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Chapter 18: Problem 53

Consider the following Bronstead acid-base reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{HF}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{HCl}(a q)+\mathrm{F}^{-}(a q) $$. (a) Predict whether \(K\) will be greater or smaller than unity, (b) Does \(\Delta S^{\circ}\) or \(\Delta H^{\circ}\) make a greater contribution to \(\Delta G^{\circ} ?\) (c) Is \(\Delta H^{\circ}\) likely to be positive or negative?

Short Answer

Expert verified
The predicted values are as follows: \(K<1\), indicating the Reaction favors the Reactants; ΔH° makes a greater contribution to ΔG°; ΔH° is likely to be positive, indicating it is an Endothermic Reaction.

Step by step solution

01

Predict whether K will be greater or smaller than unity

The reaction given is a Bronsted Acid-Base Reaction. In Acid-Base Reactions, the Acid donates a proton (H+) and the Base accepts it. In this case, HF is the Acid, because it’s donating an H+. The Cl- is the Base because it’s accepting the H+ to form HCl. On the other hand, we have HCl and F- as the Products. Comparing the Reactants and the Products, HF is a Stronger Acid than HCl, and Cl- is a Weaker Base than F-. This implies that the Reactants are more stable compared to the Products. Therefore, the Reaction will favor the Reactant Side, meaning that \(K<1\).
02

Analyze whether ΔS° or ΔH° makes a greater contribution to ΔG°

We know that Gibbs Free Energy, \(ΔG\) is given by the formula \(ΔG° = ΔH° - TΔS°\). A negative ΔG indicates a Spontaneous Reaction, while a positive ΔG indicates a non-spontaneous Reaction. From Step 1, we’ve determined \(K<1\), and because ΔG and K have a logarithmic Relationship, a smaller K corresponds to a larger (positive) ΔG. Hence, the term -TΔS must be small to give a larger ΔG. Thus, ΔH° makes a greater contribution to ΔG° than ΔS° does.
03

Determining whether ΔH° is likely to be positive or negative

Enthalpy Change (ΔH) is positive when Heat is absorbed (Endothermic Reaction), and negative when Heat is released (Exothermic Reaction). From Step 2, we’ve determined that ΔH° makes a greater contribution to ΔG°. Since ΔG° is positive (as determined in Step 2), to achieve this, ΔH must also be positive. Therefore, ΔH° is likely to be positive, indicating that the Reaction is Endothermic.

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Most popular questions from this chapter

In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce \(\mathrm{Ni}(\mathrm{CO})_{4},\) which is a gas and can therefore be separated from solid impurities: $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ Given that the standard free energies of formation of \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) are \(-137.3 \mathrm{~kJ} / \mathrm{mol}\) and \(-587.4 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the equilibrium constant of the reaction at \(80^{\circ} \mathrm{C}\). Assume that \(\Delta G_{f}^{\circ}\) is temperature independent.

Define entropy. What are the units of entropy?

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\operatorname{Li}(s)\) or \(\operatorname{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g)\) (d) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g)\) (f) \(\mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)

Consider the following facts: Water freezes spontaneously at \(-5^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and ice has a more ordered structure than liquid water. Explain how a spontaneous process can lead to a decrease in entropy.

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)
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