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 22

Chapter 6, Problem 22

a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01 kcal mol^-1 K^-1, calculate the ∆G° and the equilibrium constant at:
1. 30 °C and 2. 150 °C.
b. How does ∆G° change as T increases?
c. How does K_eq change as T increases?

Verified step by step guidance

Step 1: Recall the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Here, ΔH° is the enthalpy change, ΔS° is the entropy change, and T is the temperature in Kelvin. Convert the given temperatures (30 °C and 150 °C) to Kelvin using the formula T(K) = T(°C) + 273.

Step 2: For part (a), calculate ΔG° at each temperature using the Gibbs free energy equation. Substitute the values of ΔH° (-12 kcal/mol), ΔS° (0.01 kcal/mol·K), and the respective temperatures in Kelvin into the equation. Perform the subtraction to find ΔG° for each temperature.

Step 3: To calculate the equilibrium constant (Keq), use the relationship between ΔG° and Keq: ΔG° = -RT ln(Keq), where R is the gas constant (0.001987 kcal/mol·K). Rearrange the equation to solve for Keq: Keq = e^(-ΔG°/RT). Substitute the values of ΔG°, R, and T for each temperature to find Keq.

Step 4: For part (b), analyze how ΔG° changes as T increases. Since ΔG° = ΔH° - TΔS°, observe that the term TΔS° becomes larger as T increases. If ΔS° is positive, the magnitude of ΔG° decreases (becomes less negative or more positive) as T increases.

Step 5: For part (c), analyze how Keq changes as T increases. Since Keq = e^(-ΔG°/RT), if ΔG° becomes less negative or more positive as T increases, the exponent (-ΔG°/RT) becomes smaller, leading to a decrease in Keq. Conversely, if ΔG° becomes more negative, Keq increases.

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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 measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It is calculated using the equation ∆G° = ∆H° - T∆S°, where ∆H° is the change in enthalpy, T is the temperature in Kelvin, and ∆S° is the change in entropy. A negative ∆G° indicates a spontaneous reaction, while a positive value suggests non-spontaneity.

Equilibrium Constant (Keq)

The equilibrium constant (Keq) quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible reaction. It is related to Gibbs Free Energy by the equation ∆G° = -RT ln(Keq), where R is the universal gas constant and T is the temperature in Kelvin. Changes in temperature can affect Keq, as they influence the position of equilibrium and the favorability of the reaction.

Temperature Dependence of ∆G° and Keq

The temperature dependence of Gibbs Free Energy (∆G°) and the equilibrium constant (Keq) is significant in thermodynamics. As temperature increases, the term T∆S° becomes more influential in the Gibbs Free Energy equation, potentially leading to a decrease in ∆G° if ∆S° is positive. Consequently, an increase in temperature can shift the equilibrium constant (Keq), favoring the formation of products or reactants depending on the sign of ∆H° and ∆S°.

a. For a reaction with ∆H° = -12 kcal/mol and ∆S° = 0.01 kcal mol-1 K-1, calculate the ∆G° and the equilibrium constant at: 1. 30 °C and 2. 150 °C. b. How does ∆G° change as T increases? c. How does Keq change as T increases?