This page is the stress-test for the parent derivations . We do not just repeat "plug into the formula." We enumerate every kind of question a rate law can throw at you — every order, every direction (forward: find concentration; backward: find time or k
If you have not yet met the three integrated laws, read the parent first. Everything here uses only these three lines, so let us pin them where the eye can see them.
Definition What the square brackets and "M" mean[ A ] concentration of A "—the number of moles of substance A packed into one litre of solution. Its unit is mol L − 1 M molar ). So "0.10 M 0.10 − 1 0.10 M mol L − 1
t 1/2 half-life t 1/2 exactly half its current value: [ A ] → 2 1 [ A ] 1/2 0th: t 1/2 = 2 k [ A ] 0 1st: t 1/2 = k l n 2 = k 0.693 2nd: t 1/2 = k [ A ] 0 1
Look at the pictures of all three at once before we start — they look different, and that shape is what tells you which law is acting.
Intuition What to notice in the figure aboveTrace each curve from the same starting height. The white line (0th) falls at a perfectly constant tilt — equal drop every second, and it eventually crashes into the axis. The cyan curve (1st) is steepest at the start and gets ever gentler, but never touches zero — it only approaches it. The amber curve (2nd) starts fastest of all, then flattens into an even lazier tail than the cyan one. The whole game of this page is: read the shape, name the order. Constant slope ⇒ 0th; ever-shrinking-by-a-fixed-fraction ⇒ 1st; a stubborn long tail that refuses to die ⇒ 2nd.
Intuition A word about units of timeThroughout Cells A–F, H and I we work in seconds (s) , because that is what the rate constants are given in. In the drug problem (Cell G) the data arrive in hours (h) , so we keep k h − 1 every time in hours — mixing s and h in one equation would be like adding metres to miles. The rule: k t k t
Every kinetics problem lands in exactly one cell of this grid. The last column names the worked example that covers it.
Cell
Order
What is unknown / what is weird
Covered by
A
0
forward: find [ A ] t
Ex 1
B
0
degenerate : reactant runs out before t Ex 2
C
1
backward: find the time to reach a target
Ex 3
D
1
limiting behaviour: fraction left as t → ∞
Ex 4
E
2
forward: find [ A ] t
Ex 5
F
2
backward: find k
Ex 6
G
any
word problem (real world), units of k Ex 7
H
any
exam twist : order is hidden — decide it from data, then answerEx 8
I
0 vs 1 vs 2
half-life comparison — same numbers, three different t 1/2
Ex 9
We march through them in order. Each example first asks you to Forecast the answer (guess the ballpark — this trains intuition), then works it, then Verifies .
Worked example Ex 1 — Ammonia on hot platinum2 NH 3 → N 2 + 3 H 2 [ NH 3 ] 0 = 0.10 M k = 2.5 × 1 0 − 4 M s − 1 [ NH 3 ] t = 300 s
Forecast: In zero order the concentration falls in a straight line . In 300 s we remove k ⋅ t = 2.5 × 1 0 − 4 × 300 = 0.075 M 0.10 0.025 M
Step 1. Pick the law. Zero order → [ A ] t = [ A ] 0 − k t Why this step? Because rate does not depend on concentration; the amount removed is just (rate)×(time), a constant slope.
Step 2. Substitute.
[ NH 3 ] t = 0.10 − ( 2.5 × 1 0 − 4 ) ( 300 ) = 0.10 − 0.075 = 0.025 M Why this step? Direct plug-in; the subtraction is the physics — the toll booth passed a fixed number of molecules per second.
Verify: 0.025 M 0.10 M − ( M s − 1 ) ( s ) = M − M = M
This is the trap almost everyone misses. In zero order the line keeps dropping at the same slope. A line does not know when to stop at zero. So blindly plugging a large t negative concentration, which is physically impossible.
t end zero-order reactant is completely used up: t end [ A ] t 0 derive its value in the example.
Intuition What to notice in the figure aboveThe dashed grey line is the bare formula 0.10 − k t negative territory — a place concentration can never go. The solid cyan line is the honest concentration: it tracks the grey line exactly until it meets the axis, then lies flat along [ A ] = 0 t end tool with a valid domain , and the amber dot is the edge of that domain.
