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Chapter 17: Problem 8

Is \(\Delta S_{\text { surt }}\) favorable or unfavorable for exothermic reactions? Endothermic reactions? Explain.

Short Answer

Expert verified
In summary, 螖S_surt is favorable (positive) for exothermic reactions, as they increase the disorder in the surroundings by releasing heat. On the other hand, 螖S_surt is unfavorable (negative) for endothermic reactions, as they decrease the disorder in the surroundings by absorbing heat.

Step by step solution

01

Understanding Entropy and Reaction Types

First, let's recall some key definitions: Entropy (S) is a measure of disorder or randomness in a system. Exothermic reactions are those that release heat to their surroundings (\(q < 0\)), while endothermic reactions absorb heat from their surroundings (\(q > 0\)). Now, let's analyze how 螖S_surt is affected by exothermic and endothermic reactions.
02

Exothermic Reactions and 螖S_surt

For exothermic reactions, heat is released into the surroundings, which usually leads to an increase in disorder or randomness. As a result, the entropy change 螖S_surt tends to be positive. A positive 螖S_surt means that the process is favorable from the standpoint of the surroundings' entropy.
03

Endothermic Reactions and 螖S_surt

For endothermic reactions, heat is absorbed from the surroundings, which usually results in a decrease in disorder or randomness. As a consequence, the entropy change 螖S_surt tends to be negative. A negative 螖S_surt implies that the process is unfavorable from the standpoint of the surroundings' entropy.
04

Conclusion

In conclusion, 螖S_surt is generally favorable (positive) for exothermic reactions, as they increase the disorder of the surroundings by releasing heat. Conversely, 螖S_surt is generally unfavorable (negative) for endothermic reactions, as they decrease the disorder of the surroundings by absorbing heat.

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Key Concepts

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

exothermic reactions
In the field of chemistry, understanding exothermic reactions is crucial. These reactions are characterized by the release of heat into the surroundings. When a chemical reaction occurs, it can either absorb or release energy. For exothermic reactions, this means that energy is released in the form of heat.
Here are some key points about exothermic reactions:
  • They usually lead to an increase in the temperature of the surroundings.
  • Common examples include combustion reactions, such as when wood burns in a fireplace.
  • The system's enthalpy change (\( ext{螖H}\)) for exothermic reactions is negative because heat exits the system.
As heat is released, the disorder or randomness in the surroundings increases. This means the entropy change (\( ext{螖S}_{ ext{sur}}\)) for the surroundings is generally positive, indicating a favorable process for the surroundings.
endothermic reactions
Unlike exothermic reactions, endothermic reactions absorb heat from their surroundings. This absorption leads to a decrease in the temperature around the reaction site. Energy is taken in by the system, typically making it cooler.
Key ideas about endothermic reactions include:
  • They tend to make the surroundings feel cooler as heat is absorbed.
  • Endothermic processes include simple everyday examples, like the melting of ice.
  • The enthalpy change (\( ext{螖H}\)) for endothermic reactions is positive, a reflection of the system's intake of heat.
When the reaction absorbs heat, the disorder or randomness in the surroundings decreases. Hence, the entropy change (\( ext{螖S}_{ ext{sur}}\)) is negative, suggesting that the process is generally unfavorable for the surroundings' entropy.
thermodynamics
Thermodynamics is a fundamental branch of physics and chemistry that deals with heat and temperature and their relation to energy and work. It is a key concept for understanding chemical reactions, particularly with enthalpy and entropy.
Some basic principles of thermodynamics include:
  • The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed, only transformed.
  • The second law of thermodynamics introduces the concept of entropy, explaining why certain processes occur spontaneously.
  • Entropy is often associated with the disorder within a system. Systems tend to move towards greater entropy.
By studying thermodynamics, scientists can predict whether a reaction will be exothermic or endothermic, and understand why certain reactions are more favorable than others. The interplay of enthalpy, entropy, and free energy ultimately determines the spontaneity and favorability of chemical processes.

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Most popular questions from this chapter

Consider the following reaction at \(25.0^{\circ} \mathrm{C} :\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03\) kJ/mol and \(-176.6 \mathrm{J} / \mathrm{K}\) . mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\) . Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\) .

The third law of thermodynamics states that the entropy of a perfect crystal at 0 \(\mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\) . b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K}\) , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K)\) . Realizing what \(\Delta G^{\circ}\) and K mean, how can you figure out the correct sign?
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