The unit of equilibrium constant `K_C` of a reaction is `"mol"^(-2) l^2`.For this reaction, the product concentration increases by

Options

A. increasing the pressure
B. Lowering the temperature
C. Lowering the pressure
D. Both B and C

Explanation

To solve the problem, we need to analyze the information given about the equilibrium constant \( K_C \) and how it relates to the concentrations of reactants and products in a chemical reaction.

### Step-by-Step Solution:

1. **Understanding the Unit of \( K_C \)**:
The unit of the equilibrium constant \( K_C \) is given as \( \text{mol}^{-2} \, \text{l}^2 \). This indicates that the equilibrium constant is related to the concentrations of products and reactants in the reaction.

2. **Relating \( K_C \) to Reaction Stoichiometry**:
The general expression for the equilibrium constant \( K_C \) for a reaction can be written as:
\[ K_C = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}} \]
Here, the concentration of products is raised to the power of their coefficients in the balanced equation, and the same for reactants.

3. **Determining \( \Delta n \)**:
The unit \( \text{mol}^{-2} \, \text{l}^2 \) can be interpreted to find \( \Delta n \), which is the change in the number of moles of gas during the reaction. The relationship between the unit of \( K_C \) and \( \Delta n \) is:
\[ K_C \propto \left( \frac{1}{[\text{Reactants}]} \right)^{\Delta n} \]
Rearranging gives:
\[ K_C \propto \text{mol}^{\Delta n} \, \text{l}^{-2} \]
From the given unit \( \text{mol}^{-2} \, \text{l}^2 \), we can deduce:
\[ \Delta n = -2 \]

4. **Interpreting \( \Delta n \)**:
A \( \Delta n \) of -2 means that the number of moles of gaseous products is less than that of gaseous reactants. For example, if we have 4 moles of A gas converting into 2 moles of B gas, we have:
\[ 4 \, \text{A} \rightarrow 2 \, \text{B} \]
Here, the number of gaseous molecules decreases, indicating that the reaction shifts towards the products when pressure is increased.

5. **Applying Le Chatelier's Principle**:
According to Le Chatelier's Principle, if we increase the pressure of a system at equilibrium, the system will shift in the direction that reduces the pressure. Since the product side has fewer moles of gas (2 moles of B) compared to the reactant side (4 moles of A), increasing the pressure will favor the formation of products.

6. **Conclusion**:
Therefore, if the product concentration increases, it is due to the increase in pressure, which shifts the equilibrium towards the products.

### Final Answer:
The product concentration increases by increasing the pressure.