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:

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

$$
LATEXBLOCK0
$$

Determine the sign and magnitude of surroundings entropy change

Using the Second Law of Thermodynamics knowledge point

$$
LATEXBLOCK1
$$

Answer:

• \(\Delta S_{\text{surr}} < 0\) 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 \(> \Delta S_{\text{sys}}\). (Correct answer)
• \(\Delta S_{\text{surr}} < 0\) and its magnitude is \(= \Delta S_{\text{sys}}\).