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

Short Answer

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For exothermic reactions, \(\Delta S_{\text{surr}}\) is positive, indicating an increase in the entropy of the surroundings, which is considered favorable. On the other hand, for endothermic reactions, \(\Delta S_{\text{surr}}\) is negative, indicating a decrease in the entropy of the surroundings, which is considered unfavorable.

Step by step solution

01

Define entropy

Entropy, represented as \(S\), is a measure of the degree of disorder or randomness in a system. An increase in entropy generally means a higher degree of disorder, while a decrease in entropy indicates a lower degree of disorder.
02

Calculate the change in entropy of the surroundings

For chemical reactions, the change in entropy of the surroundings, denoted as \(\Delta S_{\text{surr}}\), can be calculated using the relation: \begin{equation} \Delta S_{\text{surr}} = - \dfrac{q_{\text{sys}}}{T} \end{equation} where \(q_{\text{sys}}\) is the heat flow of the system, which is positive if the system absorbs heat (endothermic) and negative if the system releases heat (exothermic), and \(T\) is the temperature in Kelvin.
03

Analyze the change in entropy of surroundings for exothermic reactions

In an exothermic reaction, heat is released into the surroundings, so \(q_{\text{sys}}\) is negative. Therefore, when we plug this into the equation: \begin{equation} \Delta S_{\text{surr}} = - \dfrac{-q_{\text{sys}}}{T} \end{equation} Since both the numerator and denominator are positive, \(\Delta S_{\text{surr}}\) will be positive, meaning the entropy of the surroundings increases.
04

Analyze the change in entropy of surroundings for endothermic reactions

In an endothermic reaction, heat is absorbed from the surroundings, so \(q_{\text{sys}}\) is positive. Therefore, when we plug this into the equation: \begin{equation} \Delta S_{\text{surr}} = - \dfrac{q_{\text{sys}}}{T} \end{equation} Since the numerator is positive and the denominator is positive, \(\Delta S_{\text{surr}}\) will be negative, meaning the entropy of the surroundings decreases.
05

Conclude whether the change in entropy of the surroundings is favorable or unfavorable

From the analysis in Steps 3 and 4, we can conclude that: - For exothermic reactions, \(\Delta S_{\text{surr}}\) is positive, which indicates an increase in the entropy of the surroundings. Entropy increasing is generally considered favorable because it leads to a more dispersed and disordered state. - For endothermic reactions, \(\Delta S_{\text{surr}}\) is negative, indicating a decrease in the entropy of the surroundings, which is generally considered unfavorable because it leads to a less dispersed and more ordered state.

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