For the reaction `H_(2)O_((s))iff H_(2)O_((l))" at "0^(@)C` and normal pressure

Options

A. `DeltaH gt TDeltaS`
B. `DeltaH = TDeltaS`
C. `DeltaH=DeltaG`
D. `DeltaH lt TDeltaS`

Explanation

To solve the problem regarding the equilibrium between solid water (ice) and liquid water at 0°C and normal pressure, we can follow these steps:

### Step 1: Write the Reaction
The equilibrium reaction can be expressed as:
\[ H_2O(s) \rightleftharpoons H_2O(l) \]

### Step 2: Understand the Conditions
This reaction occurs at a temperature of 0°C (273.15 K) and at normal atmospheric pressure (1 atm). Under these conditions, ice and liquid water can coexist in equilibrium.

### Step 3: Identify the Thermodynamic Concept
This reaction is an example of a phase transition, specifically the melting of ice. At the melting point, the solid and liquid phases of water are in equilibrium.

### Step 4: Apply the Gibbs Free Energy Concept
At equilibrium, the change in Gibbs free energy (ΔG) for the reaction is zero:
\[ \Delta G = \Delta H - T\Delta S = 0 \]
Where:
- ΔH is the enthalpy change of the reaction,
- T is the temperature in Kelvin,
- ΔS is the entropy change of the reaction.

### Step 5: Analyze the Enthalpy and Entropy Changes
1. **Enthalpy Change (ΔH)**: The enthalpy change for the melting of ice is positive because heat is absorbed when ice melts.
2. **Entropy Change (ΔS)**: The entropy increases as ice (solid) transitions to water (liquid) because the liquid state has greater disorder than the solid state.

### Step 6: Conclude the Equilibrium Condition
At 0°C and 1 atm, the system is at equilibrium, meaning that the rate of melting of ice equals the rate of freezing of water.

### Final Answer
The equilibrium between solid water (ice) and liquid water at 0°C and normal pressure can be represented by the reaction:
\[ H_2O(s) \rightleftharpoons H_2O(l) \]
This equilibrium is characterized by a specific enthalpy and entropy change, which are balanced at this temperature and pressure.

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