The equilibrium constant for the reaction:
`2SO_2(g) +O_2(g) hArr 2SO_3(g)` is 256 at 1000 K. What is the value of K for the reaction:
`SO_3(g)hArr SO_2(g) + 1/2O_2(g)` at the same temperature.

Explanation

To find the equilibrium constant \( K' \) for the reaction \( SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \) given that the equilibrium constant \( K \) for the reaction \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \) is 256 at 1000 K, we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
The equilibrium constant for this reaction is:
\[ K = 256 \]

### Step 2: Reverse the reaction
To find the equilibrium constant for the reaction we need, we first reverse the given reaction:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
When a reaction is reversed, the new equilibrium constant \( K' \) is given by:
\[ K' = \frac{1}{K} \]

### Step 3: Calculate \( K' \)
Substituting the value of \( K \):
\[ K' = \frac{1}{256} \]

### Step 4: Adjust for the stoichiometry
The reaction we need is:
\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \]
This reaction is half of the reversed reaction. When the stoichiometry of a reaction is multiplied by a factor \( n \), the equilibrium constant is raised to the power of \( n \):
\[ K'' = (K')^{\frac{1}{2}} \]

### Step 5: Calculate the final equilibrium constant
Now, substituting \( K' \):
\[ K'' = \left(\frac{1}{256}\right)^{\frac{1}{2}} = \frac{1}{16} \]

### Conclusion
Thus, the value of the equilibrium constant \( K \) for the reaction \( SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \) at 1000 K is:
\[ K = \frac{1}{16} \]

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