`P hArr Q, K_1`
`1/3RhArr 1/3Q,K_2`
`2R hArr 2S, K_3`
`P hArr S, K_4 =?`
Here, `K_1`,` K_2`, `K_3` and `K_4` are equilibrium constants:-

Options

A. `(K_1 K_3^2)/K_2`
B. `(K_1(K_3)^(1//2))/K_2^3`
C. `(K_1 K_3^2)/K_2^3`
D. `K_1(K_3)^(1//2)K_2^3`

Answer

To find the equilibrium constant \( K_4 \) for the reaction \( P \rightleftharpoons S \), we will use the provided reactions and their equilibrium constants. Let's break down the steps to derive \( K_4 \).

### Step 1: Write down the given reactions and their equilibrium constants.
1. \( P \rightleftharpoons Q \), with equilibrium constant \( K_1 \)
2. \( \frac{1}{3} R \rightleftharpoons \frac{1}{3} Q \), with equilibrium constant \( K_2 \)
3. \( 2R \rightleftharpoons 2S \), with equilibrium constant \( K_3 \)

### Step 2: Manipulate the reactions to derive \( P \rightleftharpoons S \).
To derive the reaction \( P \rightleftharpoons S \), we will:
- Use the first reaction as is.
- Divide the third reaction by 2.
- Subtract the second reaction multiplied by 3.

### Step 3: Divide the third reaction by 2.
Dividing the third reaction \( 2R \rightleftharpoons 2S \) by 2 gives:
\[ R \rightleftharpoons S \]
The equilibrium constant for this reaction becomes:
\[ K_3' = K_3^{1/2} \]

### Step 4: Multiply the second reaction by 3.
Multiplying the second reaction \( \frac{1}{3} R \rightleftharpoons \frac{1}{3} Q \) by 3 gives:
\[ R \rightleftharpoons Q \]
The equilibrium constant for this reaction becomes:
\[ K_2' = K_2^3 \]

### Step 5: Combine the reactions.
Now we can combine the reactions:
1. From the first reaction: \( P \rightleftharpoons Q \) (with \( K_1 \))
2. From the modified third reaction: \( R \rightleftharpoons S \) (with \( K_3^{1/2} \))
3. From the modified second reaction: \( R \rightleftharpoons Q \) (with \( K_2^3 \))

By subtracting the second reaction from the sum of the first and modified third reactions, we get:
\[ P \rightleftharpoons S \]

### Step 6: Calculate the equilibrium constant \( K_4 \).
Using the relationship for equilibrium constants:
- When adding reactions, we multiply their equilibrium constants.
- When dividing a reaction by a coefficient, we take the root of the equilibrium constant.

Thus, we have:
\[ K_4 = K_1 \cdot K_3^{1/2} \cdot K_2^{-3} \]

### Final Expression for \( K_4 \):
\[ K_4 = \frac{K_1 \cdot K_3^{1/2}}{K_2^3} \]