An endotthermic reaction is non-spontaneous at freezing point of water and becomes feasible at its boiling point, then:

Options

A. `DeltaH` is `-ve, DeltaS` is `+ve`
B. `DeltaH` and `DeltaS` both are `+ve`
C. `DeltaH` and `DeltaS` both are `-ve`
D. `DeltaH` is `+ve`, `DeltaS` is -ve

Explanation

To solve the problem, we need to analyze the conditions under which an endothermic reaction can become spontaneous. Let's break down the steps:

### Step 1: Understand the spontaneity of reactions
For a reaction to be spontaneous, the change in Gibbs free energy (ΔG) must be negative. The relationship between ΔG, enthalpy change (ΔH), and entropy change (ΔS) is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

### Step 2: Analyze the given conditions
The problem states that the reaction is endothermic, which means:
- ΔH > 0 (positive enthalpy change)

At the freezing point of water (0°C or 273 K), the reaction is non-spontaneous, which implies:
\[ \Delta G > 0 \]

At the boiling point of water (100°C or 373 K), the reaction becomes feasible (spontaneous), which implies:
\[ \Delta G < 0 \]

### Step 3: Apply the Gibbs free energy equation
At the freezing point (T = 273 K):
\[ \Delta G = \Delta H - (273 \, \text{K}) \Delta S > 0 \]
This can be rearranged to:
\[ \Delta H < (273 \, \text{K}) \Delta S \]

At the boiling point (T = 373 K):
\[ \Delta G = \Delta H - (373 \, \text{K}) \Delta S < 0 \]
This can be rearranged to:
\[ \Delta H < (373 \, \text{K}) \Delta S \]

### Step 4: Compare the two inequalities
From the two inequalities, we have:
1. At freezing point: \( \Delta H < (273 \, \text{K}) \Delta S \)
2. At boiling point: \( \Delta H < (373 \, \text{K}) \Delta S \)

Since the reaction is endothermic (ΔH > 0), for the reaction to become spontaneous at higher temperatures, ΔS must be positive. This means that the entropy of the system increases during the reaction.

### Conclusion
Thus, we conclude that for the endothermic reaction to be non-spontaneous at the freezing point and spontaneous at the boiling point, we must have:
- ΔH > 0 (endothermic)
- ΔS > 0 (positive entropy change)

### Final Answer
The correct option is that both ΔH is positive and ΔS is positive.

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