The equilibrium concentration of `C_(2)H_(4)` in the gas phase reaction `C_(2)H_(4)+H_(2)hArr C_(2)H_(6), DeltaH=-32.7` kcal/mole, can be increased by
I. Removal of `C_(2)H_(6)` II. Removal of `H_(2)` III. Increasing temperature IV. Increasing pressure
The correct choice is :

Options

A. I, II are correct
B. I, III are correct
C. II, III are correct
D. I, II, III are correct

Answer

To determine how to increase the equilibrium concentration of \( C_2H_4 \) in the reaction

\[ C_2H_4 + H_2 \rightleftharpoons C_2H_6, \quad \Delta H = -32.7 \text{ kcal/mole} \]

we will analyze each of the given options based on Le Chatelier's principle.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- The reaction is exothermic (as indicated by the negative \( \Delta H \)), meaning it releases heat.
- The forward reaction produces \( C_2H_6 \) from \( C_2H_4 \) and \( H_2 \).

2. **Option I: Removal of \( C_2H_6 \)**:
- If \( C_2H_6 \) is removed, the equilibrium will shift to the right to produce more \( C_2H_6 \) to replace what was removed.
- This means that the concentration of \( C_2H_4 \) will not increase; instead, it will decrease as more \( C_2H_4 \) is converted to \( C_2H_6 \).
- **Conclusion**: This option does not help in increasing the concentration of \( C_2H_4 \).

3. **Option II: Removal of \( H_2 \)**:
- Removing \( H_2 \) (a reactant) will shift the equilibrium to the left to produce more \( H_2 \).
- As a result, the concentration of \( C_2H_4 \) will increase.
- **Conclusion**: This option helps in increasing the concentration of \( C_2H_4 \).

4. **Option III: Increasing Temperature**:
- For an exothermic reaction, increasing the temperature shifts the equilibrium to the left (toward the reactants) to absorb the added heat.
- This will lead to an increase in the concentration of \( C_2H_4 \).
- **Conclusion**: This option also helps in increasing the concentration of \( C_2H_4 \).

5. **Option IV: Increasing Pressure**:
- In this reaction, there are 2 moles of gas on the reactant side (\( C_2H_4 + H_2 \)) and 1 mole on the product side (\( C_2H_6 \)).
- Increasing pressure will shift the equilibrium towards the side with fewer moles of gas, which is the product side (\( C_2H_6 \)).
- This means that the concentration of \( C_2H_4 \) will decrease.
- **Conclusion**: This option does not help in increasing the concentration of \( C_2H_4 \).

### Final Answer:
The correct choices that can increase the equilibrium concentration of \( C_2H_4 \) are:
- II. Removal of \( H_2 \)
- III. Increasing temperature