/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 The equilibrium constant \(K\) e... [FREE SOLUTION] | 91Ó°ÊÓ

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The equilibrium constant \(K\) equals 56 for the reaction shown. Complete the following statement so that it correctly describes this reaction. The sign of: A. \(\Delta G\) is \(+\) at equilibrium B. \(\Delta G\) is \(-\) at equilibrium C. \(\Delta G^{\circ}\) is \(+\) D. \(\Delta G^{\circ}\) is -

Short Answer

Expert verified
The correct option is D: \( \Delta G^{\circ} \) is \(-\).

Step by step solution

01

Understanding the Relationship

The standard Gibbs free energy change, \( \Delta G^{\circ} \), is related to the equilibrium constant \( K \) by the equation:\[ \Delta G^{\circ} = -RT \ln K \]where \( R \) is the gas constant \(8.314 \ J/(mol \cdot K)\), and \( T \) is the temperature in Kelvin. A positive \( K \) greater than 1 indicates that the reaction favors products at equilibrium.
02

Analyzing \( \Delta G^{\circ} \)

Since \( K > 1 \), the natural logarithm \( \ln K \) is positive. Thus, \( \Delta G^{\circ} = -RT \ln K \) will be negative because multiplying a positive number by \(-1\) yields a negative number. Therefore, \( \Delta G^{\circ} \) is negative.
03

Understanding \( \Delta G \) at Equilibrium

At equilibrium, the free energy change \( \Delta G \) is always 0. This is because the system is in a state of balance where no net change occurs in the concentrations of reactants and products.
04

Conclusion

From our analysis, \( \Delta G^{\circ} \) is \(-\) and \( \Delta G \) at equilibrium is always \(0\). Therefore, the correct choice is that \( \Delta G^{\circ} \) is \(-\).

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Key Concepts

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

Gibbs Free Energy
Gibbs free energy is a magnificent concept in chemistry that helps us to predict the spontaneity of a reaction. It is a way to determine whether a chemical reaction will take place on its own without any additional energy. The formula for the standard Gibbs free energy change, \( \Delta G^{\circ} \), is
  • \( \Delta G^{\circ} = -RT \ln K \),
where \( R \) is the gas constant and \( T \) is the temperature in Kelvin. This formula ties together the equilibrium constant and the Gibbs free energy. When the reaction equilibrium constant \( K \) is greater than 1, it means that the products are favored in the equilibrium state. This results in \( \Delta G^{\circ} \) being negative. Negative \( \Delta G^{\circ} \) indicates the reaction is spontaneous under standard conditions. On the contrary, if \( K \) is less than 1, \( \Delta G^{\circ} \) would be positive, suggesting the reaction is non-spontaneous under those conditions.
Ultimately, understanding the sign of \( \Delta G^{\circ} \) helps in predicting the direction and feasibility of chemical reactions under set conditions.
Equilibrium Constant
The equilibrium constant, \( K \), is a crucial aspect that determines the equilibrium position of a chemical reaction. It is a ratio of the concentrations of products over reactants, each raised to the power of their stoichiometric coefficients. It provides insight into which direction a reaction must shift to reach equilibrium.
  • When \( K > 1 \), it indicates that the products are favored.
  • If \( K < 1 \), it means reactants are favored.
  • And when \( K = 1 \), it signifies a state where neither products nor reactants are favored.
In this particular exercise, with \( K = 56 \), it shows a strong tendency towards forming products. This high \( K \) value signifies that at equilibrium, the concentration of products is much higher compared to reactants. This directly affects the standard Gibbs free energy change \( \Delta G^{\circ} \). Understanding \( K \) is essential for predicting how a change in conditions might shift the equilibrium position.
Reaction Equilibrium
Understanding reaction equilibrium is fundamental to grasp how chemical reactions behave. At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This means that there is no net change in the concentration of reactants and products over time. Importantly, at equilibrium, the Gibbs free energy change \( \Delta G \) is always zero.
This zero value signifies that the system is in a state of minimal Gibbs free energy, meaning it is energetically stable. It does not favor moving spontaneously in either direction because any shift either way will require an energy input. Equilibrium does not mean that the reactants and products are present in equal amounts, but rather their proportions remain constant. At this point, monitoring \( \Delta G^{\circ} \) and \( K \) helps in understanding if a change in conditions (like temperature or pressure) would cause the system to move away from equilibrium and towards the formation of either more products or reactants.

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Most popular questions from this chapter

What combination of ester and Grignard reagent could you use to prepare each of the following tertiary alcohols?

Compound A serves as a prodrug for the analgesic benzocaine. (A prodrug is a pharmacologically inactive compound that is converted in the body to an active drug, usually by a metabolic transformation.) The enzyme amidase catalyzes the hydrolysis of compound A into benzocaine. Write the structures of the possible products of A that might be formed by hydrolysis in aqueous \(\mathrm{HCl}\).

In a similar experiment, unlabeled 4-butanolide was allowed to stand in an acidic solution in which the water had been labeled with \({ }^{18} \mathrm{O}\). When the lactone was extracted from the solution after four days, it was found to contain \({ }^{18} \mathrm{O} .\) Which oxygen of the lactone do you think became isotopically labeled?

(a) Unlike other esters which react with Grignard reagents to give tertiary alcohols, ethyl formate \(\left(\mathrm{HCOCH}_{2} \mathrm{CH}_{3}\right)\) yields a different class of alcohols on treatment with Grignard reagents. What kind of alcohol is formed in this case and why? (b) Diethyl carbonate \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCOCH}_{2} \mathrm{CH}_{3}\right)\) reacts with excess Grignard reagent to yield alcohols of a particular type. What is the structural feature that characterizes alcohols prepared in this way?

Unlike esters, which can be prepared by acid-catalyzed condensation of an alcohol and a carboxylic acid, amides cannot be prepared by an acid-catalyzed condensation of an amine and a carboxylic acid. Why?

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