The equilibrium constant of the reaction `SO_(2)(g) + 1//2O_(2)(g)hArrSO_(3)(g)` is `4xx10^(-3)atm^(-1//2)`. The equilibrium constant of the reaction
`2SO_(3)(g)hArr2SO_(2)(g) + O_(2)(g)` would be:

Options

A. `250atm`
B. `4xx10^(3) atm`
C. `0.25xx10^(4)atm`
D. `6.25xx10^(4)atm`

Explanation

To find the equilibrium constant for the reaction

\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]

we start from the given equilibrium constant for the reaction

\[ SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g) \]

with an equilibrium constant \( K = 4 \times 10^{-3} \, \text{atm}^{-1/2} \).

### Step 1: Reverse the Reaction
First, we reverse the original reaction. The equilibrium constant for the reverse reaction is given by:

\[ K' = \frac{1}{K} \]

### Step 2: Calculate \( K' \)
Substituting the value of \( K \):

\[ K' = \frac{1}{4 \times 10^{-3}} = \frac{1}{4} \times 10^{3} = 0.25 \times 10^{3} = 2.5 \times 10^{2} \, \text{atm}^{1/2} \]

### Step 3: Multiply the Reaction by 2
Next, we multiply the reversed reaction by 2:

\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]

When we multiply a reaction by a coefficient, the equilibrium constant is raised to the power of that coefficient:

\[ K'' = (K')^2 \]

### Step 4: Calculate \( K'' \)
Substituting the value of \( K' \):

\[ K'' = (2.5 \times 10^{2})^2 = 6.25 \times 10^{4} \, \text{atm} \]

### Final Answer
Thus, the equilibrium constant for the reaction

\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]

is

\[ K'' = 6.25 \times 10^{4} \, \text{atm} \]

---