Zero And First Order Graphs
Learn and apply Zero And First Order Graphs in the published Chemistry course sequence.
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The core idea
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H1 Zero- and First-order Graphs and Half-life: Orientation
Concentration–time evidence distinguishes zero and first order through shape, gradient and successive half-lives. H1 uses graphical reasoning without integrated rate equations.
- Justify zero- or first-order behaviour and recognise concentration-independent first-order half-life.
- Do not use integrated forms, ln plots, or calculate k from t_(1/2).
Definitions (Must Know)
- A concentration–time graph shows how a reactant concentration changes during a reaction.
- Half-life, t_(1/2), is the time for a reactant concentration to fall to half its current value.
- A zero-order reactant is consumed at a constant rate under the stated conditions.
- A first-order reactant has a rate proportional to its concentration.
Detailed Explanations
A. Zero-order evidence
A straight concentration–time line has constant negative gradient. Equal amounts are consumed in equal times, so successive halvings take less time as concentration falls.
B. First-order evidence
The curve becomes less steep because rate falls with concentration. Each halving takes the same time, so half-life is independent of the concentration from which the interval begins.
The reason is proportionality: for a first-order reactant, rate is proportional to concentration. Starting from half the concentration, there is half as much reactant to lose before the concentration halves, and at every matching point along the way the rate is also half as large. The two effects cancel, so the halving takes the same time.
C. Reading a graph fairly
Use clear horizontal and vertical guide lines and quote approximate intervals with sensible precision. Allow for experimental scatter.
D. Keep this at H1 depth
H1 explains these signatures directly. Integrated equations, logarithmic plots and k = 0.693/t_(1/2) belong outside this prescribed treatment.
Worked Examples
Modelled example 1
First-order evidence
Problem
Study the worked solution
Identify the intervals
Method
Recognise both changes as successive halvings.Reason
0.40 is half of 0.80, and 0.20 is half of 0.40.Working
First half-life = 30 s; second half-life = 30 s.Infer the order
Method
Use the approximately constant successive half-lives.Reason
Concentration-independent half-life is the prescribed first-order signature.Working
The evidence supports first order in A.
Guided practice 2
Zero-order evidence
Problem
Try this before viewing the solution
Hints
Hint 1: use the straight line
Hint 2: repeat the decrease
View solution step by step
Compare concentration changes
Method
Calculate the first decrease and the remaining decrease.Reason
Both are 0.30 mol dm⁻³.Working
0.60 → 0.30 and 0.30 → 0 are equal changes.Extend the constant gradient
Method
Add another 40 s to the first 40 s.Reason
Equal concentration changes take equal times on a straight line.Working
Zero is reached at 40 + 40 = 80 s.
Common misconception 3
Correct a half-life timing error
Learner calculation
Choose the second interval
View solution step by step
Locate successive endpoints
Method
Use 30 s as the start of the second halving and 60 s as its end.Reason
Each half-life is an interval between a concentration and half that concentration.Working
Second interval: 60-30 = 30 s.Correct the conclusion
Method
State that both half-lives are 30 s.Reason
The learner confused a graph time coordinate with an elapsed interval.Working
Successive constant half-lives support first-order behaviour.
Examiner practice 4
Classify two concentration–time traces
Problem
Try this before viewing the solution
View solution step by step
Classify P
1 markMethod
Identify equal absolute decreases in equal times.Reason
P falls by 0.20 mol dm⁻³ every 20 s.Working
P has constant negative gradient.Conclude for P
1 markMethod
State that P supports zero order.Reason
Constant concentration change per unit time is the zero-order signature.Working
P: zero order.Classify Q
1 markMethod
Identify successive halvings in equal times.Reason
Q halves every 30 s.Working
Q has constant half-life.Conclude for Q
1 markMethod
State that Q supports first order.Reason
Concentration-independent half-life is the first-order signature.Working
Q: first order.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Credit each graphical feature and its linked order conclusion.
Challenge 5
Continue a shifted first-order trace
Problem
Try this before viewing the solution
Hints
Hint 1: read the half-life
Hint 2: continue from 55 seconds
View solution step by step
Find the half-life
Method
Subtract the graph times for 0.60 and 0.30.Reason
Those concentrations differ by one halving.Working
t_(1/2) = 55-15 = 40 s.Count further halvings
Method
Continue 0.30 → 0.15 → 0.075.Reason
First-order half-life remains constant.Working
Two additional half-lives require 2(40) = 80 s.Return to graph time
Method
Add 80 s to the 55 s coordinate.Reason
The question asks for the graph time, not only elapsed time after 55 s.Working
A reaches 0.075 mol dm⁻³ at 135 s.
Mind Stretchers
Attempt each unfamiliar application before opening the hint, then compare your reasoning with the solution.
Mind stretcher 1: Distinguishing two claimsExtension
Question. A graph has a constant gradient. Does it also have a constant half-life?
Show Hint
Compare the time for 0.80→0.40 with 0.40→0.20 on a straight line.
Show Answer
No. Equal concentration decreases take equal time, so the smaller second halving takes less time. Constant gradient supports zero order, not constant half-life.
Mind stretcher 2: Evidence with scatterExtension
Question. Successive half-lives are 41, 39 and 42 s. Give a defensible conclusion.
Show Hint
Treat realistic measurements as approximate.
Show Answer
They are approximately constant within experimental scatter, supporting first-order behaviour.