Question

05 question (10 points)

the nonspontaneous reaction

\\d + e \
ightarrow f\\

decreases the system entropy by \\(37\text{ j}/(\text{k}\cdot\text{mol})\\).

what is the maximum value of the entropy change of the surroundings?

Explanation:

Identify the given thermodynamic values

We are given a nonspontaneous reaction:

$$ D + E ightarrow F $$

The change in entropy of the system is:

$$ \Delta S_{\text{sys}} = -37 \text{ J/(K}\cdot\text{mol)} $$

Apply the Second Law of Thermodynamics

For a process to be spontaneous, the total entropy change of the universe must be positive:

$$ \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0 $$

Since the reaction is specified as nonspontaneous, the total entropy change of the universe must be less than or equal to zero:

$$ \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \le 0 $$

Solve for the maximum entropy change of the surroundings

Using the inequality for a nonspontaneous process:

$$ \Delta S_{\text{sys}} + \Delta S_{\text{surr}} \le 0 $$

Substitute the given system entropy change:

$$ -37 \text{ J/(K}\cdot\text{mol)} + \Delta S_{\text{surr}} \le 0 $$

Solve for the surroundings entropy change:

$$ \Delta S_{\text{surr}} \le 37 \text{ J/(K}\cdot\text{mol)} $$

Thus, the maximum possible value is \(37 \text{ J/(K}\cdot\text{mol)}\).

Answer:

What is the maximum value of the entropy change of the surroundings? <blank>37</blank> \(\text{J/(K}\cdot\text{mol)}\)