Question
For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in terms of the half-life $\left(t_{1 / 2}\right)$ and in terms of the rate constant $k$.
Step 1
In a first-order reaction, the rate of the reaction is directly proportional to the concentration of one of the reactants. The half-life of a first-order reaction is the time it takes for the concentration of the reactant to decrease to half of its initial value. Show more…
Show all steps
Your feedback will help us improve your experience
Will Li and 89 other Chemistry 101 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
For a first-order reaction, how long will it take for the concentration of reactant to fall to one-eighth its original value? Express your answer in terms of the half-life $\left(t_{1}\right)$ and in terms of the rate constant $k$.
In a first-order decomposition reaction, $\mathrm{A} \rightarrow$ products, the amount of substance $\mathrm{A}$ at time $t$ is $$ x(t)=x(0) e^{-k t} $$ where $x(0)$ is the initial amount of $\mathrm{A}$, and $k$ is the rate constant. The time taken for the amount of A to fall to half of its initial value is called the half-life, $\tau_{1 / 2^{2}}$ of the reaction. Find the half-life for rate constants: (i) $k=3 \mathrm{~s}^{-1}$, (ii) $k=10^{-5} \mathrm{~s}^{-1}$.
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD