The equilibrium constant of the reaction `(K_c)` when the reaction is conducted in a one litre vessel was found to be `2.5 xx 10^(-3).` If the reaction is conducted at the same temperature in a 2 litre vessel then the value of `K_c` is

Options

A. `6.25 xx 100^(-4)`
B. `1.25 xx 10^(-3)`
C. `2.5 xx 10^(-3)`
D. `5 xx 10^(-3)`

Answer

To solve the problem, we need to understand the concept of the equilibrium constant \( K_c \) and how it behaves under different conditions, particularly volume changes.

### Step-by-Step Solution:

1. **Understanding the Equilibrium Constant**:
The equilibrium constant \( K_c \) is defined for a specific reaction at a given temperature. It is a measure of the ratio of the concentrations of products to reactants at equilibrium.

2. **Given Information**:
The equilibrium constant \( K_c \) for the reaction in a 1-litre vessel is given as \( 2.5 \times 10^{-3} \).

3. **Change of Volume**:
The problem states that the same reaction is conducted in a 2-litre vessel at the same temperature.

4. **Effect of Volume on \( K_c \)**:
It is important to note that the value of \( K_c \) is dependent only on the temperature and not on the volume of the container. According to Le Chatelier's Principle, changing the volume of the container does not affect the equilibrium constant as long as the temperature remains constant.

5. **Conclusion**:
Since the temperature is the same and we are only changing the volume from 1 litre to 2 litres, the value of \( K_c \) remains unchanged. Therefore, the value of \( K_c \) in the 2-litre vessel is still \( 2.5 \times 10^{-3} \).

### Final Answer:
The value of \( K_c \) when the reaction is conducted in a 2-litre vessel is \( 2.5 \times 10^{-3} \).

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