For the hypothetical reactions, the equilibrium constant `(K)` value are given
`AhArrB,K_(1)=2,BhArrC,K_(2)=4,`
`ChArrD,K_(3)=3`
The equilibrium constant (K) for the reaction
`AhArrD` is

Options

A. `48`
B. `6`
C. `27`
D. `24`

Explanation

To find the equilibrium constant \( K \) for the reaction \( A \rightleftharpoons D \), we can use the given equilibrium constants for the individual reactions.

1. **Identify the given reactions and their equilibrium constants:**
- Reaction 1: \( A \rightleftharpoons B \), with \( K_1 = 2 \)
- Reaction 2: \( B \rightleftharpoons C \), with \( K_2 = 4 \)
- Reaction 3: \( C \rightleftharpoons D \), with \( K_3 = 3 \)

2. **Write the overall reaction:**
- We need to combine these reactions to find the overall reaction \( A \rightleftharpoons D \).
- The sequence of reactions can be represented as:
\[ A \rightleftharpoons B \quad (K_1) \]
\[ B \rightleftharpoons C \quad (K_2) \]
\[ C \rightleftharpoons D \quad (K_3) \]

3. **Combine the reactions:**
- To get the overall reaction \( A \rightleftharpoons D \), we add the three reactions:
\[ A \rightleftharpoons B \quad (K_1) \]
\[ B \rightleftharpoons C \quad (K_2) \]
\[ C \rightleftharpoons D \quad (K_3) \]
- When we add these reactions, we multiply their equilibrium constants:
\[ K = K_1 \times K_2 \times K_3 \]

4. **Substitute the values of the equilibrium constants:**
- Now we can substitute the values:
\[ K = 2 \times 4 \times 3 \]

5. **Calculate the value of \( K \):**
- Performing the multiplication:
\[ K = 2 \times 4 = 8 \]
\[ K = 8 \times 3 = 24 \]

6. **Conclusion:**
- The equilibrium constant \( K \) for the reaction \( A \rightleftharpoons D \) is \( 24 \).

### Final Answer:
The equilibrium constant \( K \) for the reaction \( A \rightleftharpoons D \) is \( 24 \).

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