The equilibrium constant for the reaction,
`SO_3(g)hArr SO_2(g)+1/2O_2(g)` is `K_C=4.9xx10^(-2)` . The value of `K_C` for the reaction `2SO_2(g) + O_2 hArr 2SO_(3) (g)` will be :

Options

A. 416
B. `2.40xx10^(-3)`
C. `9.8xx10^(-2)`
D. `4.9xx10^(-2)`

Answer

To find the equilibrium constant \( K_C \) for the reaction

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

given that the equilibrium constant for the reaction

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

is \( K_C = 4.9 \times 10^{-2} \), we can follow these steps:

### Step 1: Write down the given reaction and its equilibrium constant
The given reaction is:

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

with

\[ K_C = 4.9 \times 10^{-2} \]

### Step 2: Reverse the reaction
To find \( K_C \) for the reaction \( 2SO_2(g) + O_2 \rightleftharpoons 2SO_3(g) \), we first reverse the given reaction. The equilibrium constant for the reverse reaction is the reciprocal of the original equilibrium constant:

\[ K_C' = \frac{1}{K_C} = \frac{1}{4.9 \times 10^{-2}} \]

### Step 3: Calculate the reciprocal
Now, we calculate \( K_C' \):

\[ K_C' = \frac{1}{4.9 \times 10^{-2}} \approx 20.41 \]

### Step 4: Multiply the reaction by 2
Next, we need to multiply the entire reversed reaction by 2:

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

When we multiply the coefficients in the balanced equation by 2, the equilibrium constant for the new reaction is given by:

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

### Step 5: Calculate \( K_C'' \)
Now we calculate \( K_C'' \):

\[ K_C'' = (20.41)^2 \approx 416.56 \]

### Final Answer
Thus, the value of \( K_C \) for the reaction \( 2SO_2(g) + O_2 \rightleftharpoons 2SO_3(g) \) is approximately:

\[ K_C \approx 416 \]