Question

In which of the following reactions does entropy decrease? (i) $2Pb(NO_{3})_{2(s)}\rightarrow 2PbO_{(s)}+4NO_{2(g)}+O_{2(g)}$ (ii) $H_{2}O_{(g)}\rightarrow H_{2}O_{(l)}$ (iii) $Br_{2(l)}\rightarrow 2Br_{(g)}$ (iv) $C_{6}H_{6(l)}\rightarrow C_{6}H_{6(s)}$

• A. (ii), (iii) and (iv)
• B. (i) and (iii)
• C. (i) and (iii)
• D. (i) and (iv)
• E. (ii) and (iv)

Hint:

Phase change toward a solid state always decreases entropy.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
Entropy (S) is a measure of the disorder or randomness of a system. The second law of thermodynamics states that the entropy of an isolated system tends to increase. We need to identify the reactions where the system becomes more ordered, which corresponds to a decrease in entropy (\( \Delta S<0 \)).
Step 2: Key Formula or Approach:
We can predict the sign of the entropy change (\( \Delta S \)) by looking at changes in the state of matter and the number of moles of gas. - Phase Changes: Entropy generally increases in the order: solid \(<\) liquid \(<\) gas. So, a process like freezing (liquid to solid) or condensation (gas to liquid) will have a negative \( \Delta S \). - Number of Gas Moles: For reactions involving gases, an increase in the number of moles of gas usually leads to an increase in entropy. A decrease in the number of moles of gas leads to a decrease in entropy. Step 3: Detailed Explanation:
Let's analyze the entropy change for each reaction: - (i) \( 2\text{Pb}(\text{NO}_3)_{2(s)} \rightarrow 2\text{PbO}_{(s)} + 4\text{NO}_{2(g)} + \text{O}_{2(g)} \): A solid reactant decomposes to produce a solid and two different gases. The number of moles of gas increases from 0 to (4+1) = 5. The creation of gas from a solid represents a massive increase in disorder. So, \( \Delta S>0 \) (entropy increases). - (ii) \( \text{H}_2\text{O}_{(g)} \rightarrow \text{H}_2\text{O}_{(l)} \): This is condensation, a phase transition from a more disordered state (gas) to a more ordered state (liquid). So, \( \Delta S<0 \) (entropy decreases). - (iii) \( \text{Br}_{2(l)} \rightarrow 2\text{Br}_{(g)} \): A liquid reactant turns into a gas. This represents an increase in disorder. Also, 1 mole of liquid produces 2 moles of gas particles, further increasing disorder. So, \( \Delta S>0 \) (entropy increases). - (iv) \( \text{C}_6\text{H}_{6(l)} \rightarrow \text{C}_6\text{H}_{6(s)} \): This is freezing, a phase transition from a more disordered state (liquid) to a more ordered state (solid). So, \( \Delta S<0 \) (entropy decreases). Step 4: Final Answer:
Entropy decreases in reactions (ii) and (iv).