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For a 1st order reaction if concentratio...

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

A

doubles

B

half

C

four times

D

remains same.

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To solve the question 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**: The rate of a first-order reaction can be expressed as: \[ \text{Rate} = k[A] \] where \( k \) is the rate constant and \([A]\) is the concentration of the reactant. 2. **Define the Initial Concentration**: Let’s assume the initial concentration of the reactant \( A \) is \([A] = a\). Therefore, the initial rate of the reaction (\( R \)) can be written as: \[ R = k \cdot a \] 3. **Double the Concentration**: If the concentration of the reactant is doubled, the new concentration becomes: \[ [A] = 2a \] 4. **Calculate the New Rate**: The new rate of the reaction (\( R' \)) with the doubled concentration can be calculated as: \[ R' = k \cdot (2a) = 2 \cdot (k \cdot a) = 2R \] 5. **Conclusion**: Therefore, when the concentration of the reactant is doubled, the rate of the reaction also doubles. The final answer is: \[ \text{Rate becomes: } 2R \] ### Final Answer: The rate of reaction becomes double when the concentration is doubled for a first-order reaction. ---
To solve the question 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**: The rate of a first-order reaction can be expressed as: \[ \text{Rate} = k[A] ...
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