For a 1st order reaction if concentration is doubled then rate of reaction becomes

Options

A. doubles
B. half
C. four times
D. remains same.

Explanation

To solve the problem regarding the effect of doubling the concentration on the rate of a first-order reaction, we can follow these steps:

### Step-by-Step Solution

1. **Understand the Rate Law for a First-Order Reaction**:
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k[A]^1 \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

2. **Define the Initial Rate**:
Let’s denote the initial concentration of the reactant as \([A]\). The initial rate of the reaction can be written as:
\[ \text{Rate}_1 = k[A] \]

3. **Double the Concentration**:
If the concentration of the reactant is doubled, the new concentration becomes \(2[A]\).

4. **Calculate the New Rate**:
The rate with the doubled concentration can be expressed as:
\[ \text{Rate}_2 = k[2A] = k \cdot 2[A] = 2k[A] \]

5. **Compare the Two Rates**:
Now, we can compare the two rates:
\[ \text{Rate}_2 = 2 \cdot \text{Rate}_1 \]
This shows that the new rate is double the initial rate.

6. **Conclusion**:
Therefore, if the concentration is doubled, the rate of the reaction also doubles.

### Final Answer:
The rate of the reaction becomes double when the concentration is doubled.

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