Which of the following proces is (are) expected to be spontaneous at higher temperature but non-spontaneous at lower temperature?

Options

A. `N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g)`
B. `H_(2)(g) to 2H(g)`
C. `CO_(2)(g) + C(s) to 2CO(g)` `[DeltaH_(f)^(Theta),m, CO_(2) =-394 kJ//mol, CO = -212 kJ//mol]`
D. `Na(s) to Na^(+)(g) + e^(-)`

Explanation

To determine which processes are expected to be spontaneous at higher temperatures but non-spontaneous at lower temperatures, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- \(\Delta G\) is the change in Gibbs free energy.
- \(\Delta H\) is the change in enthalpy.
- \(T\) is the temperature in Kelvin.
- \(\Delta S\) is the change in entropy.

### Step-by-Step Solution:

1. **Understand Spontaneity**: For a process to be spontaneous, \(\Delta G\) must be less than zero (\(\Delta G < 0\)). Conversely, for a process to be non-spontaneous, \(\Delta G\) is greater than zero (\(\Delta G > 0\)).

2. **Analyze Conditions for Non-Spontaneity**:
- A process is non-spontaneous when:
\[ \Delta G > 0 \implies \Delta H - T \Delta S > 0 \]
- Rearranging gives:
\[ \Delta H > T \Delta S \]
- This indicates that at lower temperatures, if \(\Delta H\) is greater than \(T \Delta S\), the process is non-spontaneous.

3. **Identify Conditions for Spontaneity at Higher Temperatures**:
- For a process to become spontaneous at higher temperatures, we need:
\[ \Delta H < T \Delta S \]
- This means that as temperature increases, the \(T \Delta S\) term can outweigh \(\Delta H\), making \(\Delta G\) negative.

4. **Evaluate Each Option**:
- **Option A**: Haber Process (exothermic, \(\Delta H < 0\), \(\Delta S < 0\))
- This process is not suitable as both \(\Delta H\) and \(\Delta S\) are negative.
- **Option B**: H₂(g) → 2H(g) (atomization, \(\Delta H > 0\), \(\Delta S > 0\))
- This process is suitable as it has positive entropy change and positive enthalpy, which can become spontaneous at higher temperatures.
- **Option C**: CO₂(g) + C(s) → 2CO(g) (calculate \(\Delta H\))
- If \(\Delta H < 0\), then it is not suitable since it does not meet the criteria for spontaneity at high temperatures.
- **Option D**: Na(s) → Na⁺(g) + e⁻ (ionization, \(\Delta H > 0\), \(\Delta S > 0\))
- This process is also suitable as it has both positive enthalpy and positive entropy, which can become spontaneous at higher temperatures.

5. **Conclusion**: The processes that are spontaneous at higher temperatures but non-spontaneous at lower temperatures are:
- **Option B**: H₂(g) → 2H(g)
- **Option D**: Na(s) → Na⁺(g) + e⁻

### Final Answer:
The correct options are **B and D**.