| English | 의미 |
|---|---|
| half-life/hɑːf laɪf/ | 반감기 |
How amounts fade with the clock
- Watch a reactant disappear and plot it against the clock.
- The shape of that curve reveals the reaction's order.
- A straight line, of the right kind, pins it down.
- One special time, the half-life, sums up the pace.
Which plot is a straight line
- First order: a plot of $\ln[A]$ versus time is straight.
- Second order: a plot of $1/[A]$ versus time is straight.
- Zero order: a plot of $[A]$ versus time is straight.
If a plot of $\ln[A]$ versus time is a straight line, the reaction is...
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
A linear $\ln[A]$-versus-time plot indicates first order.
A straight-line plot of $1/[A]$ versus time indicates which order?
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
The $1/[A]$ linear plot is the signature of second order.
The half-life
- The half-life 半衰期 is the time for half the reactant to be used up.
- For a first-order reaction, the half-life is constant.
- Every half-life, the amount halves again.
Only a first-order reaction has a constant half-life.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
For other orders the half-life changes as concentration drops.
Each half-life, the amount of reactant is cut in ____.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
A half-life halves whatever is present.
Finding the order from data
- Try each plot; whichever is straight gives the order.
- The slope of that line relates to the rate constant $k$.
- This is how experiments reveal the reaction order.
Concentration over time
A first-order reactant decays exponentially: it falls by the same fraction in each equal time step.
The slope of the straight-line plot is related to the rate constant.
난이도 1/5 · 100 코인 + 오늘 첫 완료 시 소액 보너스
The magnitude of the slope gives $k$.
A first-order reaction has a half-life of $20\ \text{s}$, starting at $8\ \text{M}$.
- After $20\ \text{s}$ it is $4\ \text{M}$; after $40\ \text{s}$, $2\ \text{M}$; after $60\ \text{s}$, $1\ \text{M}$.
- It halves every $20\ \text{s}$, no matter the amount.
A first-order reaction starts at $16\ \text{M}$ with a $10\ \text{s}$ half-life. Concentration after $20\ \text{s}$ (in M)?
난이도 2/5 · 140 코인 + 오늘 첫 완료 시 소액 보너스
Two half-lives give $16 \to 8 \to 4\ \text{M}$.
Only a first-order reaction has a constant half-life -- for other orders the half-life changes as the concentration drops. Match the straight-line plot to the order ($\ln[A]$, $1/[A]$, or $[A]$). And a steeper line means a bigger rate constant.
The order shows up in which plot is a straight line: $\ln[A]$ (first), $1/[A]$ (second), or $[A]$ (zero) versus time. The half-life is the time to use up half the reactant, and it is constant only for a first-order reaction, halving the amount again each period.