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 61

Chapter 6, Problem 61

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?

Verified step by step guidance

Step 1: Recall the relationship between the Gibbs free energy change (ΔG°) and the equilibrium constant (K). The equation is ΔG° = -RT ln(K), where R is the gas constant (1.987 cal mol⁻¹ K⁻¹), T is the temperature in Kelvin, and K is the equilibrium constant.

Step 2: Use the thermodynamic relationship ΔG° = ΔH° - TΔS° to calculate ΔG° at each temperature. For part (a), substitute ΔH° = 20 kcal/mol, ΔS° = 5.0 × 10⁻² kcal mol⁻¹ K⁻¹, and T = 298 K into the equation. For part (b), use T = 125 °C, which must first be converted to Kelvin (T = 125 + 273 = 398 K).

Step 3: Convert ΔH° and ΔS° into consistent units if necessary. Since R is given in cal mol⁻¹ K⁻¹, convert ΔH° and ΔS° from kcal to cal by multiplying by 1000 (1 kcal = 1000 cal). This ensures all units are consistent for calculations.

Step 4: Calculate ΔG° for each temperature using the equation ΔG° = ΔH° - TΔS°. Substitute the values for ΔH°, T, and ΔS° into the equation for both temperatures (298 K and 398 K).

Step 5: Rearrange the equation ΔG° = -RT ln(K) to solve for K. Use the calculated ΔG° values for each temperature and substitute them into the equation ln(K) = -ΔG° / (RT). Finally, exponentiate both sides to solve for K: K = e^(-ΔG° / (RT)).

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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 measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. The change in Gibbs Free Energy (∆G) is related to the equilibrium constant (K) of a reaction through the equation ∆G = -RT ln(K), where R is the universal gas constant and T is the temperature in Kelvin. A negative ∆G indicates a spontaneous reaction, while a positive ∆G suggests non-spontaneity.

Equilibrium Constant (K)

The equilibrium constant (K) is a dimensionless value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is derived from the standard Gibbs Free Energy change (∆G°) of the reaction, and it can be calculated using the equation K = e^(-∆G°/RT). The value of K provides insight into the position of equilibrium, indicating whether products or reactants are favored.

Van 't Hoff Equation

The Van 't Hoff equation relates the change in the equilibrium constant (K) of a reaction to the change in temperature (T) and the enthalpy change (∆H°) of the reaction. It is expressed as ln(K2/K1) = -∆H°/R(1/T2 - 1/T1). This equation allows for the calculation of the equilibrium constant at different temperatures, providing a way to understand how temperature influences the position of equilibrium in a chemical reaction.

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Related Practice

Textbook Question

a. Which of the following reactions has the larger ∆S° value?

b. Is the ∆S° value positive or negative?

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Textbook Question

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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Textbook Question

Given the reaction coordinate diagram for the reaction of A to form G, answer the following questions:

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a. How many intermediates are formed in the reaction?

b. Which letters represent transition states?

c. What is the fastest step in the reaction?

d. Which is more stable: A or G?

e. Does A or E form faster from C?

f. Which is the more stable intermediate?

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Textbook Question

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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Textbook Question

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?

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Textbook Question

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

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