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Find the sign of \(\Delta S^{\circ}\) for the reaction $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(s) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ The reaction is endothermic and spontaneous at \(25^{\circ} \mathrm{C}\). Explain the spontaneity of the reaction in terms of enthalpy and entropy changes.

Short Answer

Expert verified
\( \Delta S^{\circ} \) is positive, allowing the reaction to be spontaneous despite being endothermic.

Step by step solution

01

Define Key Concepts

To solve this problem, we need to understand how Gibbs free energy (\( \Delta G \)) is related to enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)). The relation is given by the formula \[ \Delta G = \Delta H - T \Delta S \]. For a reaction to be spontaneous at a given temperature, \( \Delta G \) must be negative.
02

Identify Endothermic Reaction

The problem states that the reaction is endothermic and spontaneous. This means \( \Delta H \) is positive because heat is absorbed during the reaction.
03

Understand Requirements for Spontaneity

Since \( \Delta G < 0 \) for spontaneity and \( \Delta H > 0 \) due to the endothermic nature of the reaction, the term \( T \Delta S \) must be large enough to ensure that \( \Delta G \) remains negative. This implies that \( \Delta S \) must be positive and significant.
04

Analyze Reaction Components

In the given reaction, solid \( \mathrm{N}_2\mathrm{O}_5 \) changes to gaseous \( \mathrm{NO}_2 \) and \( \mathrm{O}_2 \). The increase in the number of moles of gas suggests an increase in entropy, as gases have higher entropy than solids.
05

Conclusion on \( \Delta S^{\circ} \) Sign

Given the increase in gaseous products, we conclude that the standard entropy change \( \Delta S^{\circ} \) for the reaction is positive. This large positive entropy change offsets the positive enthalpy and ensures the reaction is spontaneous.

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Key Concepts

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

Endothermic Reaction
An endothermic reaction is one where the system absorbs heat from its surroundings. This process involves a positive change in enthalpy (\( \Delta H \gt 0 \) ) since energy is added to the reactants to convert them into products. Although it might seem counterintuitive, endothermic reactions can still be spontaneous under certain conditions. For example, in the reaction: \[ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(s) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \], the positive enthalpy indicates that the reaction requires energy input. One might wonder how such reactions proceed spontaneously. The answer lies in other factors, such as entropy changes, which play a crucial role in determining spontaneity. In summary, while endothermic reactions absorb heat, their occurrence depends on the interplay between enthalpy and entropy changes.
Entropy Change
Entropy (\( \Delta S \) ) is a measure of the disorder or randomness in a system. In chemical reactions, changes in entropy (\( \Delta S \) ) occur when there is a change in the number of gas molecules, as gases are more disordered compared to solids. In the given reaction, solid \( \mathrm{N}_{2}\mathrm{O}_{5} \) transforms into gaseous \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \).Such a transformation leads to a significant increase in entropy because:
  • Gases have higher entropy than solids due to their greater freedom of movement.
  • The number of moles of gas increases, indicating higher disorder.
A large positive entropy change (\( \Delta S^{\circ} \gt 0 \) ) helps offset the positive enthalpy change in the reaction. This increase contributes to making the reaction spontaneous at room temperature, explained further through Gibbs free energy.
Spontaneity Criterion
The spontaneity criterion for a chemical reaction is determined through Gibbs free energy (\( \Delta G \)), a thermodynamic potential that predicts whether a process will occur without outside intervention. The equation: \[ \Delta G = \Delta H - T \Delta S \] represents the relationship between Gibbs free energy, enthalpy (\( \Delta H \)), and entropy (\( \Delta S \) ). For a reaction to be spontaneous, \( \Delta G \) must be negative (\( \Delta G \lt 0 \)).Notably, in the discussed endothermic reaction:
  • Enthalpy (\( \Delta H \)) is positive, suggesting the input of heat.
  • Entropy (\( \Delta S \)) is positive, facilitating a distribution towards greater disorder.
  • The temperature (\( T \)) is a crucial factor, as higher temperatures promote spontaneous endothermic reactions.
The positive entropy change is sufficiently large, multiplied by temperature (\( T \Delta S \)), such that it outweighs the positive \( \Delta H \), ensuring \( \Delta G \) becomes negative. This explanation clarifies why some reactions that might not intuitively seem spontaneous, such as endothermic ones, actually can be under the right temperature conditions.

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Most popular questions from this chapter

The free energy of formation of one mole of compound refers to a particular chemical equation. For each of the following, write that equation. a. \(\operatorname{KBr}(s)\) b. \(\mathrm{CH}_{3} \mathrm{Cl}(l)\) c. \(\mathrm{H}_{2} \mathrm{~S}(g)\) d. \(\mathrm{AsH}_{3}(g)\)

Silver carbonate, \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\), is a light yellow compound that decomposes when heated to give silver oxide and carbon dioxide: $$ \mathrm{Ag}_{2} \mathrm{CO}_{3}(s) \rightleftharpoons \mathrm{Ag}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) $$ A researcher measured the partial pressure of carbon dioxide over a sample of silver carbonate at \(220^{\circ} \mathrm{C}\) and found that it was \(1.37\) atm. Calculate the partial pressure of carbon dioxide at \(25^{\circ} \mathrm{C}\). The standard enthalpies of formation of silver carbonate and silver oxide at \(25^{\circ} \mathrm{C}\) are \(-505.9 \mathrm{~kJ} / \mathrm{mol}\) and \(-31.05 \mathrm{~kJ} / \mathrm{mol}\), respectively. Make any reasonable assumptions in your calculations. State the assumptions that you make, and note why you think they are reasonable.

Describe what you would look for in a reaction involving gases in order to predict the sign of \(\Delta S^{\circ}\). Explain.

What is the useful work obtained in the ideal situation in which a chemical reaction with free-energy change \(\Delta G\) is run so that it produces no entropy?

Hydrogen gas and iodine vapor react to produce hydrogen iodide gas: $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ Calculate the free-energy change \(\Delta G\) for the following two conditions, at \(25^{\circ} \mathrm{C}\). Which one is closer to equilibrium? Explain. a. The partial pressures of \(\mathrm{H}_{2}(g), \mathrm{I}_{2}(g)\), and \(\mathrm{HI}(g)\) are each \(\frac{1}{3}\) atm. b. The partial pressures of \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) are each \(\frac{1}{3} \mathrm{~atm} ;\) the partial pressure of \(\mathrm{HI}(g)\) is \(\frac{1}{90}\) atm.

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