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Describe how the signs of \(\Delta H\) and \(\Delta S\) determine whether a reaction is spontaneous or nonspontaneous at constant temperature and pressure.

Short Answer

Expert verified
Spontaneity depends on the signs of \(\Delta H\) and \(\Delta S\): negative \(\Delta H\) and positive \(\Delta S\) always favor spontaneity.

Step by step solution

01

Understand Spontaneity Formula

A reaction is spontaneous if the change in Gibbs free energy (\(\Delta G\)) is negative. The formula for Gibbs free energy is \(\Delta G = \Delta H - T\Delta S\) where \(\Delta H\) is the change in enthalpy, \(T\) is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.
02

Case 1: Negative \(\Delta H\) and Positive \(\Delta S\)

When \(\Delta H\) is negative, this indicates an exothermic reaction, and when \(\Delta S\) is positive, it means that the disorder of the system increases. Both of these contribute to a negative \(\Delta G\), making the reaction spontaneous at all temperatures.
03

Case 2: Positive \(\Delta H\) and Negative \(\Delta S\)

If \(\Delta H\) is positive, the reaction is endothermic, and if \(\Delta S\) is negative, the disorder of the system decreases. Both factors contribute to a positive \(\Delta G\), making the reaction nonspontaneous at all temperatures.
04

Case 3: Negative \(\Delta H\) and Negative \(\Delta S\)

When \(\Delta H\) is negative, the reaction is exothermic, which favors spontaneity. However, if \(\Delta S\) is negative, this opposes spontaneity by making \(\Delta G\) less negative or positive as temperature increases. The reaction may be spontaneous at low temperatures but becomes nonspontaneous at high temperatures.
05

Case 4: Positive \(\Delta H\) and Positive \(\Delta S\)

If \(\Delta H\) is positive, meaning the reaction is endothermic, but \(\Delta S\) is positive, increasing entropy favors spontaneity. The reaction might be nonspontaneous at low temperatures but can become spontaneous at high temperatures.

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Key Concepts

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

Enthalpy Change (ΔH)
Enthalpy change, denoted as ΔH, represents the heat absorbed or released during a chemical reaction at constant pressure. It is an essential factor in determining the spontaneity of a reaction.

When ΔH is negative, it indicates that the reaction is exothermic. This means that energy is being released to the surroundings, typically favoring the spontaneity of the reaction. Imagine the chemical reaction as a warm embrace that releases energy—this warmth can drive the reaction to occur more readily.

Conversely, a positive ΔH signifies an endothermic process. Here, energy is absorbed from the surroundings, often making the reaction less spontaneous unless there is a sufficient accompanying increase in disorder within the system. In simple terms, think of an endothermic reaction as feeling a chill that requires absorbing warmth, which needs extra motivation to happen spontaneously.
  • **ΔH < 0:** Exothermic, favors spontaneity.
  • **ΔH > 0:** Endothermic, may hinder spontaneity without other favorable factors.
Entropy Change (ΔS)
Entropy change, ΔS, measures the degree of disorder or randomness in a system. It indicates whether a process leads to increased chaos or order.

A positive ΔS signifies an increase in disorder, making the system more chaotic, which is naturally preferred in the context of entropy and often enhances reaction spontaneity. Think of it as a crowd of people starting to spread out freely; this preference for disorder can drive the process.

On the other hand, a negative ΔS means the system becomes more ordered. This reduction in randomness typically works against reaction spontaneity unless compensated by other factors like heat release. Envision it as trying to arrange that crowd into neat lines—this requires effort, often countering spontaneity.
  • **ΔS > 0:** Increase in disorder, enhances spontaneity.
  • **ΔS < 0:** Decrease in disorder, opposes spontaneity.
Gibbs Free Energy (ΔG)
Gibbs free energy, symbolized as ΔG, is the crux of determining reaction spontaneity at constant temperature and pressure. It combines the influences of enthalpy and entropy through the relation:

\[ \Delta G = \Delta H - T\Delta S \]

This equation illustrates that a reaction will occur spontaneously if the calculated ΔG is negative. In simple terms, a negative ΔG means the process can happen without additional input, much like a sled naturally gliding down a hill.

The interplay of ΔH and ΔS greatly influences ΔG:
  • **Negative ΔH and Positive ΔS:** These conditions strongly favor a negative ΔG, ensuring sponteneity.
  • **Positive ΔH and Negative ΔS:** Here, both factors drive ΔG positive, opposing spontaneity.
  • **Negative ΔH and Negative ΔS:** ΔG may be negative at lower temperatures, favoring spontaneity, but could become positive at higher temperatures.
  • **Positive ΔH and Positive ΔS:** ΔG may be positive at low temperatures but can turn negative, allowing spontaneity as temperature increases.
Understanding these relationships helps predict whether a reaction will energetically move forward without needing external energy input.

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Most popular questions from this chapter

The following reaction, sometimes used in the laboratory to generate small quantities of oxygen gas, has \(\Delta G^{\circ}=-224.4 \mathrm{~kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\) : $$2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)$$ Use the following additional data at \(25^{\circ} \mathrm{C}\) to calculate the standard molar entropy \(S^{\circ}\) of \(\mathrm{O}_{2}\) at \(25^{\circ} \mathrm{C}: \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{KClO}_{3}\right)=-397.7 \mathrm{~kJ} / \mathrm{mol}\), \(\Delta H_{\mathrm{f}}^{\circ}(\mathrm{KCl})=-436.5 \mathrm{~kJ} / \mathrm{mol}, S^{\circ}\left(\mathrm{KClO}_{3}\right)=143.1 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\) and \(S^{\circ}(\mathrm{KCl})=82.6 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})\).

Give an equation that relates the entropy change in the surroundings to the enthalpy change in the system. What is the sign of \(\Delta S_{\text {surr }}\) for the following? (a) An exothermic reaction (b) An endothermic reaction

What is the entropy change when the volume of \(1.6 \mathrm{~g}\) of \(\mathrm{O}_{2}\) increases from \(2.5 \mathrm{~L}\) to \(3.5 \mathrm{~L}\) at a constant temperature of \(75^{\circ} \mathrm{C} ?\) Assume that \(\mathrm{O}_{2}\) behaves as an ideal gas.

When rolling a pair of dice, there are two ways to get a point total of \(3(1+2 ; 2+1)\) but only one way to get a point total of \(2(1+1)\). How many ways are there of getting point totals of 4-12? What is the most probable point total?

Consider the distribution of ideal gas molecules among three bulbs (A, B, and C) of equal volume. For each of the following states, determine the number of ways \((W)\) that the state can be achieved, and use Boltzmann's equation to calculate the entropy of the state: (a) 2 molecules in bulb \(A\) (b) 2 molecules randomly distributed among bulbs \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) (c) 3 molecules in bulb \(\mathrm{A}\) (d) 3 molecules randomly distributed among bulbs \(A, B\), and \(C\) (e) 1 mol of molecules in bulb \(A\) (f) \(1 \mathrm{~mol}\) of molecules randomly distributed among bulbs \(\mathrm{A}, \mathrm{B}\), and \(C\) What is \(\Delta S\) on going from state (e) to state (f)? Compare your result with \(\Delta S\) calculated from the equation \(\Delta S=n R \ln \left(V_{\text {final }} / V_{\text {initial }}\right)\).

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