Worked example Ex 2 — When does the formula break?Same reaction: [ NH 3 ] 0 = 0.10 M k = 2.5 × 1 0 − 4 M s − 1 t = 500 s t end
Forecast: At 500 s we remove 2.5 × 1 0 − 4 × 500 = 0.125 M more than we started with. So the honest answer is "gone before then." The reactant hits zero somewhere before 500 s.
Step 1. Naively substitute.
[ NH 3 ] 500 = 0.10 − 0.125 = − 0.025 M ( impossible! ) Why this step? To see the failure. Negative concentration means the model has left its valid region — the reaction already finished.
Step 2. Find the run-out time t end [ A ] t = 0 0 = [ A ] 0 − k t end ⇒ t end = k [ A ] 0 = 2.5 × 1 0 − 4 0.10 = 400 s Why this step? By definition t end [ A ] = 0 t x [ A ] = 0 0
Step 3. State the real answer piecewise.
[ NH 3 ] t = { 0.10 − k t 0 t ≤ t end = 400 s t > 400 s Why this step? The single formula 0.10 − k t while there is still ammonia . Past t end = 0 t end
Verify: At t = 400 0.10 − 2.5 × 1 0 − 4 × 400 = 0.10 − 0.10 = 0 400 2 × t 1/2 t 1/2 = [ A ] 0 /2 k = 200 amount each half-life, so two half-lives empties it completely. ✓
Common mistake The zero-order cliffFirst and second order approach zero asymptotically — they never truly reach it. Only zero order actually hits zero at a finite time t end Always check whether k t > [ A ] 0
Worked example Ex 3 — How long to fall to a target?A first-order isomerisation has k = 3.0 × 1 0 − 3 s − 1 [ A ] 0 = 0.80 M [ A ] t = 0.20 M
Forecast: We are dropping to one-quarter of the start. One-quarter = two halvings, so about two half-lives . Half-life = 0.693/ k ≈ 231 460 – 470
Step 1. Choose the log form and isolate t ln [ A ] 0 [ A ] t = − k t ⇒ t = − k 1 ln [ A ] 0 [ A ] t Why this step? Time is trapped inside the exponential; the natural log is the tool that undoes e − k t t ln log 10 e
Step 2. Substitute.
t = − 3.0 × 1 0 − 3 1 ln 0.80 0.20 = − 3.0 × 1 0 − 3 1 ln ( 0.25 ) ln ( 0.25 ) = − 1.386 t = − 3.0 × 1 0 − 3 − 1.386 = 462 s Why this step? ln ( 0.25 ) − k
Verify: Plug back: [ A ] t = 0.80 e − 3.0 × 1 0 − 3 × 462 = 0.80 e − 1.386 = 0.80 × 0.25 = 0.20 M 462 ≈ 2 × 231
Worked example Ex 4 — Fraction remainingFor 14 C t 1/2 = 5730 4 half-lives ? (b) What fraction remains as t → ∞
Forecast: Each half-life multiplies by 2 1 2 1 × 2 1 × 2 1 × 2 1 = 16 1 t → ∞ e − k t → 0
Step 1. Use k = ln 2/ t 1/2 k = 5730 0.693 = 1.21 × 1 0 − 4 yr − 1 Why this step? The half-life formula for first order is t 1/2 = ln 2/ k k
Step 2. Four half-lives means t = 4 × 5730 = 22920 [ A ] 0 [ A ] t = e − k t = e − 1.21 × 1 0 − 4 × 22920 = e − 2.773 = 0.0625 = 16 1 Why this step? We could also just do ( 1/2 ) 4 e − 4 l n 2
Step 3. Limit. As t → ∞ − k t → − ∞ e − k t → 0 Why this step? This is the defining feature of first order — asymptotic approach, never a hard stop (contrast Cell B).
