Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. \(\operatorname{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Short Answer

Expert verified
Reactions b, c, and d are expected to have a negative value for ΔS° due to the following reasons: (b) the number of gas molecules decreases and a solid forms from gas molecules, (c) the number of gas molecules decreases and no solid and/or liquid forms, and (d) the number of gas molecules decreases, and a liquid forms from gas molecules.

Step by step solution

01

Reaction (a) analysis

a. SiF6(aq) + H2(g) -> 2 HF(g) + SiF4(g) We have one gas reactant and two gas products in this reaction. The number of gas molecules increases (from 1 to 3), and no solid and/or liquid forms. Therefore, this reaction is less likely to have a negative ΔS°.
02

Reaction (b) analysis

b. 4 Al(s) + 3 O2(g) -> 2 Al2O3(s) In this reaction, three gas molecules react to form solid products, and the number of gas molecules decreases (from 3 to 0). Since the number of gas molecules decreases and a solid forms from gas molecules, this reaction is more likely to have a negative ΔS°.
03

Reaction (c) analysis

c. CO(g) + Cl2(g) -> COCl2(g) We have two gas reactants and one gas product in this reaction. The number of gas molecules decrease (from 2 to 1) and no solid and/or liquid forms, so this reaction is more likely to have a negative ΔS°.
04

Reaction (d) analysis

d. C2H4(g) + H2O(l) -> C2H5OH(l) In this reaction, we have one gas reactant and a gas and liquid reactant forming a liquid product. The number of gas molecules decreases (from 1 to 0), and a liquid forms from gas molecules. Therefore, this reaction is more likely to have a negative ΔS°.
05

Reaction (e) analysis

e. H2O(s) -> H2O(l) In this process, we have a solid turning into a liquid. The system is becoming more disordered, so this process is less likely to have a negative ΔS°. Based on the analysis of each reaction, reactions b, c, and d are expected to have a negative value for ΔS°.

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. In chemistry, it plays a crucial role in understanding how chemical reactions occur and what factors influence them.

In the context of entropy change in chemical reactions, thermodynamics tells us that systems tend to move towards a state of greater disorder or randomness, which is quantified by a property called entropy (symbolized as S). The change in entropy (ΔS) during a chemical reaction gives us insights into the degree of disorder or randomness of the system before and after the reaction.
Spontaneity of Reactions
The spontaneity of a chemical reaction can often be predicted by the sign of the entropy change (ΔS). A positive value for ΔS indicates that the disorder increases, which usually favors the spontaneity of a reaction. Conversely, a negative ΔS suggests a decrease in disorder, indicating that the reaction may not be spontaneous without some input of energy.

However, entropy alone does not determine the spontaneity of a reaction. It must be considered alongside enthalpy change and temperature to fully understand the likely behavior of a chemical process. This is where the concept of Gibbs free energy comes into play.
Gibbs Free Energy
Gibbs free energy (G) is a thermodynamic quantity that is a very useful predictor for the spontaneity of a chemical reaction. It incorporates both the entropy (S) and enthalpy (H) of a system, as described by the equation G = H - TS, where T is the temperature in Kelvin.

A negative ΔG indicates a spontaneous process, while a positive ΔG suggests a non-spontaneous process. Essentially, for a chemical reaction to be spontaneous at constant temperature and pressure, the Gibbs free energy change must be negative. ΔG becomes more negative either by a decrease in enthalpy (ΔH < 0), an increase in entropy (ΔS > 0), or both.
States of Matter in Chemical Reactions
The states of matter involved in chemical reactions—solid (s), liquid (l), gas (g), and aqueous (aq)—have a significant impact on the entropy change (ΔS). Typically, gases have higher entropy than liquids, which, in turn, have higher entropy than solids.

Hence, reactions in which gases are produced from liquids or solids tend to increase in entropy (ΔS > 0), indicating a more disordered system. Conversely, reactions where gases condense into liquids or solids generally result in a negative entropy change (ΔS < 0), reflecting increased order within the system. This intrinsic relationship between the states of matter and entropy guides chemists in predicting the direction in which a reaction will naturally proceed.

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Most popular questions from this chapter

Consider the following reaction at \(25.0^{\circ} \mathrm{C}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ} / \mathrm{mol}\) and -176.6 \(\text{J/K}\) \cdot mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

For rubidium \(\Delta H_{\text {vap }}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) at \(686^{\circ} \mathrm{C},\) its boiling point. Calculate \(\Delta S^{\circ}, q, w,\) and \(\Delta E\) for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure.

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text {surr}}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K}\) mol at 298 K. Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

From data in Appendix \(4,\) calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for each of the following reactions at \(25^{\circ} \mathrm{C}.\) a. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)\) c. \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(s)\) d. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at 298 K) for this reaction. Would this reaction be favored at a high or low temperature?

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