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Is it possible for a reaction to be nonspontaneous yet exothermic? Explain.

Short Answer

Expert verified
Yes, if entropy decreases and temperature is high enough.

Step by step solution

01

Understanding Exothermic Reactions

An exothermic reaction is one where heat is released. This implies that the enthalpy change for the reaction, denoted as \( \Delta H \), is negative. This is because the products of the reaction have lower energy than the reactants.
02

Spontaneity and Gibbs Free Energy

The spontaneity of a reaction at constant temperature and pressure is governed by the Gibbs free energy change, \( \Delta G \). A reaction is spontaneous if \( \Delta G < 0 \). Gibbs free energy is calculated using the formula \( \Delta G = \Delta H - T\Delta S \), where \( T \) is the temperature and \( \Delta S \) is the change in entropy.
03

Examining Reaction Conditions

A reaction can be nonspontaneous (\( \Delta G > 0 \)) if the term \( T\Delta S \) is significant enough and negative to outweigh a negative \( \Delta H \). This would occur if the change in entropy \( \Delta S \) is negative, indicating a decrease in disorder, and the temperature is high enough to make \( T\Delta S > \Delta H \).
04

Conclusion on Possibility

Given this information, it is possible for an exothermic reaction (negative \( \Delta H \)) to be nonspontaneous if it results in a decrease in entropy (negative \( \Delta S \)) and is conducted at a high temperature, making \( \Delta G \) positive.

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Key Concepts

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

Exothermic Reactions
Exothermic reactions are chemical processes that release heat energy into their surroundings. When we talk about a reaction being exothermic, we mean that the change in enthalpy, denoted as \( \Delta H \), is negative. During such reactions, the energy of the products is less than that of the reactants.

Think of a burning match. It's releasing heat, a visible sign of an exothermic process. In any exothermic reaction, like combustion, the system loses energy, making the surroundings warmer.

The negative \( \Delta H \) implies energy is released, not absorbed. This distinguishes exothermic reactions from endothermic ones, which absorb energy from their surroundings.
Reaction Spontaneity
Spontaneity in chemical reactions means whether a process can occur on its own without external influence. It is not solely reliant on heat; instead, it's governed by the Gibbs Free Energy, represented as \( \Delta G \).

For a reaction to be spontaneous at constant temperature and pressure, \( \Delta G \) must be less than zero. But keep in mind, a negative \( \Delta G \) doesn't necessarily mean fast; spontaneity relates to potential, not speed.
  • Spontaneous reactions have \( \Delta G < 0 \).
  • Nonspontaneous reactions have \( \Delta G > 0 \).
Gibbs Free Energy is calculated using the formula: \( \Delta G = \Delta H - T\Delta S \). Both the enthalpy change (\( \Delta H \)) and the entropy change (\( \Delta S \)) influence \( \Delta G \). By understanding these changes, we can predict spontaneity.
Entropy Change
Entropy, symbolized as \( \Delta S \), is a measure of disorder or randomness in a system. When we talk about an increase in entropy, it means that the system has become more disordered.

In a reaction, if \( \Delta S \) is positive, the disorder increases. Conversely, a negative \( \Delta S \) indicates a decrease in disorder.
  • Positive \( \Delta S \): The system becomes more random.
  • Negative \( \Delta S \): The system becomes more ordered.
Entropy change plays a crucial role in determining if a reaction is spontaneous. Even if a reaction is exothermic (negative \( \Delta H \)), a negative and large enough \( \Delta S \) can make it nonspontaneous at high temperatures, as the term \( T\Delta S \) becomes substantial. Understanding these dynamics helps in predicting how a reaction will proceed.

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Most popular questions from this chapter

What does entropy measure?

Hess's law can be used to calculate reaction enthalpies for hypothetical processes that can't be carried out in the laboratory. Set up a Hess's law cycle that will let you calculate \(\Delta H^{\circ}\) for the conversion of methane to ethylene: \(2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+2 \mathrm{H}_{2}(g)\) You can use the following information: \(2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\circ}=-3120.8 \mathrm{~kJ}\) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{\circ}=-890.3 \mathrm{~kJ}\) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) \quad \Delta H^{\circ}=-136.3 \mathrm{~kJ}\) \(\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H_{\mathrm{f}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol}\)

Citric acid has three dissociable hydrogens. When \(5.00 \mathrm{~mL}\) of 0.64 M citric acid and \(45.00 \mathrm{~mL}\) of \(0.77 \mathrm{M} \mathrm{NaOH}\) are mixed at an initial temperature of \(26.0^{\circ} \mathrm{C}\), the temperature rises to \(27.9^{\circ} \mathrm{C}\) as the citric acid is neutralized. The combined mixture has a mass of \(51.6 \mathrm{~g}\) and a specific heat of \(4.0 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\). Assuming that no heat is transferred to the surroundings, calculate the enthalpy change for the reaction of \(1.00 \mathrm{~mol}\) of citric acid in \(\mathrm{kJ}\). Is the reaction exothermic or endothermic?

Indicate the direction of heat transfer between the system and the surroundings, classify the following processes as endo- or exothermic, and give the sign of \(\Delta H^{\circ}\). (a) \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H^{\circ}=+182.6 \mathrm{~kJ}\) (b) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=+483.6 \mathrm{~kJ}\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g) \quad \Delta H^{\circ}=-184.6 \mathrm{~kJ}\)

A piece of dry ice (solid \(\mathrm{CO}_{2}\) ) is placed inside a balloon and the balloon is tied shut. Over time, the carbon dioxide sublimes, causing the balloon to increase in volume. Give the sign of the enthalpy change and the sign of work for the sublimation of \(\mathrm{CO}_{2}\).

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