Question

question 15

during a spontaneous chemical reaction, it is found that \\(\delta s_{\text{sys}} < 0\\). this means _______

\\(\delta s_{\text{surr}} < 0\\) and its magnitude is \\(< \delta s_{\text{sys}}\\).
\\(\delta s_{\text{surr}} < 0\\) and its magnitude is \\(> \delta s_{\text{sys}}\\).
an error has been made, because \\(s_{\text{sys}} > 0\\) by necessity for a spontaneous process.
\\(\delta s_{\text{surr}} > 0\\) and its magnitude is \\(< \delta s_{\text{sys}}\\).
\\(\delta s_{\text{surr}} < 0\\) and its magnitude is \\(= \delta s_{\text{sys}}\\).

Explanation:

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<reasoning>

State the condition for spontaneity

Using the Second Law of Thermodynamics and Spontaneous Processes knowledge points
\[
\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0
\]

Analyze the given system entropy change

Using the Spontaneous Processes knowledge point
\[

$$\begin{aligned} &\Delta S_{\text{sys}} < 0 \\ &\Delta S_{\text{sys}} = -|\Delta S_{\text{sys}}| \end{aligned}$$

\]

Determine the sign and magnitude of surroundings entropy change

Using the Second Law of Thermodynamics knowledge point
\[

$$\begin{aligned} &\Delta S_{\text{surr}} > -\Delta S_{\text{sys}} \\ &\Delta S_{\text{surr}} > |\Delta S_{\text{sys}}| > 0 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>\(\Delta S_{\text{surr}} < 0\) and its magnitude is \(< \Delta S_{\text{sys}}\).</mcq-option>
<mcq-option>\(\Delta S_{\text{surr}} < 0\) and its magnitude is \(> \Delta S_{\text{sys}}\).</mcq-option>
<mcq-option>an error has been made, because \(S_{\text{sys}} > 0\) by necessity for a spontaneous process.</mcq-option>
<mcq-correct>\(\Delta S_{\text{surr}} > 0\) and its magnitude is \(> \Delta S_{\text{sys}}\).</mcq-correct>
<mcq-option>\(\Delta S_{\text{surr}} < 0\) and its magnitude is \(= \Delta S_{\text{sys}}\).</mcq-option>
</answer>

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]
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Answer:

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<reasoning>

State the condition for spontaneity

Using the Second Law of Thermodynamics and Spontaneous Processes knowledge points
\[
\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0
\]

Analyze the given system entropy change

Using the Spontaneous Processes knowledge point
\[

$$\begin{aligned} &\Delta S_{\text{sys}} < 0 \\ &\Delta S_{\text{sys}} = -|\Delta S_{\text{sys}}| \end{aligned}$$

\]

Determine the sign and magnitude of surroundings entropy change

Using the Second Law of Thermodynamics knowledge point
\[

$$\begin{aligned} &\Delta S_{\text{surr}} > -\Delta S_{\text{sys}} \\ &\Delta S_{\text{surr}} > |\Delta S_{\text{sys}}| > 0 \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>\(\Delta S_{\text{surr}} < 0\) and its magnitude is \(< \Delta S_{\text{sys}}\).</mcq-option>
<mcq-option>\(\Delta S_{\text{surr}} < 0\) and its magnitude is \(> \Delta S_{\text{sys}}\).</mcq-option>
<mcq-option>an error has been made, because \(S_{\text{sys}} > 0\) by necessity for a spontaneous process.</mcq-option>
<mcq-correct>\(\Delta S_{\text{surr}} > 0\) and its magnitude is \(> \Delta S_{\text{sys}}\).</mcq-correct>
<mcq-option>\(\Delta S_{\text{surr}} < 0\) and its magnitude is \(= \Delta S_{\text{sys}}\).</mcq-option>
</answer>

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