If `K_(1)` and `K_(2)` are respective equilibrium constants for two reactions `:`
`XeF_(6)(g) +H_(2)O hArr XeOF_(4)(g) +2HF_(g)`
`XeO_(4)(g)+XeF_(6)(g)hArr XeOF_(4)(g)+XeO_(3)F_(2)(g)`
Then equilibrium constant for the reaction
`XeO_(4)(g)+2HF(g) hArr XeO_(3)F_(2)(g)+H_(2)O(g)` will be

Options

A. `K_1//(K_2)^2`
B. `K_1.K_2`
C. `K_1//K_2`
D. `K_2//K_1`

Explanation

To find the equilibrium constant \( K_3 \) for the reaction

\[ \text{XeO}_4(g) + 2 \text{HF}(g) \rightleftharpoons \text{XeO}_3\text{F}_2(g) + \text{H}_2\text{O}(g), \]

we will use the equilibrium constants \( K_1 \) and \( K_2 \) from the given reactions.

### Step 1: Write the expressions for \( K_1 \) and \( K_2 \)

For the first reaction:

\[ \text{XeF}_6(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{XeOF}_4(g) + 2 \text{HF}(g) \]

The equilibrium constant \( K_1 \) is given by:

\[ K_1 = \frac{[\text{XeOF}_4][\text{HF}]^2}{[\text{XeF}_6][\text{H}_2\text{O}]} \]

For the second reaction:

\[ \text{XeO}_4(g) + \text{XeF}_6(g) \rightleftharpoons \text{XeOF}_4(g) + \text{XeO}_3\text{F}_2(g) \]

The equilibrium constant \( K_2 \) is given by:

\[ K_2 = \frac{[\text{XeOF}_4][\text{XeO}_3\text{F}_2]}{[\text{XeO}_4][\text{XeF}_6]} \]

### Step 2: Write the expression for \( K_3 \)

For the reaction we are interested in:

\[ \text{XeO}_4(g) + 2 \text{HF}(g) \rightleftharpoons \text{XeO}_3\text{F}_2(g) + \text{H}_2\text{O}(g) \]

The equilibrium constant \( K_3 \) is given by:

\[ K_3 = \frac{[\text{XeO}_3\text{F}_2][\text{H}_2\text{O}]}{[\text{XeO}_4][\text{HF}]^2} \]

### Step 3: Relate \( K_3 \) to \( K_1 \) and \( K_2 \)

To express \( K_3 \) in terms of \( K_1 \) and \( K_2 \), we can manipulate the first and second reactions.

1. **Reverse the first reaction** to express \( \text{XeF}_6 \) and \( \text{H}_2\text{O} \) in terms of products:

\[ \text{XeOF}_4(g) + 2 \text{HF}(g) \rightleftharpoons \text{XeF}_6(g) + \text{H}_2\text{O}(g) \]

The equilibrium constant for this reversed reaction is:

\[ K' = \frac{[\text{XeF}_6][\text{H}_2\text{O}]}{[\text{XeOF}_4][\text{HF}]^2} = \frac{1}{K_1} \]

2. **Add the reversed first reaction to the second reaction**:

\[ \text{XeO}_4(g) + \text{XeF}_6(g) \rightleftharpoons \text{XeOF}_4(g) + \text{XeO}_3\text{F}_2(g) \]

This gives:

\[ \text{XeO}_4(g) + 2 \text{HF}(g) \rightleftharpoons \text{XeO}_3\text{F}_2(g) + \text{H}_2\text{O}(g) \]

### Step 4: Combine the constants

The equilibrium constant for the overall reaction can be expressed as:

\[ K_3 = K_2 \cdot K' \]

Substituting \( K' = \frac{1}{K_1} \):

\[ K_3 = K_2 \cdot \frac{1}{K_1} = \frac{K_2}{K_1} \]

### Final Answer

Thus, the equilibrium constant \( K_3 \) for the reaction

\[ \text{XeO}_4(g) + 2 \text{HF}(g) \rightleftharpoons \text{XeO}_3\text{F}_2(g) + \text{H}_2\text{O}(g) \]

is

\[ K_3 = \frac{K_2}{K_1}. \]