To determine how the entropy changes in the reaction [tex]\(2 \text{C}_3\text{H}_6(g) + 9 \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 6 \text{H}_2\text{O}(g)\)[/tex], we can examine the number of gas molecules on the reactant and product sides.
1. Count the gas molecules on the reactant side:
- There are 2 molecules of [tex]\( \text{C}_3\text{H}_6(g) \)[/tex].
- There are 9 molecules of [tex]\( \text{O}_2(g) \)[/tex].
- Total gas molecules on the reactant side: [tex]\(2 + 9 = 11\)[/tex].
2. Count the gas molecules on the product side:
- There are 6 molecules of [tex]\( \text{CO}_2(g) \)[/tex].
- There are 6 molecules of [tex]\( \text{H}_2\text{O}(g) \)[/tex].
- Total gas molecules on the product side: [tex]\(6 + 6 = 12\)[/tex].
3. Compare the number of gas molecules:
- Reactant side has 11 gas molecules.
- Product side has 12 gas molecules.
Since the number of gas molecules increases from 11 to 12 during the reaction, the entropy of the system increases. Recall that an increase in the number of gas molecules typically leads to an increase in entropy because there are more possible ways to arrange the gas molecules, hence higher randomness or disorder.
Thus, the correct answer is:
D. The entropy increases.