Verify: ( 1/2 ) 4 = 0.0625 e − 2.773 = 0.0625 k yr − 1 k t
2 2 NO 2 → N 2 O 4 2 [ NO 2 ] 0 = 0.0100 M k = 0.543 M − 1 s − 1 t = 300 s [ NO 2 ] t
Forecast: Second order decays fast at first, then crawls . Since 1/ [ A ] 0 = 100 k t = 163 k t × 1 0 − 3 M
Step 1. Reciprocal law.
[ NO 2 ] t 1 = [ NO 2 ] 0 1 + k t Why this step? Second order is linear in 1/ [ A ] [ A ]
Step 2. Substitute.
[ NO 2 ] t 1 = 0.0100 1 + ( 0.543 ) ( 300 ) = 100 + 162.9 = 262.9 M − 1 Why this step? 1/0.0100 = 100 ( 0.543 ) ( 300 ) = 162.9 1/ [ A ] linearly with time.
Step 3. Flip back.
[ NO 2 ] t = 262.9 1 = 3.80 × 1 0 − 3 M Why this step? The law lives in reciprocal space; the final flip returns us to real concentration.
Verify: 1/ ( 3.80 × 1 0 − 3 ) = 263 ≈ 262.9 M − 1 + ( M − 1 s − 1 ) ( s ) = M − 1 ≈ 38% k
Worked example Ex 6 — Extractk [ A ] 0 = 0.050 M [ A ] t = 0.020 M t = 250 s k
Forecast: We rose in reciprocal space from 1/0.050 = 20 1/0.020 = 50 30 M − 1 k ≈ 30/250 = 0.12 M − 1 s − 1
Step 1. Rearrange the law for k k = t 1 ( [ A ] t 1 − [ A ] 0 1 ) Why this step? k slope of the 1/ [ A ] t 1/ [ A ] t
Step 2. Substitute.
k = 250 1 ( 0.020 1 − 0.050 1 ) = 250 1 ( 50 − 20 ) = 250 30 = 0.12 M − 1 s − 1 Why this step? 1/0.020 = 50 1/0.050 = 20 30 M − 1 250 is k
Verify: Forward-check with this k 1/ [ A ] t = 20 + 0.12 × 250 = 20 + 30 = 50 [ A ] t = 0.020 M k ( M − 1 ) / ( s ) = M − 1 s − 1
Worked example Ex 7 — Drug clearance in the bodyA drug is eliminated by first-order kinetics. A patient is given 500 mg 6.0 hours 125 mg k ≤ 5%
Forecast: 500 → 125 1/16 = 6.25%
Part (a) — find k
Step 1. Amount is proportional to concentration, so the ratio law works on masses directly.
ln 500 125 = − k ( 6.0 ) ⇒ ln ( 0.25 ) = − 1.386 = − 6.0 k Why this step? First-order depends only on the ratio of amounts; the volume cancels, so we may use mg instead of M. All times here are in hours , so k
Step 2. Solve for k k = 6.0 1.386 = 0.231 h − 1 Why this step? Units are h − 1 hours , matching the hours in the data) because first-order k
Part (b) — find the half-life.
Step 3. Use the first-order half-life formula.
t 1/2 = k 0.693 = 0.231 0.693 = 3.0 h Why this step? First-order half-life is ln 2/ k k h − 1 hours — no conversion needed since k 500 → 125
Part (c) — time to reach 5%.
Step 4. Set the remaining fraction to 0.05 ln ( 0.05 ) = − k t ⇒ t = − k 1 ln ( 0.05 ) = − 0.231 − 2.996 = 12.97 h ≈ 13 h Why this step? Same "solve for time" move as Cell C — the natural log undoes the exponential to release t 0.05 [ A ] t / [ A ] 0 = 0.05
Verify: Forecasts confirmed: t 1/2 = 3.0 500 e − 0.231 × 6 = 500 × 0.25 = 125 500 e − 0.231 × 12.97 = 500 × 0.0500 = 25 = 5% 500 12.97 6.25%
The nastiest exam questions do not tell you the order. You must decide it from the data , then answer. The cleanest tell: check which quantity changes by a constant amount per equal time step .
