Which are true for the reaction: `A_(2) hArr 2C+D`?

Options

A. if `DeltaH=0, K_(p)` and increases with temperature and dissociation.
B. if `DeltaH=+ve, K_(p)` increases with temperature and dissociation of `A_(2)` increases.
C. if `DeltaH=-ve, K_(p)` decreases with temperature and dissociation of `A_(2)` decreases..
D. `K_(p)=4alpha^(3)[P/(1+2alpha)]^(2)`

Explanation

To determine which statements are true for the reaction \( A_2 \rightleftharpoons 2C + D \), we will analyze the thermodynamic properties of the reaction and the implications of the changes in enthalpy (\( \Delta H \)) on the equilibrium constant (\( K_p \)) and the dissociation of \( A_2 \).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction is \( A_2 \rightleftharpoons 2C + D \). Here, one mole of \( A_2 \) dissociates into two moles of \( C \) and one mole of \( D \).

2. **Analyzing the First Statement**:
- **Statement**: If \( \Delta H = 0 \), \( K_p \) and the dissociation of \( A_2 \) do not change with temperature.
- **Explanation**: When \( \Delta H = 0 \), the reaction is neither endothermic nor exothermic. Therefore, the equilibrium constant \( K_p \) remains constant with temperature changes. Thus, this statement is **true**.

3. **Analyzing the Second Statement**:
- **Statement**: If \( \Delta H > 0 \), \( K_p \) increases with temperature and the dissociation of \( A_2 \) increases.
- **Explanation**: A positive \( \Delta H \) indicates that the reaction is endothermic. According to Le Chatelier's principle, increasing the temperature will shift the equilibrium to favor the formation of products, thereby increasing \( K_p \) and the dissociation of \( A_2 \). Thus, this statement is **true**.

4. **Analyzing the Third Statement**:
- **Statement**: If \( \Delta H < 0 \), \( K_p \) decreases with temperature and the dissociation of \( A_2 \) decreases.
- **Explanation**: A negative \( \Delta H \) indicates that the reaction is exothermic. Increasing the temperature will shift the equilibrium to favor the reactants, leading to a decrease in \( K_p \) and a decrease in the dissociation of \( A_2 \). Thus, this statement is **true**.

5. **Analyzing the Fourth Statement**:
- **Statement**: The expression for \( K_p \) is derived correctly.
- **Derivation**: At equilibrium, if \( \alpha \) moles of \( A_2 \) dissociate, we have:
- Moles of \( A_2 \) at equilibrium = \( 1 - \alpha \)
- Moles of \( C \) at equilibrium = \( 2\alpha \)
- Moles of \( D \) at equilibrium = \( \alpha \)
- Total moles at equilibrium = \( 1 + 2\alpha \)
- The expression for \( K_p \) can be derived as:
\[ K_p = \frac{(2\alpha)(\alpha)}{(1 - \alpha)} = \frac{2\alpha^2}{(1 - \alpha)} \text{ (after substituting for partial pressures)} \]
- After simplification, we find that \( K_p \) is indeed a function of \( \alpha \) and the total pressure, confirming the derivation is correct. Thus, this statement is **true**.

### Conclusion:
All four statements regarding the reaction \( A_2 \rightleftharpoons 2C + D \) are true.