For the reaction `2HgO(s) rarr 2Hg(l) + O_(2)(g)`

Options

A. `Delta H gt 0 and Delta S lt 0`
B. `Delta H gt 0 and Delta S gt 0`
C. `Delta H lt 0 and Delta S lt 0`
D. `Delta H lt 0 and Delta S gt 0`

Answer

To solve the problem regarding the reaction \( 2 \text{HgO}(s) \rightarrow 2 \text{Hg}(l) + \text{O}_2(g) \), we need to analyze the changes in enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) during the reaction.

### Step-by-Step Solution:

1. **Identify the States of Reactants and Products**:
- The reactant is \( \text{HgO} \) in solid state (s).
- The products are \( \text{Hg} \) in liquid state (l) and \( \text{O}_2 \) in gaseous state (g).

2. **Analyze the Change in Entropy (\( \Delta S \))**:
- Entropy is a measure of disorder or randomness in a system.
- In this reaction, we are converting 2 moles of solid \( \text{HgO} \) into 2 moles of liquid \( \text{Hg} \) and 1 mole of gas \( \text{O}_2 \).
- The transition from solid to liquid and gas indicates an increase in randomness because solids have closely packed particles, while liquids and gases have more freedom of movement.
- Therefore, we conclude that the entropy change (\( \Delta S \)) is positive:
\[ \Delta S > 0 \]

3. **Analyze the Change in Enthalpy (\( \Delta H \))**:
- The reaction involves breaking bonds in \( \text{HgO} \) to form \( \text{Hg} \) and \( \text{O}_2 \).
- Breaking bonds requires energy input, which means that the reaction is endothermic.
- Hence, the change in enthalpy (\( \Delta H \)) is also positive:
\[ \Delta H > 0 \]

4. **Summarize the Results**:
- From our analysis, we have:
- \( \Delta S > 0 \) (positive change in entropy)
- \( \Delta H > 0 \) (positive change in enthalpy)

5. **Select the Correct Option**:
- Based on the signs of \( \Delta H \) and \( \Delta S \), we can conclude that the correct option that matches these conditions is option (b).