Match Column-I with Column -II,
`{:("Column -I","Column-II"),(A_2(g)+3B_2(g)hArr2AB_3(g),(RT)^(-2)),(A_2(g)+B_2(g)hArr2AB(g),(RT)^0),(A_2(s)+3/2B_2(g)hArrAB_3(g),(RT)^(1//2)),(AB_2(g)hArrAB(g)+1/2B_2(g),(RT)^(-1//2)),("","Favours forward reaction by rising pressure"):}`

Answer

To solve the problem of matching Column-I with Column-II, we will analyze each reaction in Column-I and determine the corresponding value of \( (RT)^{\Delta n} \) in Column-II based on the change in the number of moles of gas during the reaction.

### Step-by-Step Solution:

1. **Identify the reactions and calculate \( \Delta n \)**:
- For each reaction, we need to find \( \Delta n \), which is defined as:
\[ \Delta n = n_p - n_r \]
where \( n_p \) is the sum of the stoichiometric coefficients of the products (in gaseous phase) and \( n_r \) is the sum of the stoichiometric coefficients of the reactants (in gaseous phase).

2. **Reaction A**: \( A_2(g) + 3B_2(g) \rightleftharpoons 2AB_3(g) \)
- Products: \( 2 \) moles of \( AB_3 \) (gaseous) → \( n_p = 2 \)
- Reactants: \( 1 + 3 = 4 \) moles of \( A_2 \) and \( B_2 \) (gaseous) → \( n_r = 4 \)
- \( \Delta n = 2 - 4 = -2 \)
- Thus, \( K_p/K_c = (RT)^{-2} \).

3. **Reaction B**: \( A_2(g) + B_2(g) \rightleftharpoons 2AB(g) \)
- Products: \( 2 \) moles of \( AB \) (gaseous) → \( n_p = 2 \)
- Reactants: \( 1 + 1 = 2 \) moles of \( A_2 \) and \( B_2 \) (gaseous) → \( n_r = 2 \)
- \( \Delta n = 2 - 2 = 0 \)
- Thus, \( K_p/K_c = (RT)^{0} \).

4. **Reaction C**: \( A_2(s) + \frac{3}{2}B_2(g) \rightleftharpoons AB_3(g) \)
- Products: \( 1 \) mole of \( AB_3 \) (gaseous) → \( n_p = 1 \)
- Reactants: \( 0 + \frac{3}{2} = \frac{3}{2} \) moles of \( B_2 \) (gaseous) → \( n_r = \frac{3}{2} \)
- \( \Delta n = 1 - \frac{3}{2} = -\frac{1}{2} \)
- Thus, \( K_p/K_c = (RT)^{-1/2} \).

5. **Reaction D**: \( AB_2(g) \rightleftharpoons AB(g) + \frac{1}{2}B_2(g) \)
- Products: \( 1 + \frac{1}{2} = \frac{3}{2} \) moles of \( AB \) and \( B_2 \) (gaseous) → \( n_p = \frac{3}{2} \)
- Reactants: \( 1 \) mole of \( AB_2 \) (gaseous) → \( n_r = 1 \)
- \( \Delta n = \frac{3}{2} - 1 = \frac{1}{2} \)
- Thus, \( K_p/K_c = (RT)^{1/2} \).

6. **Determine which reactions favor the forward reaction by rising pressure**:
- According to Le Chatelier's principle, increasing pressure favors the reaction that produces fewer moles of gas.
- For Reaction A: \( 4 \) moles (reactants) → \( 2 \) moles (products) → Favors forward.
- For Reaction B: \( 2 \) moles (reactants) → \( 2 \) moles (products) → No effect.
- For Reaction C: \( \frac{3}{2} \) moles (reactants) → \( 1 \) mole (products) → Favors forward.
- For Reaction D: \( 1 \) mole (reactants) → \( \frac{3}{2} \) moles (products) → Favors backward.

### Final Matches:
- **A**: \( (RT)^{-2} \)
- **B**: \( (RT)^{0} \)
- **C**: \( (RT)^{-1/2} \)
- **D**: \( (RT)^{1/2} \)