Which of the following reactions is asssociated with the most negative change in entropy?

Options

A. `2SO_(2)(g)+O_(2)(g)rarr2SO_(3)(g)`
B. `C_(2)H_(4)(g)+H_(2)(g)rarrC_(2)H_(6)(g)`
C. `C(s,"graphite")+O_(2)(g)rarrCO_(2)(g)`
D. `3C_(2)H_(2)(g)rarrC_(6)H_(6)(l)`

Explanation

To determine which of the given reactions is associated with the most negative change in entropy (ΔS), we will analyze each reaction based on the change in the number of moles of gaseous reactants and products. Here’s a step-by-step solution:

### Step 1: Understand the relationship between ΔG and ΔS
The change in Gibbs free energy (ΔG) is related to the change in entropy (ΔS) through the equation:
\[ \Delta G = \Delta H - T\Delta S \]
However, for our purpose, we can simplify our analysis by focusing on the change in the number of moles of gaseous species.

### Step 2: Calculate ΔnG for each reaction
ΔnG is defined as:
\[ \Delta nG = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]

We will analyze each reaction to find ΔnG.

### Step 3: Analyze each option
1. **Option 1**:
- Products: 2 moles (gaseous)
- Reactants: 3 moles (gaseous)
- ΔnG = 2 - 3 = -1

2. **Option 2**:
- Products: 1 mole (gaseous)
- Reactants: 2 moles (gaseous)
- ΔnG = 1 - 2 = -1

3. **Option 3**:
- Products: 1 mole (gaseous)
- Reactants: 1 mole (gaseous)
- ΔnG = 1 - 1 = 0

4. **Option 4**:
- Products: 0 moles (liquid, not counted)
- Reactants: 3 moles (gaseous)
- ΔnG = 0 - 3 = -3

### Step 4: Determine the most negative ΔnG
From our calculations:
- Option 1: ΔnG = -1
- Option 2: ΔnG = -1
- Option 3: ΔnG = 0
- Option 4: ΔnG = -3

The most negative ΔnG is from **Option 4**, which has a ΔnG of -3.

### Step 5: Conclusion
Since a more negative ΔnG corresponds to a more negative ΔS, we conclude that **Option 4** is associated with the most negative change in entropy.