The combustion of acetylene is used in welder's torches because it produces a very hot flame:

\[ \text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + \text{H}_2\text{O}(g), \quad \Delta H° = -1255.5 \, \text{kJ} \]

Use the data provided to calculate $\Delta S°$ for this reaction.

Hint:
- $ S°_{\text{C}_2\text{H}_2(g)} = 200.8 \, \text{J/mol} \cdot \text{K} $
- $ S°_{\text{O}_2(g)} = 205.0 \, \text{J/mol} \cdot \text{K} $
- $ S°_{\text{CO}_2(g)} = 213.6 \, \text{J/mol} \cdot \text{K} $
- $ S°_{\text{H}_2\text{O}(g)} = 188.7 \, \text{J/mol} \cdot \text{K} $

Options:
1) 194.8 J/K
2) -194.8 J/K
3) -97.4 J/K
4) 97.4 J/K

The change in entropy (∆S°) for the combustion of acetylene is calculated using the standard entropy values for the reactants and products of the reaction, and the answer is -97.4 J/mol·K.

Explanation:

To calculate the change in entropy (∆S°) for the combustion of acetylene, we use the standard entropy values (S°) provided for each reactant and product and apply the equation ∆S° = ∑S°(products) - ∑S°(reactants).

First, we calculate the total entropy of the products:

For CO₂ (g): 2 × 213.6 J/mol·K = 427.2 J/mol·K
For H₂O (g): 188.7 J/mol·K

Sum of entropy of products = 427.2 J/mol·K + 188.7 J/mol·K = 615.9 J/mol·K

Then, the total entropy of the reactants:

For C₂H₂ (g): 200.8 J/mol·K
For O₂ (g): ⅒O₂ requires multiplying the entropy by 5/2, so 5/2 × 205.0 J/mol·K = 512.5 J/mol·K

Sum of entropy of reactants = 200.8 J/mol·K + 512.5 J/mol·K = 713.3 J/mol·K

Therefore, ∆S° = 615.9 J/mol·K - 713.3 J/mol·K = -97.4 J/mol·K

The change in entropy for the combustion of acetylene is -97.4 J/K.