If `K_(1)` and `K_(2)` are the equilibrium constant for the two reations
`XeF_(4(g))+H_(2)O_((g))hArrXeOF_(4(g))+2HF_((g))`
`XeO_(4(g))+XeF_(6(g))hArrXeOF_(4(g))+XeO_(3)F_(2(g))`
The equilibrium constant for the reaction
`XeO_(4(g))+2HF_((g))hArrXeO_(3)F_(2(g))+H_(2)O_((g))` is

Options

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

Answer

To find the equilibrium constant for the reaction

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

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

### Step 1: Write down the given reactions and their equilibrium constants.

1. **First Reaction:**
\[ \text{XeF}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{XeOF}_4(g) + 2\text{HF}(g) \quad (K_1) \]

2. **Second Reaction:**
\[ \text{XeO}_4(g) + \text{XeF}_6(g) \rightleftharpoons \text{XeOF}_4(g) + \text{XeO}_3\text{F}_2(g) \quad (K_2) \]

### Step 2: Reverse the second reaction to find the equilibrium constant.

When we reverse the second reaction, the equilibrium constant becomes the reciprocal of \( K_2 \):

\[ \text{XeOF}_4(g) + \text{XeO}_3\text{F}_2(g) \rightleftharpoons \text{XeO}_4(g) + \text{XeF}_6(g) \quad \left( \frac{1}{K_2} \right) \]

### Step 3: Combine the reactions.

Now we will combine the reversed second reaction with the first reaction.

- From the first reaction:
\[ \text{XeF}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{XeOF}_4(g) + 2\text{HF}(g) \]

- From the reversed second reaction:
\[ \text{XeOF}_4(g) + \text{XeO}_3\text{F}_2(g) \rightleftharpoons \text{XeO}_4(g) + \text{XeF}_6(g) \]

### Step 4: Cancel out common species.

When we add these two reactions, we can cancel out the common species:

- \(\text{XeOF}_4(g)\) cancels out.
- The overall reaction becomes:
\[ \text{XeO}_4(g) + 2\text{HF}(g) \rightleftharpoons \text{XeO}_3\text{F}_2(g) + \text{H}_2\text{O}(g) \]

### Step 5: Determine the equilibrium constant for the overall reaction.

The equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions:

\[ K = K_1 \cdot \left( \frac{1}{K_2} \right) = \frac{K_1}{K_2} \]

### Final Answer:

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

\[ K = \frac{K_1}{K_2} \]