For the reactions,
`AhArrB, K_(c)=1,BhArrC,K_(c)=3,ChArrD,K_(c)=5`. `K_(c)` for the reaction `AhArrD` is

Options

A. `15`
B. `5`
C. `3`
D. `1`

Explanation

To find the equilibrium constant \( K_c \) for the reaction \( A \rightleftharpoons D \), we can use the equilibrium constants provided for the intermediate reactions:

1. **Given Reactions and Their Equilibrium Constants:**
- \( A \rightleftharpoons B \), \( K_{c1} = 1 \)
- \( B \rightleftharpoons C \), \( K_{c2} = 3 \)
- \( C \rightleftharpoons D \), \( K_{c3} = 5 \)

2. **Finding \( K_c \) for the Overall Reaction \( A \rightleftharpoons D \):**
To find \( K_c \) for the reaction \( A \rightleftharpoons D \), we can express it in terms of the other reactions:

\[ K_c = K_{c1} \times K_{c2} \times K_{c3} \]

This is because the overall equilibrium constant for a series of reactions is the product of the equilibrium constants for the individual reactions.

3. **Substituting the Values:**
\[ K_c = K_{c1} \times K_{c2} \times K_{c3} = 1 \times 3 \times 5 \]

4. **Calculating the Result:**
\[ K_c = 1 \times 3 = 3 \]
\[ K_c = 3 \times 5 = 15 \]

5. **Final Answer:**
Therefore, the equilibrium constant \( K_c \) for the reaction \( A \rightleftharpoons D \) is \( 15 \).