Consider the following two equilibrium established together in a closed container
`A(s)hArr2B(g)+3C(g) , K_(P_(1)`
`A(s)hArr3D(g) , K_(P_(2)`
Starting with only `A(s)`, molar ratio of `B(g)` & `D(g)` at equilibrium is found to be in a ratio `1:6` determine `(K_(P_(2)))/(8K_(P_(1)))`

Explanation

To solve the problem, we need to analyze the two equilibrium reactions and their respective equilibrium constants. Let's break it down step by step.

### Step 1: Write the reactions and define variables
We have two reactions:
1. \( A(s) \rightleftharpoons 2B(g) + 3C(g) \) with equilibrium constant \( K_{P1} \)
2. \( A(s) \rightleftharpoons 3D(g) \) with equilibrium constant \( K_{P2} \)

Let:
- \( P_B \) = partial pressure of \( B \)
- \( P_C \) = partial pressure of \( C \)
- \( P_D \) = partial pressure of \( D \)

### Step 2: Establish the relationships at equilibrium
From the problem, we know the molar ratio of \( B \) to \( D \) at equilibrium is \( 1:6 \). This can be expressed as:
\[ \frac{P_B}{P_D} = \frac{1}{6} \]

### Step 3: Express partial pressures in terms of a common variable
Let’s denote the partial pressure of \( B \) as \( P_B = 2P_1 \) (from the first reaction) and the partial pressure of \( D \) as \( P_D = 3P_2 \) (from the second reaction). Thus, we can write:
\[ \frac{2P_1}{3P_2} = \frac{1}{6} \]

### Step 4: Solve for \( P_1 \) in terms of \( P_2 \)
Cross-multiplying gives:
\[ 2P_1 \cdot 6 = 3P_2 \implies 12P_1 = 3P_2 \implies P_1 = \frac{P_2}{4} \]

### Step 5: Write expressions for \( K_{P1} \) and \( K_{P2} \)
For the first reaction:
\[ K_{P1} = \frac{(P_B)^2 (P_C)^3}{1} = \frac{(2P_1)^2 (3P_1)^3}{1} = 4P_1^2 \cdot 27P_1^3 = 108P_1^5 \]

For the second reaction:
\[ K_{P2} = \frac{(P_D)^3}{1} = (3P_2)^3 = 27P_2^3 \]

### Step 6: Substitute \( P_1 \) in terms of \( P_2 \) into \( K_{P1} \)
Now substituting \( P_1 = \frac{P_2}{4} \) into \( K_{P1} \):
\[ K_{P1} = 108\left(\frac{P_2}{4}\right)^5 = 108 \cdot \frac{P_2^5}{1024} = \frac{108P_2^5}{1024} \]

### Step 7: Calculate \( \frac{K_{P2}}{8K_{P1}} \)
Now we can find \( \frac{K_{P2}}{8K_{P1}} \):
\[ \frac{K_{P2}}{8K_{P1}} = \frac{27P_2^3}{8 \cdot \frac{108P_2^5}{1024}} = \frac{27P_2^3 \cdot 1024}{864P_2^5} \]
This simplifies to:
\[ \frac{27 \cdot 1024}{864} \cdot \frac{1}{P_2^2} = \frac{27648}{864} \cdot \frac{1}{P_2^2} = \frac{32}{P_2^2} \]

### Final Result
Thus, the final expression we derived is:
\[ \frac{K_{P2}}{8K_{P1}} = \frac{32}{P_2^2} \]