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))`

Answer

To solve the problem, we need to find the equilibrium constant for the reaction:

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

using the given equilibrium constants \( K_1 \) and \( K_2 \) for the two reactions:

1. \[ \text{XeF}_6(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{XeOF}_4(g) + 2\text{HF}(g) \] with equilibrium constant \( K_1 \)

2. \[ \text{XeO}_4(g) + \text{XeF}_6(g) \rightleftharpoons \text{XeOF}_4(g) + \text{XeO}_3\text{F}_2(g) \] with equilibrium constant \( K_2 \)

### Step-by-Step Solution:

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

For the first reaction:
\[ K_1 = \frac{[\text{XeOF}_4][\text{HF}]^2}{[\text{XeF}_6][\text{H}_2\text{O}]} \]

For the second reaction:
\[ K_2 = \frac{[\text{XeOF}_4][\text{XeO}_3\text{F}_2]}{[\text{XeO}_4][\text{XeF}_6]} \]

**Step 2: Rearrange the second reaction to isolate \(\text{XeF}_6\)**

From the second reaction, we can express \([\text{XeF}_6]\) in terms of the other species:
\[ [\text{XeF}_6] = \frac{[\text{XeOF}_4][\text{XeO}_3\text{F}_2]}{K_2[\text{XeO}_4]} \]

**Step 3: Substitute \([\text{XeF}_6]\) into the \( K_1 \) expression**

Now, we substitute \([\text{XeF}_6]\) into the expression for \( K_1 \):
\[ K_1 = \frac{[\text{XeOF}_4][\text{HF}]^2}{\left(\frac{[\text{XeOF}_4][\text{XeO}_3\text{F}_2]}{K_2[\text{XeO}_4]}\right)[\text{H}_2\text{O}]} \]

**Step 4: Simplify the expression**

Rearranging gives:
\[ K_1 = \frac{K_2[\text{XeO}_4][\text{HF}]^2}{[\text{XeOF}_4][\text{XeO}_3\text{F}_2][\text{H}_2\text{O}]} \]

**Step 5: Rearranging for \( K_3 \)**

Now we want to express \( K_3 \) for the desired reaction:
\[ K_3 = \frac{[\text{XeO}_3\text{F}_2][\text{H}_2\text{O}]}{[\text{XeO}_4][\text{HF}]^2} \]

From the rearranged \( K_1 \), we can see that:
\[ K_3 = \frac{K_2}{K_1} \]

**Step 6: Final expression for \( K_3 \)**

Thus, the equilibrium constant 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:
\[ K_3 = \frac{K_2}{K_1} \]

### Conclusion

The answer to the question is:
\[ K_3 = \frac{K_2}{K_1} \]