What combination of \(\Delta H\) and \(\Delta S\) values always produces a negative free-energy change?

Short Answer

Expert verified
\Delta H < 0 and \Delta S > 0 always produce a negative \(\Delta G\).

Step by step solution

01

Understand the Question

To determine the combination of \(\Delta H\) (change in enthalpy) and \(\Delta S\) (change in entropy) that always results in a negative free-energy change \(\Delta G\), use the Gibbs free energy equation: \(\Delta G = \Delta H - T\Delta S\).
02

Recall the Gibbs Free Energy Equation

The Gibbs free energy change \(\Delta G\) is calculated with the formula: \(\Delta G = \Delta H - T\Delta S\). For \(\Delta G\) to be negative, the term \(\Delta H - T\Delta S\) must be less than zero.
03

Consider Values of \(\Delta H\) and \(\Delta S\)

To achieve a negative \(\Delta G\), let's analyze the signs of \(\Delta H\) and \(\Delta S\). Both positive \(\Delta H\) with large negative \(T\Delta S\) and negative \(\Delta H\) with positive \(T\Delta S\) can yield a negative \(\Delta G\).
04

Combination That Always Yields Negative \(\Delta G\)

If \(\Delta H\) is negative and \(\Delta S\) is positive, the subtraction \(\Delta G = \Delta H - T\Delta S\) will always result in a negative \(\Delta G\) regardless of temperature.

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Key Concepts

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

enthalpy
Enthalpy, symbolized as \(\backslashDelta H\), represents the total heat content of a system under constant pressure. It is a key concept in thermodynamics, as it helps determine how much energy is absorbed or released during a process. When a reaction releases heat to its surroundings, \(\backslashDelta H\) is negative, indicating an exothermic process. In contrast, when a reaction absorbs heat, \(\backslashDelta H\) is positive, signifying an endothermic process. Understanding enthalpy is pivotal for predicting how reactions will behave and what conditions are necessary for them to proceed.
entropy
Entropy, denoted as \(\backslashDelta S\), measures the level of disorder or randomness in a system. It is an essential concept in the study of thermodynamics because it helps predict the direction of spontaneous processes. A higher entropy means greater disorder. Generally, natural processes tend to increase entropy. When considering the Gibbs free energy equation, if \(\backslashDelta S\) increases (positive entropy change), it contributes to making \(\backslashDelta G\) more negative. This insight is crucial for understanding why certain reactions favor a higher state of disorder and thus proceed spontaneously.
thermodynamics
Thermodynamics is the science that studies the relationships between heat, work, temperature, and energy. It encompasses several laws and principles that help us understand how energy is transferred and transformed in physical and chemical processes. One of the essential principles is the Gibbs free energy equation: \(\backslashDelta G = \backslashDelta H - T\backslashDelta S\), which combines enthalpy, entropy, and temperature to predict reaction spontaneity. Negative \(\backslashDelta G\) values indicate reactions that can occur spontaneously without additional energy input. By mastering thermodynamic principles, students can predict the feasibility and direction of chemical reactions accurately.

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Most popular questions from this chapter

Performance Design a simple calorimeter investigation to determine the molar enthalpy of fusion of water. Use the following materials: a large plastic-foam cup with cover, a thermometer, a balance, water at room temperature, and an ice cube. Allow your teacher to review your design. Then carry out the investigation, and write a laboratory report including your calculations and a comparison of your quantitative results with known values. Try to account for any disagreements between the experimental and actual values.

Relating ldeas Given the entropy change for the first two reactions below, calculate the entropy change for the third reaction below. \(\mathrm{S}_{8}(s)+8 \mathrm{O}_{2}(s) \longrightarrow 8 \mathrm{SO}_{2}(g) \qquad \Delta S=89 \mathrm{J} / \mathrm{K}\) \(2 \mathrm{SO}_{2}(s)+\mathrm{O}_{2}(s) \longrightarrow 2 \mathrm{SO}_{3}(g) \qquad \Delta S=-188 \mathrm{J} / \mathrm{K}\) \(\mathrm{S}_{8}(s)+12 \mathrm{O}_{2}(s) \longrightarrow 8 \mathrm{SO}_{3}(g) \qquad \Delta S=?\)

The standard enthalpy of formation for sulfur dioxide gas is \(-296.8 \mathrm{kJ} / \mathrm{mol}\) . Calculate the amount of energy given off in \(\mathrm{kJ}\) when 30.0 \(\mathrm{g}\) of \(\mathrm{SO}_{2}(g)\) is formed from its elements.

The reaction to synthesize methanol (CH_ \(_{3} \mathrm{OH} )\) industrially is $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ The \(\Delta H_{\text { reaction }}^{0}=-90.7 \mathrm{kJ}\) and the \(\Delta S_{\text {reaction}}^{0}=\) \(-220.8 \mathrm{J} / \mathrm{K}\) . At what temperatures is the reaction nonspontaneous?

Rewrite each equation below with the \(\Delta H\) value included with either the reactants or the products, and identify the reaction as endothermic or exothermic. a. \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) \(\Delta H^{0}=-285.83 \mathrm{kJ}\) b. \(2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s)\) \(\Delta H^{0}=-1200 \mathrm{kJ}\) c. \(\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g) ; \Delta H^{0}=+62.4 \mathrm{kJ}\) d. \(3 \mathrm{CO}(g)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g)$$\Delta H^{0}=-24.7 \mathrm{kJ}\)

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