Consider the following equilibrium:
`SO_(2)(g)+(1)/(2)O_(2)(g)overset(K_(1))(hArr)SO_(3)(g),`
`2SO_(3)(g) overset(K_(2))(hArr)` `2SO_(2)(g) + O_(2) (g)`
What is the relation between `K_(1)` and `K_(2)`?

Options

A. (a) `K_(1)=1/K_(2)`
B. (b) `K_(1)=1/sqrt(K_(2))`
C. (c) `K_(1)=K_(2)`
D. (d) `K_(1)=1/K_(2)^(2)`

Explanation

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we can follow these steps:

### Step 1: Write the reactions and their equilibrium constants
1. The first reaction is:
\[ \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \quad (K_1) \]
2. The second reaction is:
\[ 2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g) \quad (K_2) \]

### Step 2: Reverse the first reaction
When we reverse the first reaction, we have:
\[ \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \]
The equilibrium constant for this reversed reaction, which we can call \( K_3 \), is given by:
\[ K_3 = \frac{1}{K_1} \]

### Step 3: Multiply the reversed reaction by 2
Next, we multiply the entire reversed reaction by 2:
\[ 2 \text{SO}_3(g) \rightleftharpoons 2 \text{SO}_2(g) + \text{O}_2(g) \]
The equilibrium constant for this new reaction, which we can call \( K_4 \), is:
\[ K_4 = (K_3)^2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

### Step 4: Relate \( K_4 \) to \( K_2 \)
Since the reaction we obtained after reversing and multiplying is exactly the second reaction, we can equate \( K_4 \) to \( K_2 \):
\[ K_2 = \frac{1}{K_1^2} \]

### Step 5: Rearranging the equation
From the equation \( K_2 = \frac{1}{K_1^2} \), we can express \( K_1 \) in terms of \( K_2 \):
\[ K_1^2 = \frac{1}{K_2} \implies K_1 = \frac{1}{\sqrt{K_2}} \]

### Conclusion
Thus, the relationship between \( K_1 \) and \( K_2 \) is:
\[ K_2 = \frac{1}{K_1^2} \]