For 2A + 2B `hArr 2C + 2D, K_(c) = (1)/(16) `then `K_(c) ` for C + D`hArr` A + B is _______

Answer

To solve the problem, we need to find the equilibrium constant \( K_c \) for the reaction \( C + D \rightleftharpoons A + B \) based on the given information about the reaction \( 2A + 2B \rightleftharpoons 2C + 2D \) with \( K_c = \frac{1}{16} \).

### Step-by-Step Solution:

1. **Write the given reaction and its equilibrium constant**:
The reaction provided is:
\[ 2A + 2B \rightleftharpoons 2C + 2D \]
The equilibrium constant for this reaction is given as:
\[ K_c = \frac{1}{16} \]

2. **Express the equilibrium constant in terms of concentrations**:
The equilibrium constant expression for the reaction is:
\[ K_c = \frac{[C]^2[D]^2}{[A]^2[B]^2} \]
Given that \( K_c = \frac{1}{16} \), we can write:
\[ \frac{[C]^2[D]^2}{[A]^2[B]^2} = \frac{1}{16} \]

3. **Simplify the expression**:
We can rearrange this expression:
\[ [C]^2[D]^2 = \frac{1}{16} [A]^2[B]^2 \]
Taking the square root of both sides gives:
\[ [C][D] = \frac{1}{4} [A][B] \]

4. **Reverse the reaction**:
Now, we need to find the equilibrium constant for the reverse reaction:
\[ C + D \rightleftharpoons A + B \]
The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. Therefore:
\[ K_c' = \frac{1}{K_c} \]

5. **Calculate the new equilibrium constant**:
Since \( K_c = \frac{1}{16} \), we find:
\[ K_c' = \frac{1}{\frac{1}{16}} = 16 \]

### Final Answer:
Thus, the equilibrium constant \( K_c \) for the reaction \( C + D \rightleftharpoons A + B \) is:
\[ \boxed{16} \]