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For which of the following reaction is the standard entropy of reaction \(\Delta \mathrm{S}^{\circ}\) positive? a. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) c. \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) d. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

Short Answer

Expert verified
Reactions b and d have a positive \( \Delta \mathrm{S}^{\circ} \).

Step by step solution

01

Analyze Reaction A

For reaction a. \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(l) \): The reactants are 1 liquid and 3 gases, while the products are 2 gases and 3 liquids. The number of gaseous molecules decreases, which typically implies a decrease in entropy as gases have higher entropy than liquids.
02

Analyze Reaction B

For reaction b. \( 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \): The reactants include 2 gaseous water molecules, and the products include 3 gaseous molecules (2 hydrogen and 1 oxygen). Since there is an increase in the number of gaseous molecules, the entropy should increase, meaning \( \Delta \mathrm{S}^{\circ} \) is positive.
03

Analyze Reaction C

For reaction c. \( \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) \): The number of gas molecules changes from 3 in the reactants to 1 in the product. This decrease in the number of gas molecules usually corresponds to a decrease in entropy.
04

Analyze Reaction D

For reaction d. \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \): Both reactants and products are gases, with 4 moles of gas on the reactant side and 5 moles of gas on the product side. The increase in the number of gaseous moles suggests an increase in entropy.

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Key Concepts

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

Standard Entropy Change (ΔS°)
In the realm of thermodynamics, the standard entropy change, denoted as \( \Delta S^{\circ} \), is a key concept to understand. It measures the change in entropy (or disorder) of a system during a reaction under standard conditions.
Entropy itself is a thermodynamic quantity that represents the unavailability of a system’s energy to do work, often interpreted as the degree of disorder or randomness in the system.
The symbol \( \Delta S^{\circ} \) helps chemists to predict whether a reaction leads to an increase or decrease in entropy.
  • When \( \Delta S^{\circ} \) is positive, it means that the system has become more disordered.
  • Conversely, a negative \( \Delta S^{\circ} \) indicates a decrease in disorder (i.e., the system has become more ordered).

For instance, in reaction b from our original exercise, \( 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \), the standard entropy change would be positive since more gaseous molecules are formed, leading to an increase in randomness.
Gaseous Reaction Products
Gases, due to their inherent nature, possess higher entropy compared to liquids and solids because of their increased molecular motion and free energy.
When analyzing chemical reactions, the number of gaseous products is crucial in determining the overall entropy change.
In general, if the reaction results in more gaseous products than reactants, we anticipate an increase in entropy.
Consider reaction d: \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \). Here, both reactants and products are gases.
The reactant side has 4 moles of gas, while the product side has 5. The increase in the number of gaseous moles signifies higher entropy, indicating a positive \( \Delta S^{\circ} \).
Entropy and Phase State
The phase of a substance greatly influences its entropy.
Gases have the highest entropy compared to liquids and solids because gas particles move freely and are less ordered.
  • Solid state: Highly ordered structure with low entropy.
  • Liquid state: Less ordered than solids with moderate entropy.
  • Gas state: Highly disordered with high entropy.

In the context of our example reactions, this principle is clearly evident in reaction a: \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \).
Here, the transition from gaseous \( \mathrm{O}_{2} \) to liquid \( \mathrm{H}_{2} \mathrm{O} \), along with the reduction in gaseous molecules, typically results in reduced entropy, explaining a negative or less positive \( \Delta\mathrm{S}^{\circ} \). Understanding that gases generally have more entropy helps to intuitively predict the entropy changes as reactions proceed.

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Most popular questions from this chapter

\(4.8 \mathrm{~g}\) of \(\mathrm{C}\) (diamond) on complete combustion evolves \(1584 \mathrm{~kJ}\) of heat. The standard heat of formation of gaseous carbon is \(725 \mathrm{~kJ} / \mathrm{mol}\). The energy required for the process (i) \(\mathrm{C}\) (graphite) \(\rightarrow \mathrm{C}\) (gas) (ii) \(\mathrm{C}\) (diamond) \(\rightarrow \mathrm{C}\) (gas) are a. 725,723 b. 725,727 c. 727,725 d. None of these

Match the following: List I List II 1\. Exothermic substance (i) Less stable 2\. Endothermic substance (ii) More stable 3\. Heat of formation (iii) Exothermic (iv) May be exothermic 4\. Heat of combustion or endothermic (v) Endothermic

Consider the reaction: \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{~F}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{NF}_{3}(\mathrm{~g})\) \(\Delta \mathrm{H}^{\circ}=-249 \mathrm{~kJ}\) and \(\Delta \mathrm{S}^{\circ}=-278 \mathrm{JK}^{-1}\) at \(25^{\circ} \mathrm{C}\) Calculate \(\Delta \mathrm{G}^{\circ}\) and state whether the equilibrium composition should favour reactants or products at standard conditions. a. \(\Delta \mathrm{G}^{\circ}=-166 \mathrm{~kJ} ;\) the equilibrium composition should favour reactants. b. \(\Delta \mathrm{G}^{\circ}=-166 \mathrm{~kJ} ;\) the equilibrium composition should favour products. c. \(\Delta \mathrm{G}^{\circ}=-322 \mathrm{~kJ}\); the equilibrium composition should favour products. d. \(\Delta G^{\circ}=-166 \mathrm{~kJ} ;\) the equilibrium composition should favour reactants.

The signs of \(\Delta \mathrm{G}, \Delta \mathrm{H}\) and \(\Delta \mathrm{S}\) at \(25^{\circ} \mathrm{C}\) are shown below for three reactions. Reaction \(\Delta \mathrm{G}\) \(\Delta \mathrm{H} \quad \Delta \mathrm{S}\) I. \(+\quad+\) II. \(+\) III. - - - Which reaction could go in the reverse direction at high temperature? a. II only b. III only c. I and II d. II and III

Which of the following is in accord with the three laws of thermodynamics? (I) The entropy of the universe never decreases. (II) A non-spontaneous process can occur if it is coupled to a process that is sufficiently spontaneous. a. I is true and II is false. b. I is false and II is true. c. Both I and II are true. d. Both I and II are false.

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