Answer :
To determine which reaction results in decreased entropy, let's recall that entropy is a measure of disorder or randomness in a system. Entropy tends to increase when a system becomes more disordered, such as when a substance transitions from solid to liquid to gas. Conversely, entropy decreases when a substance transitions from gas to liquid to solid.
Let's evaluate each reaction step-by-step:
1. Option A: [tex]\( 2 \text{O}_3(g) \rightarrow 3 \text{O}_2(g) \)[/tex]
- This reaction involves two ozone molecules (O3) in the gaseous form converting into three oxygen molecules (O2), also in the gaseous form. Since both reactants and products are gases and the number of gas molecules increases, this reaction likely increases entropy due to increased disorder.
2. Option B: [tex]\( \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \)[/tex]
- Here, water transitions from a gaseous state (vapor) to a liquid state. This transition decreases entropy because the liquid state is more ordered compared to the gaseous state. Therefore, this reaction results in decreased entropy.
3. Option C: [tex]\( \text{CO}_2(s) \rightarrow \text{CO}_2(g) \)[/tex]
- In this reaction, solid carbon dioxide (dry ice) transitions to gaseous carbon dioxide. This transition increases entropy because the gas state is more disordered than the solid state.
4. Option D: [tex]\( \text{N}_2\text{O}_4(g) \rightarrow 2 \text{NO}_2(g) \)[/tex]
- This reaction involves a single dinitrogen tetroxide (N2O4) gas molecule dissociating into two nitrogen dioxide (NO2) gas molecules. The number of gas molecules increases, indicating an increase in disorder and thus an increase in entropy.
From the above analysis, only Option B [tex]\( \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \)[/tex] results in a decrease in entropy because it involves a change from a less ordered gaseous state to a more ordered liquid state. Therefore, the correct answer is:
Option B.
Let's evaluate each reaction step-by-step:
1. Option A: [tex]\( 2 \text{O}_3(g) \rightarrow 3 \text{O}_2(g) \)[/tex]
- This reaction involves two ozone molecules (O3) in the gaseous form converting into three oxygen molecules (O2), also in the gaseous form. Since both reactants and products are gases and the number of gas molecules increases, this reaction likely increases entropy due to increased disorder.
2. Option B: [tex]\( \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \)[/tex]
- Here, water transitions from a gaseous state (vapor) to a liquid state. This transition decreases entropy because the liquid state is more ordered compared to the gaseous state. Therefore, this reaction results in decreased entropy.
3. Option C: [tex]\( \text{CO}_2(s) \rightarrow \text{CO}_2(g) \)[/tex]
- In this reaction, solid carbon dioxide (dry ice) transitions to gaseous carbon dioxide. This transition increases entropy because the gas state is more disordered than the solid state.
4. Option D: [tex]\( \text{N}_2\text{O}_4(g) \rightarrow 2 \text{NO}_2(g) \)[/tex]
- This reaction involves a single dinitrogen tetroxide (N2O4) gas molecule dissociating into two nitrogen dioxide (NO2) gas molecules. The number of gas molecules increases, indicating an increase in disorder and thus an increase in entropy.
From the above analysis, only Option B [tex]\( \text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \)[/tex] results in a decrease in entropy because it involves a change from a less ordered gaseous state to a more ordered liquid state. Therefore, the correct answer is:
Option B.