`{:(List - I , List - II),((A) N_(2) + O_(2) hArr 2NO,(P) (atm)^(-2)),((B) 2SO_(2)(g) + O_(2) hArr 2SO_(3)(g),(Q) "No unit"),((C) H_(2) + I_(2) hArr 2HI,(R)"No cffect of pressure"),((D) N_(2)(g) + 3H_(2) (g) hArr 2NH_(3)(g), (S) "High pressure favours forward reaction"):}`

Answer

To solve the given problem, we need to match the reactions from List I with their corresponding effects or characteristics from List II. Let's analyze each reaction step by step.

### Step 1: Analyze Reaction A
**Reaction:** \( N_2 + O_2 \rightleftharpoons 2NO \)

- **Moles of Reactants:** 2 (1 mole of \( N_2 \) + 1 mole of \( O_2 \))
- **Moles of Products:** 2 (2 moles of \( NO \))
- **Effect of Pressure:** Since the number of moles of reactants equals the number of moles of products, there is no change in volume when pressure is applied. Therefore, there is **no effect of pressure**.
- **Unit of Equilibrium Constant (K):** Since the number of moles is the same on both sides, the equilibrium constant has **no unit**.

**Conclusion for A:** Matches with (Q) "No unit" and (R) "No effect of pressure".

### Step 2: Analyze Reaction B
**Reaction:** \( 2SO_2(g) + O_2 \rightleftharpoons 2SO_3(g) \)

- **Moles of Reactants:** 3 (2 moles of \( SO_2 \) + 1 mole of \( O_2 \))
- **Moles of Products:** 2 (2 moles of \( SO_3 \))
- **Effect of Pressure:** Since the number of moles decreases from 3 to 2, increasing pressure will favor the forward reaction (towards fewer moles).
- **Unit of Equilibrium Constant (K):** The equilibrium constant will have units of \( atm^{-1} \) due to the change in moles.

**Conclusion for B:** Matches with (S) "High pressure favors forward reaction".

### Step 3: Analyze Reaction C
**Reaction:** \( H_2 + I_2 \rightleftharpoons 2HI \)

- **Moles of Reactants:** 2 (1 mole of \( H_2 \) + 1 mole of \( I_2 \))
- **Moles of Products:** 2 (2 moles of \( HI \))
- **Effect of Pressure:** Similar to Reaction A, since the number of moles is the same on both sides, there is **no effect of pressure**.
- **Unit of Equilibrium Constant (K):** The equilibrium constant has **no unit**.

**Conclusion for C:** Matches with (Q) "No unit" and (R) "No effect of pressure".

### Step 4: Analyze Reaction D
**Reaction:** \( N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \)

- **Moles of Reactants:** 4 (1 mole of \( N_2 \) + 3 moles of \( H_2 \))
- **Moles of Products:** 2 (2 moles of \( NH_3 \))
- **Effect of Pressure:** Since the number of moles decreases from 4 to 2, increasing pressure will favor the forward reaction (towards fewer moles).
- **Unit of Equilibrium Constant (K):** The equilibrium constant will have units of \( atm^{-2} \) due to the change in moles.

**Conclusion for D:** Matches with (P) \( (atm)^{-2} \) and (S) "High pressure favors forward reaction".

### Final Matching:
- A: (Q) "No unit", (R) "No effect of pressure"
- B: (S) "High pressure favors forward reaction"
- C: (Q) "No unit", (R) "No effect of pressure"
- D: (P) \( (atm)^{-2} \), (S) "High pressure favors forward reaction"

### Summary of Answers:
- A → Q, R
- B → S
- C → Q, R
- D → P, S