A drug with concentration 0.70 M decomposes via acid-catalysed hydrolysis with a rate constant of 0.013 M−1 h−1. What is the drug concentration after 61 hours? Report your answer to 2 decimal places.

Question

BlackTom AI · Accepted Answer

The units of the rate constant, M−1 h−1, indicate second-order kinetics (overall second order or bimolecular). For a second-order reaction with a single reactant A following 2nd-order kinetics, the integrated rate law is:
1/[A]t = kt + 1/[A]0

Step-by-step calculation:
1. Identify given values:
- Initial concentration [A]0 = 0.70 M
- Rate constant k = 0.013 M−1 h−1
- Time t = 61 h

2. Compute kt:
k × t = 0.013 M−1 h−1 × 61 h = 0.793 M−1

3. Compute the initial reciprocal concentration:
1/[A]0 = 1 / 0.70 M = 1.428571... M−1

4. Sum kt and 1/[A]0:
1/[A]t = kt + 1/[A]0 = 0.793 + 1.428571... = 2.221571... M−1

5. Invert to find [A]t:
[A]t = 1 / 2.221571... = 0.450049... M

Rounding to two decimal places gives 0.45 M.

Why this is correct:
- The rate constant units (M−1 h−1) match the second-order integrated law used. Using a first-order law (ln[A] = −kt + ln[A]0) would be incorrect because that form requires k in h−1; plugging the given k with first-order formula would yield a wrong numerical result and is inconsistent with the units.

No alternative choices apply (this is a numerical question), but any approach assuming first-order kinetics is incorrect due to unit mismatch.

Conclusion: The drug concentration after 61 hours is 0.45 M, which matches the provided answer.

A drug with concentration 0.70 M decomposes via acid-catalysed hydrolysis with a rate constant of 0.013 M−1 h−1. What is the drug concentration after 61 hours? Report your answer to 2 decimal places.数值题

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