In the following equilibria
`I:A+2B hArr C, K_(eq)=K_(1)`
`II: C+DhArr 3A, K_(eq)=K_(2)`
`III, 6B+D hArr 2C, K_(eq)=K_(3)`
Hence,

Options

A. `3K_(1)xxK_(2)=K_(3)`
B. `K_(1)^(3)xxK_(2)^(2)=K_(3)`
C. `3K_(1)xxK_(2)^(2)=K_(3)`
D. `K_(1)^(3)xx K_(2)=K_(3)`

Answer

To solve the problem, we need to find the equilibrium constant \( K_3 \) for the reaction:

\[ 6B + D \rightleftharpoons 2C \]

using the given equilibria:

1. \( A + 2B \rightleftharpoons C \) with \( K_{eq} = K_1 \)
2. \( C + D \rightleftharpoons 3A \) with \( K_{eq} = K_2 \)
3. \( 6B + D \rightleftharpoons 2C \) with \( K_{eq} = K_3 \)

### Step 1: Analyze the first equilibrium
The first equilibrium is:

\[ A + 2B \rightleftharpoons C \]

The equilibrium constant for this reaction is given as \( K_1 \).

### Step 2: Multiply the first equilibrium by 3
To obtain \( 6B \) and \( 2C \), we need to multiply the first reaction by 3:

\[ 3(A + 2B \rightleftharpoons C) \implies 3A + 6B \rightleftharpoons 3C \]

When we multiply a reaction by a factor \( n \), the equilibrium constant is raised to the power of \( n \):

\[ K_1' = K_1^3 \]

### Step 3: Analyze the second equilibrium
The second equilibrium is:

\[ C + D \rightleftharpoons 3A \]

The equilibrium constant for this reaction is given as \( K_2 \).

### Step 4: Reverse the second equilibrium
To use the second equilibrium, we need to reverse it to get \( A \) on the left side:

\[ 3A \rightleftharpoons C + D \]

The equilibrium constant for the reversed reaction is:

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

### Step 5: Combine the modified first and reversed second equilibria
Now we can add the modified first equilibrium and the reversed second equilibrium:

1. From the modified first equilibrium:
\[ 3A + 6B \rightleftharpoons 3C \]
with \( K_1' = K_1^3 \)

2. From the reversed second equilibrium:
\[ 3A \rightleftharpoons C + D \]
with \( K_2' = \frac{1}{K_2} \)

Adding these two reactions gives:

\[ 6B + D \rightleftharpoons 2C \]

### Step 6: Calculate the equilibrium constant for the combined reaction
When adding two reactions, the equilibrium constants multiply:

\[ K_3 = K_1' \times K_2' = K_1^3 \times \frac{1}{K_2} \]

Thus, we have:

\[ K_3 = \frac{K_1^3}{K_2} \]

### Final Answer
The equilibrium constant \( K_3 \) for the reaction \( 6B + D \rightleftharpoons 2C \) is:

\[ K_3 = \frac{K_1^3}{K_2} \]