equation determines whether a process is thermodynamically favored. A negative value of ΔG indicates a favored process.
Step 2: Ensure the units are consistent. Here, ΔH° is given in kcal/mol, while ΔS° is in cal/mol•K. Convert ΔH° to cal/mol by multiplying by 1000: ΔH° = 7.34 kcal/mol × 1000 = 7340 cal/mol.
Step 3: Substitute the given values into the Gibbs free energy equation. Use ΔH° = 7340 cal/mol, ΔS° = +43 cal/mol•K, and T = 325 K. The equation becomes: G = 7340 - (325 × 43).
Step 4: Analyze the sign of ΔG. If the result of the calculation is negative, the process is thermodynamically favored. If it is positive, the process is not favored.
Step 5: Consider the temperature dependence. Since ΔH° is positive (endothermic) and ΔS° is positive (increase in entropy), the process may become favored at higher temperatures. Evaluate whether the given temperature (325 K) is sufficient to make ΔG negative.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gibbs Free Energy
Gibbs Free Energy (G) is a thermodynamic potential that helps predict whether a process will occur spontaneously at constant temperature and pressure. It is calculated using the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. A negative ΔG indicates a spontaneous process, while a positive ΔG suggests non-spontaneity.
Recommended video:
Guided course
05:02
Breaking down the different terms of the Gibbs Free Energy equation.
Enthalpy (ΔH)
Enthalpy (ΔH) is a measure of the total heat content of a system. A positive ΔH indicates that the process is endothermic, meaning it absorbs heat from the surroundings. This can affect the spontaneity of a reaction, as higher energy input may be required to drive the process, especially if the entropy change is not favorable.
Recommended video:
Guided course
04:38Calculating Enthalpies
Entropy (ΔS)
Entropy (ΔS) is a measure of the disorder or randomness in a system. A positive ΔS indicates an increase in disorder, which generally favors spontaneity. In the context of the Gibbs Free Energy equation, a larger positive ΔS can offset a positive ΔH, potentially leading to a negative ΔG and making the process more favorable at higher temperatures.
Recommended video:
Guided course
02:46Explaining what entropy is.
Related Practice
Textbook Question
For the following equilibrium processes and the corresponding ∆G°, calculate (i) Keq and (ii) the % composition of the equilibrium mixture (% reactants, % products) at 298 K.
(a)
7
views
Textbook Question
At what temperature does the entropy change of a process not contribute to the favorability of a process?
3
views
Textbook Question
For the following values of ∆H° , ∆S°, and T, tell whether the process would be favored.
(c) ∆H° = -21.3 kcal/mol ; AS° = -51 cal/mol•K ; T = 373 K
2
views
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
5
views
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
3
views