Match List I to List II
List-I
P- `K_p=20 atm^(-1)`
Q-`H_(2(g))+I_(2(g))hArr2HI_(g)`
R- `N_2+3H_2hArr2NH_3`, `DeltaH=-ve`
S-`DeltaG_(pT)gt 0`
List II
1- No effect of pressure on the yield of the product
2-`DeltaH=DeltaE-RT`)
3- Non-spontaneous process
4- `k_p` decreases with temperature

Options

A. a) `{:(P,Q,R,S),(2,1,4,3):}`
B. b) `{:(P,Q,R,S),(2,3,1,4):}`
C. c) `{:(P,Q,R,S),(3,2,1,4):}`
D. d) `{:(P,Q,R,S),(2,4,1,3):}`

Answer

To solve the problem of matching List I to List II, we will analyze each item in List I and determine the corresponding item in List II step by step.

### Step 1: Analyze Item P
- **Given:** \( K_p = 20 \, \text{atm}^{-1} \)
- **Analysis:** The equilibrium constant \( K_p \) has units of \( \text{atm}^{-1} \). This indicates that the change in the number of moles of gas (\( \Delta n \)) is negative, specifically \( \Delta n = \text{moles of products} - \text{moles of reactants} = -1 \).
- **Conclusion:** This suggests that as the reaction proceeds, the number of moles of gas decreases, which implies that \( K_p \) decreases with temperature due to Le Chatelier's principle. Therefore, **P matches with 4**.

### Step 2: Analyze Item Q
- **Given:** \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \)
- **Analysis:** Here, we have 1 mole of \( H_2 \) and 1 mole of \( I_2 \) reacting to form 2 moles of \( HI \). Thus, \( \Delta n = 2 - 2 = 0 \).
- **Conclusion:** Since the number of moles of reactants equals the number of moles of products, there will be no effect of pressure on the yield of the product. Therefore, **Q matches with 1**.

### Step 3: Analyze Item R
- **Given:** \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \), \( \Delta H = -\text{ve} \)
- **Analysis:** This reaction is exothermic (\( \Delta H < 0 \)). The number of moles of products (2) is less than the number of moles of reactants (4), leading to a decrease in the total number of moles.
- **Conclusion:** As the reaction releases heat, increasing the temperature will shift the equilibrium to the left (towards reactants), thus decreasing \( K_p \). Therefore, **R matches with 4**.

### Step 4: Analyze Item S
- **Given:** \( \Delta G_{pT} > 0 \)
- **Analysis:** A positive Gibbs free energy change (\( \Delta G > 0 \)) indicates that the process is non-spontaneous.
- **Conclusion:** Therefore, **S matches with 3**.

### Final Matching
- **P matches with 4**
- **Q matches with 1**
- **R matches with 2**
- **S matches with 3**

### Summary of Matches
- P → 4
- Q → 1
- R → 2
- S → 3