The equilibrium constant of the reaction,
`SO_(3)(g) hArr SO_(2)(g)+1//2 O_(2)(g)`
is `0.15` at `900 K`. Calculate the equilibrium constant for
`2SO_(2)(g)+O_(2)(g) hArr 2SO_(3)(g)`

Explanation

To find the equilibrium constant for the reaction

\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]

given that the equilibrium constant for the reaction

\[ \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \]

is \( K = 0.15 \) at \( 900 \, K \), we can follow these steps:

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

\[ \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \]

with \( K = 0.15 \).

### Step 2: Invert the first reaction.
To find the equilibrium constant for the reverse reaction, we invert the first reaction:

\[ \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \]

The equilibrium constant for the reverse reaction, denoted as \( K' \), is the reciprocal of \( K \):

\[ K' = \frac{1}{K} = \frac{1}{0.15} \]

### Step 3: Calculate \( K' \).
Calculating \( K' \):

\[ K' = \frac{1}{0.15} = 6.67 \]

### Step 4: Multiply the reaction by 2.
Now we multiply the entire reaction by 2 to match the desired reaction:

\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]

When we multiply a reaction by a factor, the equilibrium constant is raised to the power of that factor. Therefore, the new equilibrium constant \( K'' \) will be:

\[ K'' = (K')^2 = (6.67)^2 \]

### Step 5: Calculate \( K'' \).
Calculating \( K'' \):

\[ K'' = 6.67^2 = 44.49 \]

### Final Answer:
Thus, the equilibrium constant for the reaction

\[ 2 \text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2 \text{SO}_3(g) \]

is approximately

\[ K'' \approx 44.49 \]

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