Consider the following reversible conversion:
Ortho`(H_(2))hArr"Para "(H_(2))`
this equilibrium will shift in forward direction

Options

A. On increasing temperature
B. On increasing ortho concentration
C. On decreasing temperature
D. On decreasing ortho concentration

Answer

To solve the problem regarding the equilibrium shift in the conversion of ortho-H₂ to para-H₂, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**: The reaction we are dealing with is:
\[ \text{Ortho-H}_2 \rightleftharpoons \text{Para-H}_2 \]
This is a reversible reaction where ortho-H₂ can convert to para-H₂ and vice versa.

2. **Apply Le Chatelier's Principle**: According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium will shift in a direction that counteracts the change.

3. **Analyze the Effects of Concentration**:
- If the concentration of ortho-H₂ is increased, the equilibrium will shift to the right (forward direction) to produce more para-H₂.
- Conversely, if the concentration of para-H₂ is increased, the equilibrium will shift to the left (backward direction).

4. **Analyze the Effects of Temperature**:
- The conversion from ortho-H₂ to para-H₂ is favored at lower temperatures. Thus, if the temperature is decreased, the equilibrium will shift to the right (forward direction) to produce more para-H₂.
- If the temperature is increased, the equilibrium will shift to the left (backward direction).

5. **Summarize the Conditions**:
- **Increasing ortho-H₂ concentration**: Shifts equilibrium to the right (forward).
- **Decreasing temperature**: Shifts equilibrium to the right (forward).

6. **Conclusion**: The equilibrium will shift in the forward direction under the conditions of increasing ortho-H₂ concentration and decreasing temperature.

### Final Answer:
The equilibrium will shift in the forward direction when:
- The concentration of ortho-H₂ is increased.
- The temperature is decreased.