If the \(\Delta G^{\circ}\) for a reaction is \(-4.5 \mathrm{kcal} / \mathrm{mol}\) at \(298 \mathrm{~K}\), what is the \(K_{\text {eq }}\) for this reaction? What is the change in entropy of this reaction if \(\Delta H^{\circ}=-3.2 \mathrm{kcal} / \mathrm{mol}\) ?

Short Answer

Expert verified
Question: Calculate the equilibrium constant (K_eq) and the change in entropy (ΔS°) for a reaction with a given ΔG° of -4.5 kcal/mol and ΔH° of -3.2 kcal/mol at 298 K. Answer: The equilibrium constant (K_eq) for this reaction is approximately 2920, and the change in entropy (ΔS°) is approximately -18.26 J/(mol·K).

Step by step solution

01

Convert the Gibbs Free Energy from \(\mathrm{kcal/mol}\) to \(\mathrm{J/mol}\)

Given \(\Delta G^\circ=-4.5\mathrm{kcal/mol}\), we know that 1 kcal = 1000 cal and 1 cal = 4.184 J, so we need to convert it into \(\mathrm{J/mol}\) by multiplying it with 1000 and 4.184: \[\Delta G^\circ=-4.5\mathrm{kcal/mol} \times 1000 \times 4.184 \mathrm{J/kcal}=-18828\mathrm{~J/mol}\]
02

Calculate the equilibrium constant \(K_{eq}\)

Remember the formula of Gibbs Free Energy: \(\Delta G^\circ=-RT \ln K_{eq}\). Now that we know \(\Delta G^\circ\), and the temperature \(T=298\mathrm{~K}\), we can calculate \(K_{eq}\). The gas constant \(R \approx 8.314 \mathrm{J/(mol~K)}\), thus \[-18828=-8.314 \times 298 \times \ln K_{eq}\] Now, we can solve for \(K_{eq}\): \[K_{eq}=\exp{\left(\frac{-18828}{8.314\times 298}\right)}\approx \exp{7.98} \approx2920\]
03

Calculate the change in entropy \(\Delta S^\circ\)

We know our Gibbs Free Energy formula \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\) and given \(\Delta H^\circ=-3.2\mathrm{kcal/mol}\), we can solve for \(\Delta S^\circ\). First, convert the enthalpy value to \(\mathrm{J/mol}\): \[\Delta H^\circ=-3.2\mathrm{kcal/mol} \times 1000 \times 4.184=-13388.8 \mathrm{J/mol}\] Now plug in the values of \(\Delta G^\circ\), \(\Delta H^\circ\), and \(T\) into \(\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ\): \[-18828=-13388.8-298\Delta S^\circ\] \[\Delta S^\circ=\frac{-18828+13388.8}{298}=\frac{-5439.2}{298}=-18.26 \mathrm{J/(mol~K)}\] So, the equilibrium constant \(K_{eq}\) for this reaction is approximately 2920 and the change in entropy \(\Delta S^\circ\) is approximately \(-18.26 \mathrm{J/(mol~K)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, denoted as Keq, is a number that expresses the ratio of the concentrations of the products to the reactants for a reversible chemical reaction at equilibrium. The value of Keq provides insight into the position of the equilibrium. If Keq is much greater than 1, the reaction favors the formation of products. Conversely, if it is much less than 1, the reaction favors the reactants.

Calculating Keq involves using the Gibbs Free Energy change (abla G^ablacircn) of the reaction. The formula that connects these two is abla G^ablacircn = -RT ln Keqn, where R is the universal gas constant and T is the temperature in Kelvin. In our example, the equilibrium constant was found to be approximately 2920, indicating that at 298 K, the products are heavily favored at equilibrium.
Entropy
In thermodynamics, entropy is a measure of the disorder or randomness in a system. Symbolized as S, entropy is an important factor in determining the spontaneity of a process. The second law of thermodynamics states that in an isolated system, entropy tends to increase over time.

