Match the sign of `DeltaH` and `DeltaS` given in List - I with the type of reacting given in List - II and select the correct option.
`{: ( "List-I", " List-II"),( "(P) +ve and + ve", " (1)Spontaneous at all temperatures.") , ("(Q)-ve and -ve" , "(2)Non spontaneous at all temperatures."),(" (R)+ve and -ve" , "(3)Spontaneous only at low temperature") , ( "(S)-ve and +ve" , " (4)Spontaneous only at high temperature."):}`

Options

A. P-4, Q-2, R-3,S-1
B. P-3, Q-4,R-2, S-1
C. P -3, Q-4, R-1, S-2
D. P-4,Q-3, R-2, S-1

Answer

To solve the question of matching the signs of ΔH (enthalpy change) and ΔS (entropy change) with their corresponding spontaneity conditions, we can use the Gibbs free energy equation:

\[ \Delta G = \Delta H - T \Delta S \]

Where:
- ΔG < 0 indicates a spontaneous process.
- ΔG > 0 indicates a non-spontaneous process.

Now, let's analyze each case provided in List-I:

### Step 1: Analyze Each Case

1. **Case P: ΔH > 0 (positive) and ΔS > 0 (positive)**
- Here, both ΔH and ΔS are positive.
- For spontaneity, we need ΔG < 0:
\[ \Delta G = \Delta H - T \Delta S \]
At high temperatures, TΔS can outweigh ΔH, making ΔG negative. Thus, this process is spontaneous at high temperatures.
- **Match:** P → 4 (Spontaneous only at high temperature)

2. **Case Q: ΔH < 0 (negative) and ΔS < 0 (negative)**
- Here, both ΔH and ΔS are negative.
- For spontaneity:
\[ \Delta G = \Delta H - T \Delta S \]
At low temperatures, ΔG can be negative since ΔH is negative and TΔS is also negative but less significant. Thus, this process is spontaneous at low temperatures.
- **Match:** Q → 3 (Spontaneous only at low temperature)

3. **Case R: ΔH > 0 (positive) and ΔS < 0 (negative)**
- Here, ΔH is positive and ΔS is negative.
- For spontaneity:
\[ \Delta G = \Delta H - T \Delta S \]
Both terms are positive, meaning ΔG will always be positive regardless of temperature. Thus, this process is non-spontaneous at all temperatures.
- **Match:** R → 2 (Non-spontaneous at all temperatures)

4. **Case S: ΔH < 0 (negative) and ΔS > 0 (positive)**
- Here, ΔH is negative and ΔS is positive.
- For spontaneity:
\[ \Delta G = \Delta H - T \Delta S \]
Both terms can lead to ΔG being negative, making the process spontaneous at all temperatures.
- **Match:** S → 1 (Spontaneous at all temperatures)

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

### Conclusion
Thus, the correct option is **D: P-4, Q-3, R-2, S-1**.

---