For the reactions `A Leftrightarrow B " "K_c=2 `
`B Leftrightarrow C" "K_(c)=4`
`C Leftrightarrow D " "K_(c )=6`
`K_c` for the reaction `A Leftrightarrow D` is

Options

A. (2+4+6)
B. `(2 xx 4)/6`
C. `(4 xx 6)/2`
D. `2 xx 4 xx 6`

Answer

To find the equilibrium constant \( K_c \) for the overall reaction \( A \leftrightarrow D \), we need to consider the individual reactions and their equilibrium constants.

### Step-by-Step Solution:

1. **Identify the Reactions and Their Constants**:
- The first reaction is \( A \leftrightarrow B \) with \( K_{c1} = 2 \).
- The second reaction is \( B \leftrightarrow C \) with \( K_{c2} = 4 \).
- The third reaction is \( C \leftrightarrow D \) with \( K_{c3} = 6 \).

2. **Write the Overall Reaction**:
- The overall reaction from \( A \) to \( D \) can be expressed as:
\[ A \leftrightarrow B \leftrightarrow C \leftrightarrow D \]

3. **Combine the Equilibrium Constants**:
- The equilibrium constant for the overall reaction can be calculated by multiplying the equilibrium constants of the individual steps:
\[ K_c = K_{c1} \times K_{c2} \times K_{c3} \]

4. **Substitute the Values**:
- Substitute the known values into the equation:
\[ K_c = 2 \times 4 \times 6 \]

5. **Calculate the Result**:
- Perform the multiplication:
\[ K_c = 2 \times 4 = 8 \]
\[ K_c = 8 \times 6 = 48 \]

6. **Final Answer**:
- Thus, the equilibrium constant \( K_c \) for the reaction \( A \leftrightarrow D \) is \( 48 \).