To the system,
`LaCl_(3)(s)+H_(2)O(g) hArr LaClO(s)+2HCL(g)-"Heat"` already at equilibrium, more water vapour is added without altering temperature or volume of the system. When equilibrium is re-established, the pressure of water vapour is doubled. The pressure of `HCl` present in the system increases by a factor of

Options

A. `2`
B. `2^(1//2)`
C. `2^(1//3)`
D. `2^(2)`

Answer

To solve the problem, we will analyze the equilibrium reaction and how the addition of water vapor affects the pressures of the gases involved.

### Step-by-Step Solution:

1. **Write the Equilibrium Reaction**:
The reaction given is:
\[ \text{LaCl}_3(s) + \text{H}_2O(g) \rightleftharpoons \text{LaClO}(s) + 2 \text{HCl}(g) - \text{Heat} \]
This indicates that the reaction is endothermic.

2. **Define Initial Conditions**:
Let the initial partial pressure of water vapor (H₂O) be \(Y\) and the initial partial pressure of HCl be \(X\).

3. **Write the Expression for \(K_p\)**:
The equilibrium constant \(K_p\) for the reaction can be expressed as:
\[ K_p = \frac{(P_{\text{HCl}})^2}{P_{\text{H}_2O}} = \frac{X^2}{Y} \]

4. **Change in Conditions**:
When more water vapor is added, the pressure of water vapor doubles. Therefore, the new pressure of water vapor is:
\[ P_{\text{H}_2O}' = 2Y \]
Let the new pressure of HCl be \(X'\).

5. **Write the New Expression for \(K_p\)**:
The equilibrium constant remains the same since the temperature is constant:
\[ K_p = \frac{(P_{\text{HCl}}')^2}{P_{\text{H}_2O}'} = \frac{(X')^2}{2Y} \]

6. **Set the Two Expressions for \(K_p\) Equal**:
Since \(K_p\) remains constant:
\[ \frac{X^2}{Y} = \frac{(X')^2}{2Y} \]

7. **Simplify the Equation**:
Cancel \(Y\) from both sides:
\[ X^2 = \frac{(X')^2}{2} \]

8. **Solve for \(X'\)**:
Rearranging gives:
\[ (X')^2 = 2X^2 \]
Taking the square root of both sides:
\[ X' = \sqrt{2}X \]

9. **Determine the Factor of Increase**:
The pressure of HCl has increased from \(X\) to \(\sqrt{2}X\). Therefore, the factor by which the pressure of HCl increases is:
\[ \text{Factor of Increase} = \sqrt{2} \]

### Final Answer:
The pressure of HCl present in the system increases by a factor of \(\sqrt{2}\).

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