Given that equilibrium constant for the reaction `2SO_(2)(g) + O_(2)(g)hArr2SO_(3)(g)` has a value of 278 at a particular temperature. What is the value of the equilibrium constant for the following reaction at the same temperature ? `SO_(3)(g)hArrSO_(2)(g) + (1)/(2)O_(2)(g)`

Options

A. `3.6xx10^(-3)`
B. `6.0xx10^(-2)`
C. `1.3xx10^(-5)`
D. `1.8xx10^(-3)`

Explanation

To find the equilibrium constant for the reaction \( SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \) given that the equilibrium constant for the reaction \( 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \) is 278, we can follow these steps:

### Step 1: Write the given reaction and its equilibrium constant
The given reaction is:
\[ 2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \]
with an equilibrium constant \( K = 278 \).

### Step 2: Reverse the reaction
To find the equilibrium constant for the reverse reaction:
\[ 2SO_3(g) \rightleftharpoons O_2(g) + 2SO_2(g) \]
The equilibrium constant for the reverse reaction is given by:
\[ K' = \frac{1}{K} = \frac{1}{278} \]

### Step 3: Halve the coefficients of the reversed reaction
Now, we need to halve the coefficients of the reversed reaction:
\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \]
When we halve the coefficients of a reaction, the equilibrium constant is raised to the power of \( \frac{1}{2} \):
\[ K'' = (K')^{\frac{1}{2}} = \left(\frac{1}{278}\right)^{\frac{1}{2}} \]

### Step 4: Calculate the new equilibrium constant
Now we calculate \( K'' \):
\[ K'' = \frac{1}{\sqrt{278}} \]

### Step 5: Evaluate the numerical value
Calculating \( \sqrt{278} \):
\[ \sqrt{278} \approx 16.67 \]
Thus,
\[ K'' \approx \frac{1}{16.67} \approx 0.06 \]

### Final Answer
The equilibrium constant for the reaction \( SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \) is approximately:
\[ K'' \approx 0.06 \]