For the reaction, `A iff n B` the concentration of A decreases from 0.06 to 0.03 mol `L^(-1)` and that of B rises from 0 to 0.06 mol `L^(-1)` at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are

Options

A. 2 and 0.12
B. 2 and 1.2
C. 3 and 0.12
D. 3 and 1.2

Explanation

To solve the problem, we need to determine the values of \( n \) and the equilibrium constant \( K_c \) for the reaction \( A \iff nB \).

### Step-by-step Solution:

1. **Identify the Initial and Final Concentrations**:
- Initial concentration of \( A \) = 0.06 mol/L
- Final concentration of \( A \) = 0.03 mol/L
- Initial concentration of \( B \) = 0 mol/L
- Final concentration of \( B \) = 0.06 mol/L

2. **Calculate the Change in Concentration of \( A \)**:
- Change in concentration of \( A \) = Initial concentration of \( A \) - Final concentration of \( A \)
\[ \Delta[A] = 0.06 - 0.03 = 0.03 \text{ mol/L} \]

3. **Relate the Change in Concentration of \( A \) to \( B \)**:
- According to the stoichiometry of the reaction \( A \iff nB \), for every decrease of 1 mol of \( A \), \( n \) moles of \( B \) are produced.
- Therefore, the change in concentration of \( B \) can be expressed as:
\[ \Delta[B] = n \times \Delta[A] \]
- We know that \( \Delta[B] = 0.06 \) mol/L (the final concentration of \( B \)).

4. **Set Up the Equation**:
\[ n \times 0.03 = 0.06 \]

5. **Solve for \( n \)**:
\[ n = \frac{0.06}{0.03} = 2 \]

6. **Determine the Equilibrium Constant \( K_c \)**:
- The equilibrium expression for the reaction \( A \iff nB \) is given by:
\[ K_c = \frac{[B]^n}{[A]} \]
- Substituting the known concentrations:
\[ K_c = \frac{(0.06)^2}{0.03} \]

7. **Calculate \( K_c \)**:
\[ K_c = \frac{0.0036}{0.03} = 0.12 \]

### Final Answers:
- The value of \( n \) is **2**.
- The equilibrium constant \( K_c \) is **0.12**.