Entropy also plays a crucial role in the Gibbs Free Energy formula: abla G^ablacircn = abla H^ablacircn - Tabla S^ablacircn. In this equation, T is the temperature, abla H^ablacircn is the change in enthalpy, and abla S^ablacircn is the change in entropy. A positive abla S^ablacircn value indicates increased disorder, while a negative value indicates increased order. The example provided shows a negative change in entropy, meaning the reaction results in a more ordered state.
Enthalpy
Enthalpy, represented by H, is a measure of the total energy of a thermodynamic system, including internal energy and the product of pressure and volume. It reflects the heat transferred during a chemical reaction at constant pressure. The change in enthalpy (abla H^ablacircn) is an essential component in calculating Gibbs Free Energy.

The formula for Gibbs Free Energy requires knowing the change in enthalpy: abla G^ablacircn = abla H^ablacircn - Tabla S^ablacircn. Here, a negative abla H^ablacircn implies that the reaction releases heat and is exothermic, while a positive abla H^ablacircn means it absorbs heat and is endothermic. Our step-by-step solution reveals an exothermic process with abla H^ablacircn = -13388.8 J/mol, as the reaction releases energy to the surroundings.
Chemical Thermodynamics
Chemical thermodynamics is the area of chemistry that studies the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. Gibbs Free Energy, entropy, and enthalpy are all fundamental concepts within this field.

The Gibbs Free Energy is particularly significant because it predicts the direction of chemical change and the extent to which a reaction will proceed. Our exercise shows the practical application of chemical thermodynamics in determining both the equilibrium constant and the change in entropy, highlighting the interconnectedness of these thermodynamic quantities and their importance in predicting reaction behavior.

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Most popular questions from this chapter

Following is a structural formula for guanidine, the compound by which migratory birds excrete excess metabolic nitrogen. The hydrochloride salt of this compound is a white crystalline powder, freely soluble in water and ethanol. (a) Write a Lewis structure for guanidine showing all valence electrons. (b) Does proton transfer to guanidine occur preferentially to one of its \(-\mathrm{NH}_{2}\) groups \((\) cation A) or to its = NH group (cation B)? Explain.

Explain why the hydronium ion, \(\mathrm{H}_{3} \mathrm{O}^{+}\), is the strongest acid that can exist in aqueous solution. What is the strongest base that can exist in aqueous solution?

Acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH}\), is a weak organic acid, \(\mathrm{p} K_{\mathrm{a}} 4.76\). Write an equation for the equilibrium reaction of acetic acid with each base. Which equilibria lie considerably toward the left? Which lie considerably toward the right? (a) \(\mathrm{NaHCO}_{3}\) (b) \(\mathrm{NH}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{NaOH}\)

Methyl isocyanate, \(\mathrm{CH}_{3}-\mathrm{N}=\mathrm{C}=\mathrm{O}\), is used in the industrial synthesis of a type of pesticide and herbicide known as a carbamate. As a historical note, an industrial accident in Bhopal, India, in 1984 resulted in leakage of an unknown quantity of this chemical into the air. An estimated 200,000 persons were exposed to its vapors, and over 2000 of these people died. (a) Write a Lewis structure for methyl isocyanate, and predict its bond angles. What is the hybridization of its carbonyl carbon? Of its nitrogen atom? (b) Methyl isocyanate reacts with strong acids, such as sulfuric acid, to form a cation. Will this molecule undergo protonation more readily on its oxygen or nitrogen atom? In considering contributing structures to each hybrid, do not consider structures in which more than one atom has an incomplete octet.

For each value of \(K_{\mathrm{a}}\), calculate the corresponding value of \(\mathrm{p} K_{\mathrm{a}}\). Which compound is the stronger acid? (a) Acetic acid, \(K_{\mathrm{a}}=1.74 \times 10^{-5}\) (b) Chloroacetic acid, \(K_{\mathrm{a}}=1.38 \times 10^{-3}\)

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