The equilibrium constant of the reaction `SO_2(g)+1/2O_2(g) hArr SO_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. 250 atm
B. `4xx10^(3)` atm
C. `0.25xx10^4` atm
D. `6.25xx10^4` atm

Answer

To find the equilibrium constant for the reaction \(2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)\), we can follow these steps:

### Step 1: Write down the given reaction and its equilibrium constant
The equilibrium constant for the reaction:
\[ SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g) \]
is given as:
\[ K_1 = 4 \times 10^{-3} \, \text{atm}^{-1/2} \]

### Step 2: Reverse the reaction
To find the equilibrium constant for the reaction \(2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)\), we first need to reverse the original reaction. When we reverse a reaction, the equilibrium constant for the new reaction is the reciprocal of the original equilibrium constant.

Thus, for the reaction:
\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2}O_2(g) \]
the equilibrium constant \(K'\) will be:
\[ K' = \frac{1}{K_1} = \frac{1}{4 \times 10^{-3}} = 250 \, \text{atm}^{1/2} \]

### Step 3: Double the coefficients
Next, we need to double the coefficients in the reversed reaction to get:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
When we double the coefficients of a balanced reaction, the equilibrium constant for the new reaction is the original equilibrium constant raised to the power of the coefficient change.

Thus, for our new reaction, the equilibrium constant \(K_2\) will be:
\[ K_2 = (K')^2 = (250)^{2} = 62500 \, \text{atm} \]

### Step 4: Final result
Therefore, the equilibrium constant for the reaction \(2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)\) is:
\[ K_2 = 6.25 \times 10^4 \, \text{atm} \]

### Summary
The equilibrium constant for the reaction \(2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g)\) is \(6.25 \times 10^4 \, \text{atm}\).

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