Intuition What to notice in the figure aboveThe amber dots are the raw data replotted as 1/ [ A ] . Notice they fall on a dead-straight line — each equal 100 s step lifts 1/ [ A ] 1.00 M − 1 1/ [ A ] extends that same straight line one step further to read off the prediction at t = 400 [ A ] ln [ A ]
Worked example Ex 8 — Which order is this?A reactant is monitored:
t 0
100
200
300
[ A ] 1.00
0.50
0.333
0.25
Identify the order, find k [ A ] t = 400 s
Forecast: The concentration halved in the first 100 s, but did not halve again by 200 s (it went to 0.333, not 0.25). A shrinking fall like that is the signature of second order (fast then slow).
Step 1. Test zero order: is Δ [ A ] 0.50 , 0.167 , 0.083 not constant. Reject zero order.
Why this step? Zero order falls in a straight line (equal drops per equal time), so unequal drops rule it out immediately.
Step 2. Test first order: is the ratio per step constant? 0.50/1.00 = 0.50 0.333/0.50 = 0.667 not constant. Reject first order.
Why this step? First order multiplies by the same factor each equal interval; a changing factor rules it out.
Step 3. Test second order: is 1/ [ A ] constant amount?
[ A ] 1 : 1.00 , 2.00 , 3.00 , 4.00 1.00 M − 1 Second order confirmed.
Why this step? Second order is the one that is linear in 1/ [ A ]
Step 4. Slope gives k k = Δ t Δ ( 1/ [ A ]) = 100 1.00 = 1.0 × 1 0 − 2 M − 1 s − 1 Why this step? For second order k 1/ [ A ] t 1.00 M − 1 100
Step 5. Predict t = 400 s 1/ [ A ] = 1.00 + ( 0.01 ) ( 400 ) = 5.00 [ A ] 400 = 5.00 1 = 0.20 M Why this step? Once the line's slope and intercept are known, extrapolating one more step is just reading the straight line at t = 400
Verify: Reproduce the table: 1/ [ A ] = 1 + 0.01 t t = 200 1/ [ A ] = 3 ⇒ 0.333 t = 300 = 4 ⇒ 0.25 0.20
The single most illuminating comparison: feed the same [ A ] 0 same numeric k t 1/2
Intuition What to notice in the figure aboveThree bars, same input numbers, three different heights . The white (0th) bar is tallest — a big starting stock takes longest to burn through at a fixed removal rate. The amber (2nd) bar is shortest — here a large starting concentration means abundant collision partners, so the first halving happens soonest . The cyan (1st) bar sits in the middle and, crucially, would not move at all if you changed [ A ] 0 k see the three dependences on [ A ] 0
Worked example Ex 9 — One start, three fatesTake [ A ] 0 = 2.0 M k = 0.10 t 1/2
Forecast: Zero order should feel "slow to start" because it loses a fixed amount ; second order should be fastest to reach half here because the large [ A ] 0 [ A ] 0
Step 1. Zero order: t 1/2 = 2 k [ A ] 0 = 2 ( 0.10 ) 2.0 = 10 s Why this step? Half the material at a constant removal rate — bigger start means longer half-life (proportional to [ A ] 0
Step 2. First order: t 1/2 = k 0.693 = 0.10 0.693 = 6.93 s Why this step? Exponential decay halves in a fixed time regardless of [ A ] 0
Step 3. Second order: t 1/2 = k [ A ] 0 1 = ( 0.10 ) ( 2.0 ) 1 = 5.0 s Why this step? Here a larger start actually shortens the first half-life — plenty of partners to collide with early on (inversely proportional to [ A ] 0
Verify: 10 , 6.93 , 5.0 [ A ] 0
Recall Quick self-testZero-order NH3 [ A ] 0 = 0.10 k = 2.5 × 1 0 − 4 − 1 t end = [ A ] 0 / k = 400 1/ [ A ] [ A ] 0 k t 1/2 M mol L − 1
Mnemonic Pick-the-law reflex"Amount, Ratio, Reciprocal." Constant amount lost per step → 0th. Constant ratio per step → 1st. Constant jump in the reciprocal → 2nd.
