For the reaction `2H(g)rarr H_(2)(g)`, the sign of `DeltaH` and `DeltaS` respectively are :

Options

A. `+, -`
B. `+,+`
C. `-,+`
D. `-,-`

Explanation

To determine the signs of ΔH (enthalpy change) and ΔS (entropy change) for the reaction \( 2H(g) \rightarrow H_2(g) \), we can analyze the process step by step.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction involves two hydrogen atoms (H) combining to form one molecule of hydrogen gas (H2).

2. **Analyzing ΔH (Enthalpy Change)**:
- In this reaction, energy is released when two hydrogen atoms bond to form a molecule of H2. This is because forming bonds generally releases energy.
- Therefore, since energy is released, ΔH will be negative.
- **Conclusion**: \( ΔH < 0 \) (negative).

3. **Analyzing ΔS (Entropy Change)**:
- Entropy (S) is a measure of disorder or randomness in a system.
- In this reaction, we start with 2 gaseous hydrogen atoms and end up with 1 gaseous hydrogen molecule.
- The formation of one molecule from two atoms represents a decrease in the number of gas particles, leading to a more ordered state.
- Therefore, the entropy decreases, which means ΔS will be negative.
- **Conclusion**: \( ΔS < 0 \) (negative).

4. **Final Result**:
- The signs of ΔH and ΔS for the reaction \( 2H(g) \rightarrow H_2(g) \) are:
- \( ΔH < 0 \) (negative)
- \( ΔS < 0 \) (negative)

### Summary:
- The signs of ΔH and ΔS for the reaction \( 2H(g) \rightarrow H_2(g) \) are both negative.