Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics

Bruice8th EditionOrganic ChemistryISBN: 9780135213711Not the one you use?Change textbook

Bruice 8th Edition

Ch. 5 - Alkenes: Structure, Nomenclature, and an Introduction to Reactivity • Thermodynamics and Kinetics

Problem 63

Chapter 6, Problem 63

For a reaction carried out at 25 °C with an equilibrium constant of 1 × 10^-3, to increase the equilibrium constant by a factor of 10:
a. how much must ∆G° change?
b. how much must ∆H° change if ∆S° = 0 kcal mol^-1K^-1?
c. how much must ∆S° change if ∆H° = 0 kcal mol^-1?

Verified step by step guidance

Step 1: Recall the relationship between the equilibrium constant (K) and the standard Gibbs free energy change (∆G°). The equation is:

∆G° = -RT ln(K)

, where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin.

Step 2: For part (a), determine how much ∆G° must change to increase the equilibrium constant (K) by a factor of 10. Use the equation

ln(K_2/K_1) = -∆G°_change / RT

, where K₁ = 1 × 10⁻³ and K₂ = 1 × 10⁻². Solve for ∆G°_change.

Step 3: For part (b), if ∆S° = 0 kcal/mol·K, recall the relationship between ∆G°, ∆H°, and ∆S°:

∆G° = ∆H° - T∆S°

. Since ∆S° = 0, the equation simplifies to

∆G° = ∆H°

. Use the result from part (a) to determine how much ∆H° must change.

Step 4: For part (c), if ∆H° = 0 kcal/mol, use the same relationship

∆G° = ∆H° - T∆S°

. Since ∆H° = 0, the equation simplifies to

∆G° = -T∆S°

. Use the result from part (a) to determine how much ∆S° must change.

Step 5: Substitute the known values (R = 8.314 J/mol·K, T = 298 K, and the calculated ∆G°_change from part (a)) into the equations derived in parts (b) and (c) to express the required changes in ∆H° and ∆S°.

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Key Concepts

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

Gibbs Free Energy (∆G°)

Gibbs Free Energy (∆G°) is a thermodynamic potential that indicates the spontaneity of a reaction at constant temperature and pressure. It is calculated using the equation ∆G° = -RT ln(K), where R is the gas constant, T is the temperature in Kelvin, and K is the equilibrium constant. A negative ∆G° indicates a spontaneous reaction, while a positive value suggests non-spontaneity. Understanding how changes in K affect ∆G° is crucial for predicting reaction behavior.

Enthalpy (∆H°) and Entropy (∆S°)

Enthalpy (∆H°) and Entropy (∆S°) are key thermodynamic properties that influence the Gibbs Free Energy of a reaction. Enthalpy represents the heat content of a system, while entropy measures the disorder or randomness. The relationship between these quantities is described by the Gibbs-Helmholtz equation: ∆G° = ∆H° - T∆S°. Changes in either ∆H° or ∆S° can significantly impact the equilibrium constant and the spontaneity of a reaction.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It is temperature-dependent and reflects the extent to which a reaction proceeds. A higher K value indicates a greater concentration of products at equilibrium, while a lower K suggests more reactants. Understanding how K relates to thermodynamic parameters like ∆G°, ∆H°, and ∆S° is essential for analyzing reaction conditions and predicting shifts in equilibrium.

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Draw the structure of a compound with molecular C₈H₁₄ that reacts with one equivalent of H₂ over Pd/C to form a meso compound

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Given that the free energy of the twist-boat conformer of cyclohexane is 5.3 kcal/mol greater than that of the chair conformer, calculate the percentage of twist-boat conformers present in a sample of cyclohexane at 25 °C. Does your answer agree with the statement made in Section 3.13 about the relative number of molecules in these two conformations?

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Using curved arrows, show the mechanism of the following reaction:

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Textbook Question

From the following rate constants, determined at five temperatures, calculate the experimental energy of activation and ∆G^‡, ∆H^‡, and ∆S^‡ for the reaction at 30 °C:

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Textbook Question

a. What is the equilibrium constant for a reaction that is carried out at 25 °C (298 K) with ∆H° = 20 kcal/mol and ∆S° = 5.0 × 10^-2 kcal mol^-1 K^-1?

b. What is the equilibrium constant for the same reaction carried out at 125 °C?

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For a reaction carried out at 25 °C with an equilibrium constant of 1 × 10-3, to increase the equilibrium constant by a factor of 10:a. how much must ∆G° change?b. how much must ∆H° change if ∆S° = 0 kcal mol-1 K-1?c. how much must ∆S° change if ∆H° = 0 kcal mol-1?

Verified video answer for a similar problem:

Key Concepts

Gibbs Free Energy (∆G°)

Enthalpy (∆H°) and Entropy (∆S°)

Equilibrium Constant (K)

For a reaction carried out at 25 °C with an equilibrium constant of 1 × 10^-3, to increase the equilibrium constant by a factor of 10:
a. how much must ∆G° change?
b. how much must ∆H° change if ∆S° = 0 kcal mol^-1K^-1?
c. how much must ∆S° change if ∆H° = 0 kcal mol^-1?