In a reaction with a rate constant of 0.03 s^-1, how long will it take for the c

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Question: In a reaction with a rate constant of 0.03 s^-1, how long will it take for the concentration to decrease to 25% of its initial value?

Options:

Correct Answer: 46.2 s

Solution:

For a first-order reaction, t = (ln([A]0/[A]) / k). Here, [A] = 0.25[A]0, so t = (ln(1/0.25) / 0.03) ≈ 46.2 s.

In a reaction with a rate constant of 0.03 s^-1, how long will it take for the concentration to decrease to 25% of its initial value?

In a reaction with a rate constant of 0.03 s^-1, how long will it take for the concentration to decrease to 25% of its initial value?

Step 1: Identify the rate constant (k) given in the problem, which is 0.03 s^-1.
Step 2: Understand that we are dealing with a first-order reaction.
Step 3: Recall the formula for the time (t) it takes for a concentration to change in a first-order reaction: t = (ln([A]0/[A]) / k).
Step 4: Determine the initial concentration [A]0 and the final concentration [A]. Since we want the concentration to decrease to 25% of its initial value, we have [A] = 0.25[A]0.
Step 5: Substitute [A] into the formula: t = (ln([A]0/(0.25[A]0)) / k).
Step 6: Simplify the equation: t = (ln(1/0.25) / 0.03).
Step 7: Calculate ln(1/0.25), which is ln(4).
Step 8: Use a calculator to find ln(4) ≈ 1.386.
Step 9: Substitute this value back into the equation: t = (1.386 / 0.03).
Step 10: Perform the final calculation: t ≈ 46.2 seconds.

First-Order Kinetics – Understanding the relationship between concentration and time in first-order reactions, where the rate of reaction is directly proportional to the concentration of the reactant.
Natural Logarithm in Kinetics – Applying the natural logarithm to relate initial and final concentrations in the context of reaction rates.
Rate Constant – Recognizing the significance of the rate constant (k) in determining the time required for a reaction to reach a certain concentration.