The unit of equilibrium constant `K_(c)` for the reaction `A+B hArr C` would be

Options

A. `mol^(-1) L`
B. `mol L^(-1)`
C. `mol L`
D. No unit

Explanation

To determine the unit of the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation**:
The reaction is given as:
\[ A + B \rightleftharpoons C \]

2. **Write the Expression for the Equilibrium Constant \( K_c \)**:
The equilibrium constant \( K_c \) is defined in terms of the concentrations of the products and reactants. For the reaction \( A + B \rightleftharpoons C \), the expression for \( K_c \) is:
\[ K_c = \frac{[C]}{[A][B]} \]
where \([C]\), \([A]\), and \([B]\) are the molar concentrations of the respective species at equilibrium.

3. **Identify the Units of Concentration**:
The concentration of a substance is measured in moles per liter (mol/L). Thus, the units for \([C]\), \([A]\), and \([B]\) are:
\[ [C] = \text{mol/L}, \quad [A] = \text{mol/L}, \quad [B] = \text{mol/L} \]

4. **Substitute the Units into the \( K_c \) Expression**:
Now, substituting the units into the \( K_c \) expression:
\[ K_c = \frac{[\text{C}]}{[\text{A}][\text{B}]} = \frac{\text{mol/L}}{(\text{mol/L})(\text{mol/L})} \]

5. **Simplify the Expression**:
Simplifying the units gives:
\[ K_c = \frac{\text{mol/L}}{\text{mol}^2/\text{L}^2} = \frac{\text{mol/L}}{\text{mol}^2} \cdot \text{L}^2 = \frac{\text{L}}{\text{mol}} \]

6. **Final Result**:
Therefore, the unit of the equilibrium constant \( K_c \) for the reaction \( A + B \rightleftharpoons C \) is:
\[ K_c = \text{L/mol} \]