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

Options

A. `DeltaH gt 0` and `DeltaS lt 0`
B. `DeltaH gt O` and `DeltaS gt 0`
C. `DeltaH lt 0` and `DeltaS lt 0`
D. `DeltaH lt 0` and `DeltaS gt 0`

Explanation

To solve the problem regarding the reaction \( 2 \text{HgO}(s) \rightarrow 2 \text{Hg}(l) + \text{O}_2(g) \), we need to determine the changes in enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) for this 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 Entropy Change (\( \Delta S \)):**
- Entropy (\( S \)) is a measure of randomness or disorder in a system.
- Solids have lower entropy compared to liquids and gases because their particles are more ordered.
- In this reaction, we are converting solid \( \text{HgO} \) to liquid \( \text{Hg} \) and gaseous \( \text{O}_2 \).
- Since the products (liquid and gas) are more disordered than the reactants (solid), we can conclude that:
\[ S_{\text{products}} > S_{\text{reactants}} \quad \Rightarrow \quad \Delta S > 0 \]

3. **Calculate Enthalpy Change (\( \Delta H \)):**
- The reaction involves the decomposition of \( \text{HgO} \), which requires energy to break the bonds in \( \text{HgO} \).
- Since energy is absorbed to break the bonds, the change in enthalpy (\( \Delta H \)) for this reaction is positive:
\[ \Delta H > 0 \]

4. **Summarize the Results:**
- From our analysis, we find:
\[ \Delta S > 0 \quad \text{and} \quad \Delta H > 0 \]

5. **Conclusion:**
- Both \( \Delta H \) and \( \Delta S \) are greater than zero for the reaction \( 2 \text{HgO}(s) \rightarrow 2 \text{Hg}(l) + \text{O}_2(g) \).

### Final Answer:
- \( \Delta H > 0 \)
- \( \Delta S > 0 \)