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Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Short Answer

Expert verified
a. For \(H_2(g)+\frac{1}{2}O_2(g) \rightarrow H_2O(l)\), ΔS° is negative. ΔS° = [S°(H2O(l))] - [S°(H2(g)) + 1/2 S°(O2(g))] b. For \(2 CH_3OH(g) + 3 O_2(g) \rightarrow 2 CO_2(g) + 4 H_2O(g)\), ΔS° is positive. ΔS° = [2S°(CO2(g)) + 4S°(H2O(g))] - [2S°(CH3OH(g)) + 3S°(O2(g))] c. For \(HCl(g) \rightarrow H^+(aq) + Cl^-(aq)\), ΔS° is negative. ΔS° = [S°(H+(aq)) + S°(Cl^-(aq))] - [S°(HCl(g))]

Step by step solution

01

Predict the sign of ΔS°

For each reaction, we will analyze the change in the number of particles and the phase change (if any). If the reaction results in an increase in the number of particles or a change in phase from a more ordered state to a less ordered state (e.g., solid to gas), it will have a positive ΔS°. In contrast, if the reaction leads to a decrease in the number of particles or a change in phase from a less ordered state to a more ordered state (e.g., gas to solid), it will have a negative ΔS°. a. H2(g) + 1/2 O2(g) -> H2O(l) There is a net decrease in gas molecules from the reactants to products, resulting in a decrease in randomness but no phase change. Hence, ΔS° should be negative. b. 2 CH3OH(g) + 3 O2(g) -> 2 CO2(g) + 4 H2O(g) There is an increase in gas molecules from the reactants to products, which results in an increase in randomness. ΔS° should be positive. c. HCl(g) -> H+(aq) + Cl^-(aq) Here, there is no change in the number of particles, but there is a change in the phase from gas to aqueous, which leads to a more ordered state. Thus, ΔS° should be negative.
02

Calculate ΔS° using standard entropy values

To compute ΔS° for each reaction, we will apply the formula: ΔS° = Σ [entropy of products] - Σ [entropy of reactants] For each substance, we will use its given standard entropy value (S°) from reference tables. These entropy values are usually in the units of J/mol K. a. H2(g) + 1/2 O2(g) -> H2O(l) ΔS° = [S°(H2O(l))] - [S°(H2(g)) + 1/2 S°(O2(g))] b. 2 CH3OH(g) + 3 O2(g) -> 2 CO2(g) + 4 H2O(g) ΔS° = [2S°(CO2(g)) + 4S°(H2O(g))] - [2S°(CH3OH(g)) + 3S°(O2(g))] c. HCl(g) -> H+(aq) + Cl^-(aq) ΔS° = [S°(H+(aq)) + S°(Cl^-(aq))] - [S°(HCl(g))] Calculate the values using the standard entropy values from reference tables for each substance and obtain the ΔS° for each reaction.

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Key Concepts

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

Reaction Spontaneity
Understanding reaction spontaneity is crucial for predicting whether a chemical reaction will occur on its own. A reaction is considered spontaneous if it proceeds in the forward direction naturally without needing additional energy input. One of the factors affecting spontaneity is the change in entropy, denoted as \(\Delta S^{\circ}\). In thermodynamics, we've learned that spontaneous processes often involve an increase in the system's disorder or randomness.
To predict reaction spontaneity, \(\Delta G\) or the Gibbs free energy change is often used. The relationship between Gibbs free energy, enthalpy (\(\Delta H\)), and entropy (\(\Delta S\)) is described by the equation: \[ \Delta G = \Delta H - T \Delta S \] where \(T\) is the temperature in Kelvin.
The sign of \(\Delta G\) tells us about spontaneity:
  • If \(\Delta G < 0\), the reaction is spontaneous.
  • If \(\Delta G > 0\), the reaction is non-spontaneous.
  • If \(\Delta G = 0\), the system is in equilibrium.
Changes in entropy often play a significant role when assessing the spontaneity of a reaction. A positive \(\Delta S^{\circ}\) can make a reaction more likely to be spontaneous, especially at higher temperatures, by making \(\Delta G\) more negative.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. In chemistry, it helps explain what governs the direction and extent of reactions. A key part of thermodynamics is understanding how energy changes during reactions.

Key principles of thermodynamics related to chemical reactions include:
  • **First Law of Thermodynamics**: Energy cannot be created or destroyed, only transformed.
  • **Second Law of Thermodynamics**: For any spontaneous process, the entropy of the universe increases.
  • **Third Law of Thermodynamics**: As the temperature of a system approaches absolute zero, the entropy approaches a constant minimum.

In reaction thermodynamics, both entropy (\(\Delta S\)) and enthalpy (\(\Delta H\)) influence whether a reaction will occur.
When a system transitions from reactants to products, the energy changes indicate as enthalpy change. On the other hand, entropy measures the disorder or randomness in the system. Together, these values dictate the Gibbs Free Energy, which ultimately reveals if a reaction can happen without added energy. This comprehensive understanding allows chemists to manipulate and predict reaction conditions for desired outcomes.
Standard Entropy Values
Standard entropy values, denoted as \(S^{\circ}\), provide a measure of the disorder in a system at standard conditions (1 atm and 25°C or 298 K).
These values are crucial for calculating entropy changes in reactions because they give a baseline to compare the degree of disorder in different substances.

Standard entropy values are usually listed in reference tables and have the units of joules per mole per Kelvin (J/mol·K). With this information, you can compute \(\Delta S^{\circ}\) for a reaction using the formula: \[ \Delta S^{\circ} = \Sigma \text{[entropy of products]} - \Sigma \text{[entropy of reactants]} \]

In practice:
  • For gases, remember that increasing volume generally increases entropy because particles can occupy more space.
  • Entropy values for pure elements in their most stable form at 298 K are generally non-zero due to inherent atomic/personal motion.
  • Liquids and solids typically have lower entropy values than gases due to their restricted molecular motion.
These standard entropy values allow chemists to quantitatively assess the changes in disorder as reactions proceed, directly influencing predictions about a reaction's spontaneity.

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Most popular questions from this chapter

Describe how the following changes affect the positional probability of a substance. a. increase in volume of a gas at constant T b. increase in temperature of a gas at constant V c. increase in pressure of a gas at constant T

For a liquid, which would you expect to be larger, \(\Delta S_{\text { fusion }}\) or \(\Delta S_{\text { evaporation }} ?\) Why?

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol} )\) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density \(=0.879 \mathrm{g} / \mathrm{cm}^{3} )\) The elevated boiling point of the solution was 355.4 \(\mathrm{K}\) . What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 \(\mathrm{kJ} / \mathrm{mol}\) , the entropy of vaporization is 95.95 \(\mathrm{J} / \mathrm{K} \cdot\) mol, and the boiling- point elevation constant is 2.5 \(\mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\) .

Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g) . \)\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=?\( \)\mathrm{Fe}(\mathrm{CO})_{5}(l) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) \quad \Delta S^{\circ}=107 \mathrm{J} / \mathrm{K}\( \)\mathrm{Fe}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) \quad \Delta S^{\circ}=-677 \mathrm{J} / \mathrm{K}$

The third law of thermodynamics states that the entropy of a perfect crystal at 0 \(\mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

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