Looking ahead In Chapter 5, we explain that the equilibrium constant ( K eq ) for a reaction can be calculated based on the difference in energy between reactants and products, according to the following equation: K e q = e − Δ E R T K_{eq}=e^{-\frac{\Delta E}{RT}} Using this equation, calculate the equilibrium constant for the 'reaction' shown. [ For the rest of the book, if not otherwise specified, assume room temperature (298K). ]

Hello everyone. Today we have the following problem. The equilibrium constant of a reaction can be calculated using the following equation that we have the equilibrium constant is equal to E race with the power of negative. And that is the change in the energy of our product subtracting the energy of our reactant divided by R which is a gas constant, which is multiplied by the temperature in Kelvin. And so we are being asked to calculate the equilibrium constant for the following reaction at room temperature which is 298 degrees Kelvin. Then we have the following answer choices A through D with a being six B being 60.6 C being 6000 and D being 6.66. And so we will use that exact equation to do so. So we would just rewrite it. We have E which is a variable we then have that it is raised to the power of negative. Now, our change in our energy will come from our strain energy which is given by our eclipsed group here and it is three kg per mole. We have to however convert this into cows per mole or calories, per mole. And we can do that by using the following conversion, that one kg calorie is equal to 10 to the third calories. And when we cancel our units for kilocalories, we arrive at an answer of 3000 calories per mole. And so we will use that for our change in energy. So back to our equation, we have negative 3000 calories from all, we will divide that by our gas constant, which is 1.987 calories that is divided by degrees of Kelvin multiplied by moles. And then we will multiply that by the temperature which is 298 Kelvin. Now to simplify this, we would have e raise with the power of negative calories per mole, which would be divided by 592 0. to six calories per mole because our degrees Kelvin were canceled out in the previous equation. Simplifying this even more. You would get e race of the power of negative 5. as our degrees of calories and moles would cancel out and plugging this into our calculator. We would get a final equilibrium constant of 0.6 as our final answer or answer choice B And with that, we've answered the question overall, I hope this helped. And until next time

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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

Gibbs Free Energy

Organic Chemistry

6. Thermodynamics and Kinetics

Gibbs Free Energy

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2:48 minutes

Problem 64

Textbook Question

Looking ahead In Chapter 5, we explain that the equilibrium constant (K_eq) for a reaction can be calculated based on the difference in energy between reactants and products, according to the following equation:
$K_{e q} = e^{- \frac{Δ E}{R T}}$
Using this equation, calculate the equilibrium constant for the 'reaction' shown. [For the rest of the book, if not otherwise specified, assume room temperature (298K).]

Verified step by step guidance

Step 1: Understand the equation provided. The equilibrium constant (Kₑ_q) is calculated using the formula:

Kₑ_q = e ^ (-∆E / RT)

, where ∆E is the energy difference between reactants and products, R is the gas constant, and T is the temperature in Kelvin.

Step 2: Identify the values needed for the calculation. ∆E (energy difference) should be provided or calculated based on the reaction. R, the gas constant, is typically 8.314 J/(mol·K), and T is given as room temperature, 298 K.

Step 3: Substitute the values into the equation. Replace ∆E with the energy difference, R with 8.314 J/(mol·K), and T with 298 K in the formula:

Kₑ_q = e ^ (-∆E / (8.314 × 298))

Step 4: Simplify the exponent. Calculate the value of

-∆E / (8.314 × 298)

to determine the exponent for the base e.

Step 5: Compute the final value of Kₑ_q by evaluating the exponential function

e ^ (-∆E / RT)

. This will give the equilibrium constant for the reaction.

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Key Concepts

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

Equilibrium Constant (K_eq)

The equilibrium constant (K_eq) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction. It provides insight into the extent to which a reaction proceeds, with values greater than 1 indicating a tendency towards products, and values less than 1 indicating a preference for reactants.

Gibbs Free Energy (∆E)

Gibbs free energy (∆E) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. In the context of chemical reactions, a negative change in Gibbs free energy indicates that the reaction is spontaneous, while a positive change suggests non-spontaneity.

Temperature (T) and the Gas Constant (R)

Temperature (T) in the equation is typically expressed in Kelvin, and it influences the kinetic energy of molecules, affecting reaction rates and equilibria. The gas constant (R) is a proportionality constant that relates energy scales in thermodynamics, with a value of 8.314 J/(mol·K). Together, T and R are crucial for calculating K_eq using the relationship with Gibbs free energy.

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Compounds containing deuterium (D = ²H) are useful for kinetic studies and metabolic studies with new pharmaceuticals. One way to introduce deuterium is by using the reagent LiAlD₄, equivalent in reactivity to LiAlH₄. Show how to make these deuterium-labeled compounds, using LiAlD₄ and D₂O as your sources of deuterium, and any non-deuterated starting materials you wish.

a. CH₃CHDOH

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Textbook Question

The following reaction has a value of ΔG= -2.1 kJ/mol (-0.50 kcal/mol). CH3Br + H2S ⇌ CH3SH + HBra. Calculate Keq at room temperature (25 °C) for this reaction as written.

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If ΔG° for a particular reaction is < 0 kcal/mol, which side of the reaction is favored?

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If a particular acid/base reaction has a K_eq = 10²⁷, how will ΔG° vary with temperature?

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Textbook Question

The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).

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a. Calculate K_eq at room temperature (25 °C) for this reaction as written.

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Textbook Question

The following reaction has a value of ΔG° = –2.1 kJ/mol (–0.50 kcal/mol).

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b. Starting with a 1 M solution of CH3Br and H2S, calculate the final concentrations of all four species at equilibrium.

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Textbook Question

a. Which reaction has a greater equilibrium constant: one with a rate constant of 1 × 10^-3 sec^-1 for the forward reaction and a rate constant of 1 × 10^-5sec-1 for the reverse reaction, or one with a rate constant of 1 × 10^-2 sec^-1 for the forward reaction and a rate constant of 1 × 10^-3 sec^-1 for the reverse reaction?

b. If both reactions start with a reactant concentration of 1.0 M, which reaction will form the most product when the reactions have reached equilibrium?

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Textbook Question

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:

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b. How does ∆G° change as T increases?

c. How does K_eq change as T increases?

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