Increase in the presssure for the following equilibrium results in the :
`H_(2)O(l)hArrH_(2)O(g)`
Equilibrium will shift left

Options

A. formation of more `H_(2)O` (l)
B. formation of more `H_(2)O(g)`
C. increase in b.p of `H_(2)O(l)`
D. decrease in b.p. of `H_(2)O(l)`

Answer

To solve the question regarding the effect of increased pressure on the equilibrium of the reaction \( H_2O(l) \rightleftharpoons H_2O(g) \), we will follow these steps:

### Step 1: Understand the Reaction
The equilibrium reaction is between liquid water (\( H_2O(l) \)) and gaseous water (\( H_2O(g) \)).

### Step 2: Analyze the States of Matter
In this equilibrium:
- The left side (reactants) has liquid water, which occupies a smaller volume compared to the right side (products), which has gaseous water.

### Step 3: Apply Le Chatelier's Principle
According to Le Chatelier's Principle, if an external change (like pressure) is applied to a system at equilibrium, the system will adjust to counteract that change.

### Step 4: Consider the Effect of Increased Pressure
When the pressure is increased, the equilibrium will shift towards the side with fewer moles of gas. In this case:
- The left side has 1 mole of liquid (which does not contribute to gas volume) and the right side has gaseous water, which occupies more volume.

### Step 5: Determine the Direction of the Shift
Since the left side has fewer moles of gas (in fact, zero gas), increasing the pressure will shift the equilibrium to the left, favoring the formation of more liquid water.

### Step 6: Conclusion
Thus, the equilibrium will shift to the left when the pressure is increased.

### Final Answer
The equilibrium will shift left, favoring the formation of more \( H_2O(l) \).

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