Question
It takes $15.4$ minutes for the concentration of a reactant to drop to $5.0 \%$ of its initial value in a second-order reaction. What is the rate constant for the reaction in the units of $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~min}^{-1}$ ?
Step 1
We know that the reaction is second-order, so we can use the second-order integrated rate law: $\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt$ Show more…
Show all steps
Your feedback will help us improve your experience
Ronald Prasad and 55 other Chemistry 102 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If it takes 75.0 min for the concentration of a reactant to drop to $25.0 \%$ of its initial value in a first-order reaction, what is the rate constant for the reaction in the units $\min ^{-1} ?$
'What is the rate constant of first-order reaction that takes 459 seconds for the reactant concentration to drop to half of its initial value?'
The rate constant for a first order reaction is $60 \mathrm{~s}^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its $1 / 16^{\text {th }}$ value?
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD