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Ch.18 - Free Energy and Thermodynamics
Chapter 18, Problem 96

The values of ΔG°f for the hydrogen halides become less negative with increasing atomic number. The ΔG°f of HI is slightly positive. However, the trend in ΔS°f is to become more positive with increasing atomic number. Explain.

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Key Concepts

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

Gibbs Free Energy (ΔGf°)

Gibbs Free Energy (ΔGf°) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. A negative ΔGf° indicates that a substance is thermodynamically stable and favors formation, while a positive value suggests instability. The trend of ΔGf° becoming less negative for hydrogen halides as atomic number increases reflects the increasing stability of heavier halides.
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Entropy (ΔSf°)

Entropy (ΔSf°) is a measure of the disorder or randomness in a system. In the context of hydrogen halides, the trend of ΔSf° becoming more positive with increasing atomic number indicates that the heavier halides have greater molecular complexity and thus higher degrees of freedom, leading to increased disorder. This increase in entropy contributes to the overall thermodynamic stability of the heavier halides despite their less negative Gibbs Free Energy.
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Trends in Halides

The trends in the properties of hydrogen halides, such as ΔGf° and ΔSf°, are influenced by atomic size and bond strength. As the atomic number increases from fluorine to iodine, the bond lengths increase, and bond strengths generally decrease, affecting the stability and formation energies of the compounds. This interplay between bond characteristics and molecular complexity explains the observed trends in Gibbs Free Energy and entropy for these halides.
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Related Practice
Textbook Question

These reactions are important in catalytic converters in automobiles. Calculate ΔG° for each at 298 K. Predict the effect of increasing temperature on the magnitude of ΔG°.

a. 2 CO(g) + 2 NO(g) → N2(g) + 2 CO2(g)

b. 5 H2(g) + 2 NO(g) → 2 NH3(g) + 2 H2O(g)

c. 2 H2(g) + 2 NO(g) → N2(g) + 2 H2O(g)

d. 2 NH3(g) + 2 O2(g) → N2O(g) + 3 H2O(g)

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Textbook Question

Calculate ΔG° at 298 K for these reactions and predict the effect on ΔG° of lowering the temperature.

a. NH3(g) + HBr(g) → NH4Br(s)

b. CaCO3(s) → CaO(s) + CO2(g)

c. CH4(g) + 3 Cl2(g) → CHCl3(g) + 3 HCl(g) (ΔG°f for CHCl3(g) is -70.4 kJ/mol.)

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Textbook Question

All the oxides of nitrogen have positive values of ΔG°f at 298 K, but only one common oxide of nitrogen has a positive ΔS°f. Identify that oxide of nitrogen without reference to thermodynamic data and explain.

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Open Question

Consider the reaction X2(g) → 2X(g). When a vessel initially containing 755 torr of X2 comes to equilibrium at 298 K, the equilibrium partial pressure of X is 103 torr. The same reaction is repeated with an initial partial pressure of 748 torr of X2 at 755 K; the equilibrium partial pressure of X is 532 torr. Find ΔH° for the reaction.

Textbook Question

Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g) → 2 NO2(g) ΔH°rxn = 55.3 kJ At 298 K, a reaction vessel initially contains 0.100 atm of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2. What percentage of N2O4 decomposes at 388 K? Assume that the initial pressure of N2O4 is the same (0.100 atm).

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Open Question
Is the sign of ΔSuniv for each process positive or negative? Explain for the following: b. the electrolysis of H2O(l) to H2(g) and O2(g) at 298 K c. the growth of an oak tree from a little acorn.