If concentration of reactant is increased by 2 times then K
becomes :-

Options

A. `1n(K)/(2)`
B. `(K)/(2)`
C. 2K
D. K

Explanation

To solve the question regarding the effect of increasing the concentration of a reactant on the rate constant \( K \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Constant**:
The rate constant \( K \) is a proportionality constant in the rate law expression. It relates the rate of a reaction to the concentrations of the reactants raised to their respective powers (reaction orders).

2. **Write the Rate Law Expression**:
For a general reaction:
\[ aA + bB \rightarrow \text{Products} \]
The rate law can be expressed as:
\[ \text{Rate} = k[A]^\alpha[B]^\beta \]
where \( \alpha \) and \( \beta \) are the orders of the reaction with respect to reactants \( A \) and \( B \), respectively.

3. **Effect of Concentration Change**:
If the concentration of reactant \( A \) is increased by 2 times, we can denote the new concentration as:
\[ [A]_{\text{new}} = 2[A] \]
However, the rate constant \( K \) (or \( k \)) is defined as being independent of the concentrations of the reactants.

4. **Conclusion**:
Since the rate constant \( K \) does not depend on the concentration of the reactants, even if the concentration of \( A \) is doubled, \( K \) remains unchanged. Therefore:
\[ K_{\text{new}} = K \]

### Final Answer:
The value of \( K \) remains the same, regardless of the change in concentration of the reactants.