Solved Worksheet 2: Thermodynamics Entropy and Gibbs Free | Chegg.com - Free Printable

Let's solve each part of this thermodynamics worksheet step by step.

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Question 1: Predict the Entropy Change (ΔS) – Increase (+) or Decrease (−)

Entropy (S) is a measure of disorder or randomness. Generally:
- More gas molecules → higher entropy
- Solid → liquid → gas → increasing entropy
- Fewer moles of gas → decrease in entropy
- More complex molecules may have higher entropy

#### a. \( 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g) \)

- Reactants: 2 mol NO(g) + 1 mol O₂(g) = 3 mol gas
- Products: 2 mol NO₂(g) = 2 mol gas
- Gas moles decrease → decrease in entropy

✔ Answer: − (entropy decreases)

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#### b. \( \text{N}_2\text{O}(g) + \text{NO}_2(g) \rightarrow 3\text{NO}(g) \)

- Reactants: 1 mol N₂O(g) + 1 mol NO₂(g) = 2 mol gas
- Products: 3 mol NO(g) = 3 mol gas
- Gas moles increase → increase in entropy

✔ Answer: + (entropy increases)

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#### c. \( \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \)

- Gas → Liquid: phase change from high disorder to low disorder
- Entropy decreases significantly

✔ Answer: − (entropy decreases)

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✔ Summary for Question 1:

| Reaction | ΔS Sign |
|--------|--------|
| a | − |
| b | + |
| c | − |

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Question 2: Calculate ΔS° using Standard Molar Entropies

We use:
\[
\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}
\]

Given standard molar entropies:

| Substance | \( S^\circ \) (J/mol·K) |
|----------|------------------------|
| H₂O(g) | 189 |
| NO(g) | 211 |
| NO₂(g) | 240 |
| N₂O(g) | 220 |
| H₂(g) | 131 |
| O₂(g) | 205 |
| H₂O(l) | 70 |
| O₃(g) | 239 |

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#### a. \( 2\text{NO}(g) + \text{O}_2(g) \rightarrow 2\text{NO}_2(g) \)

\[
\Delta S^\circ = [2 \times S^\circ(\text{NO}_2)] - [2 \times S^\circ(\text{NO}) + S^\circ(\text{O}_2)]
\]
\[
= [2 \times 240] - [2 \times 211 + 205] = 480 - (422 + 205) = 480 - 627 = -147 \, \text{J/K}
\]

✔ ΔS° = −147 J/K → Decrease, matches prediction (−)

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#### b. \( \text{N}_2\text{O}(g) + \text{NO}_2(g) \rightarrow 3\text{NO}(g) \)

\[
\Delta S^\circ = [3 \times S^\circ(\text{NO})] - [S^\circ(\text{N}_2\text{O}) + S^\circ(\text{NO}_2)]
\]
\[
= [3 \times 211] - [220 + 240] = 633 - 460 = +173 \, \text{J/K}
\]

✔ ΔS° = +173 J/K → Increase, matches prediction (+)

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#### c. \( \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \)

\[
\Delta S^\circ = S^\circ(\text{H}_2\text{O}(l)) - S^\circ(\text{H}_2\text{O}(g)) = 70 - 189 = -119 \, \text{J/K}
\]

✔ ΔS° = −119 J/K → Decrease, matches prediction (−)

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#### d. \( 3\text{H}_2(g) + \text{O}_3(g) \rightarrow 3\text{H}_2\text{O}(g) \)

\[
\Delta S^\circ = [3 \times S^\circ(\text{H}_2\text{O}(g))] - [3 \times S^\circ(\text{H}_2) + S^\circ(\text{O}_3)]
\]
\[
= [3 \times 189] - [3 \times 131 + 239] = 567 - (393 + 239) = 567 - 632 = -65 \, \text{J/K}
\]

✔ ΔS° = −65 J/K → Decrease in entropy

Note: This was not in question 1, but included here as per the problem.

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✔ Comparison with Predictions:

All calculated values match predictions:
- a: − → − (correct)
- b: + → + (correct)
- c: − → − (correct)
- d: − (calculated), no prediction made in Q1, but we can say it’s expected due to fewer gas moles (4 → 3), so entropy decreases.

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Question 3: What sign must G be for a reaction to be spontaneous?

G refers to Gibbs Free Energy (ΔG).

The criterion for spontaneity is:
\[
\Delta G = \Delta H - T\Delta S
\]

A reaction is spontaneous when:
\[
\boxed{\Delta G < 0} \quad \text{(negative)}
\]

✔ Answer: G must be negative (−) for a reaction to be spontaneous.

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✔ Final Answers:

#### 1. Predicted ΔS signs:
a. −
b. +
c. −

#### 2. Calculated ΔS° values:
a. −147 J/K
b. +173 J/K
c. −119 J/K
d. −65 J/K

→ All match predictions.

#### 3. Spontaneity condition:
ΔG must be negative (−) for a reaction to be spontaneous.

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Let me know if you'd like help calculating ΔG or determining spontaneity based on ΔH and ΔS!