The equilibrium constant for the reation,
`SO_3(g) hArr SO_2( g) +(1)/(2) O_2(g)`
is `K_c= 4.9 xx 10^(-2) ` The value of `K_c` for the reaction
` 2SO_2(g) + O_2(g) hArr 2SO_3(g), `
will be

Options

A. ` 416 `
B. ` 2.4 xx 10 ^(-3)`
C. ` 9.8 xx 10 ^(-2)`
D. ` 4.9 xx 10 ^(-2)`

Explanation

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

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

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 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: Multiply the reaction by 2.
To find the equilibrium constant for the reaction

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

we first multiply the entire reaction by 2:

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

### Step 3: Determine the equilibrium constant for the modified reaction.
When we multiply a reaction by a factor \( n \), the equilibrium constant for the new reaction is given by:

\[ K_1 = K_c^n \]

In our case, \( n = 2 \):

\[ K_1 = (4.9 \times 10^{-2})^2 \]

### Step 4: Calculate \( K_1 \).
Calculating \( K_1 \):

\[ K_1 = (4.9 \times 10^{-2})^2 = 2.401 \times 10^{-3} \]

### Step 5: Reverse the reaction.
Now we reverse the reaction to get:

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

### Step 6: Find the equilibrium constant for the reversed reaction.
When a reaction is reversed, the equilibrium constant becomes the reciprocal:

\[ K_2 = \frac{1}{K_1} \]

Thus,

\[ K_2 = \frac{1}{2.401 \times 10^{-3}} \]

### Step 7: Calculate \( K_2 \).
Calculating \( K_2 \):

\[ K_2 \approx 416.5 \]

### Final Answer:
The equilibrium constant \( K_c \) for the reaction

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

is approximately \( 416.5 \).

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