Decreasing in the pressure for the following equilibria: `H_(2) O_((s)) hArr H_(2)O_((l))` result in the :

Options

A. Formation of more `H_(2)O_((s))`
B. Formation more ` H_(2)O_((l))`
C. Increase in melting point of `H_(2)O_((s))`
D. Decrease in melting point of `H_(2) O_((s))`

Answer

To solve the problem regarding the effect of decreasing pressure on the equilibrium of the reaction \( H_2O_{(s)} \rightleftharpoons H_2O_{(l)} \), we can follow these steps:

### Step 1: Understand the Reaction
The reaction involves the phase change of water from solid (ice) to liquid (water). The equilibrium can be represented as:
\[ H_2O_{(s)} \rightleftharpoons H_2O_{(l)} \]

### Step 2: Analyze the Volume Changes
In this equilibrium:
- The solid state (ice) has a greater volume compared to the liquid state (water).
- Therefore, when the solid ice melts into liquid water, the volume decreases.

### Step 3: Apply Le Chatelier's Principle
According to Le Chatelier's Principle, if a system at equilibrium is subjected to a change in pressure, the equilibrium will shift in the direction that counteracts the change.

### Step 4: Consider the Effect of Decreasing Pressure
- When the pressure is decreased, the equilibrium shifts towards the side with a greater volume to counteract the change.
- Since the solid phase (ice) has a greater volume than the liquid phase (water), the equilibrium will shift to the left (towards the solid).

### Step 5: Conclusion
Thus, decreasing the pressure will result in the formation of more solid \( H_2O \) (ice). Therefore, the correct answer is:
**Formation of more \( H_2O_{(s)} \)**.

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