Consider the following two gaseous equilibria involving `SO_2` and the corresponding equilibrium constants at 298 K:
` SO_2(g) +(1)/(2) O_2(g) hArr SO_3 (g) ,K1`
` 2SO_3 (g) hArr 2SO_2 (g) + O_2(g) ,K_2`
The values of equilibrium constants are related by

Options

A. ` K_2 = K_1`
B. ` K_2= K_1^(2)`
C. ` K_2= ( 1)/(K_1^(2))`
D. ` K_2= ( 1)/(K_1)`

Answer

To find the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) for the given reactions, we will analyze each reaction step by step.

### Step 1: Write the reactions and their equilibrium constants
1. The first reaction is:
\[ SO_2(g) + \frac{1}{2} O_2(g) \rightleftharpoons SO_3(g) \quad (K_1) \]
2. The second reaction is:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \quad (K_2) \]

### Step 2: Reverse the first reaction
When we reverse the first reaction, we get:
\[ SO_3(g) \rightleftharpoons SO_2(g) + \frac{1}{2} O_2(g) \]
The equilibrium constant for the reversed reaction is given by:
\[ K' = \frac{1}{K_1} \]

### Step 3: Multiply the reversed reaction by 2
Now, we multiply the reversed reaction by 2:
\[ 2SO_3(g) \rightleftharpoons 2SO_2(g) + O_2(g) \]
When a reaction is multiplied by a coefficient, the equilibrium constant is raised to the power of that coefficient. Therefore, the equilibrium constant for this new reaction becomes:
\[ K_2 = (K')^2 = \left(\frac{1}{K_1}\right)^2 = \frac{1}{K_1^2} \]

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

### Final Answer
The correct option is:
\[ K_2 = \frac{1}{K_1^2} \]

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