This page is a self-test ladder. Each problem states its difficulty level, and the full solution hides inside a collapsible callout — try it first, then reveal. Every symbol used here is built in the parent note; if you meet one you have forgotten, look there first.
Before we start, one picture to fix the two rate ideas in your mind.
The three moves you will repeat all page:
Slope = change in timechange in concentration.
Sign flip — put a minus in front for a reactant so the rate comes out positive.
Normalize — divide the species rate by its stoichiometric coefficient to get the reaction rate r.
What we do: apply the definition of average rate for a reactant.
ravg=−ΔtΔ[A]=−60−00.50−0.80Why the minus:Δ[A]=0.50−0.80=−0.30M is negative (A is disappearing). The minus in the definition flips it to a positive speed.
ravg=−60−0.30=5.0×10−3M/sAnswer:5.0×10−3M/s.
Recall Solution 1.2
What we do: recognize which line is which (look at figure s01).
The chord (amber) gives the average rate over the whole interval. The tangent (white) touches the curve at exactly one point and gives the instantaneous rate there.
Answer: the ==tangent line at t=40 s==.
(a) Consumption rate of N₂O₅ — slope then sign flip.
−ΔtΔ[N2O5]=−500.070−0.100=−50−0.030=6.0×10−4M/s(b) Reaction rate — normalize by the coefficient of N₂O₅, which is 2.
r=21(6.0×10−4)=3.0×10−4M/sWhy divide by 2: two N₂O₅ molecules vanish per single reaction event, so the reaction "clock ticks" half as fast as N₂O₅ disappears.
Answer: (a) 6.0×10−4 M/s; (b) 3.0×10−4 M/s.
Recall Solution 2.2
What we use: the linked-rate relation
r=41dtd[NO2]=11dtd[O2].NO₂ (coefficient 4, a product so no minus needed):
dtd[NO2]=4r=4×3.0×10−4=1.2×10−3M/sO₂ (coefficient 1):
dtd[O2]=1⋅r=3.0×10−4M/sWhy multiply: for every reaction event, 4 NO₂ appear but only 1 O₂ — so NO₂ builds up four times as fast.
Answer: NO₂ forms at 1.2×10−3 M/s; O₂ at 3.0×10−4 M/s.
Recall Solution 2.3
Relation:r=−dtd[H2]=21dtd[HI] (H₂ coefficient 1, HI coefficient 2).
−dtd[H2]=21(8.0×10−4)=4.0×10−4M/sWhy halve: 2 HI form per 1 H₂ used, so HI appears twice as fast as H₂ disappears.
Answer:H2 is consumed at 4.0×10−4 M/s.
Average consumption rate of H₂O₂:−ΔtΔ[H2O2]=−12000.40−1.00=12000.60=5.0×10−4M/sAverage reaction rate (divide by coefficient 2):
ravg=21(5.0×10−4)=2.5×10−4M/sInitial instantaneous reaction rate (given consumption 1.2×10−3, divide by 2):
rinst(0)=21(1.2×10−3)=6.0×10−4M/sRatio:2.5×10−46.0×10−4=2.4.
Why the instant is bigger: the reaction slows as [H2O2] falls (fewer molecules to collide — see Collision theory). At t=0 the concentration is highest, so the tangent is steepest. The average over 0–1200 s blends fast early moments with slow late ones, landing below the peak initial rate. Look at s01: the tangent at the left is steeper than the chord.
Answer:ravg=2.5×10−4 M/s, rinst(0)=6.0×10−4 M/s; the initial instant is 2.4× larger.
Recall Solution 3.2
Why a derivative here: instantaneous rate = tangent slope = −dtd[A]. Because [A] is given as a smooth formula, we can differentiate exactly instead of measuring a slope.
The derivative of ekt is kekt, so
dtd[A]=0.80⋅(−0.010)e−0.010t=−8.0×10−3e−0.010t.At t=0:e0=1,
rinst(0)=−dtd[A]=8.0×10−3M/s.At t=100:e−1≈0.3679,
rinst(100)=8.0×10−3×0.3679=2.94×10−3M/s.Average over 0–100 s: first the endpoint concentrations.
[A]0=0.80 M; [A]100=0.80e−1=0.2943 M.
ravg=−1000.2943−0.80=1000.5057=5.06×10−3M/s.Sanity check:ravg (5.06×10−3) lies between the fast initial (8.0×10−3) and the slow later (2.94×10−3) instantaneous rates — exactly as a chord slope must sit between the steepest and shallowest tangents.
Answer:rinst(0)=8.0×10−3, rinst(100)=2.94×10−3, ravg=5.06×10−3 M/s.
Master relation:r=−dtd[N2]=−31dtd[H2]=21dtd[NH3].(a) From NH₃ (coefficient 2):
r=21(6.0×10−4)=3.0×10−4M/s.(b) N₂ (coefficient 1): −dtd[N2]=r=3.0×10−4M/s.(c) H₂ (coefficient 3): −dtd[H2]=3r=9.0×10−4M/s.Why H₂ is fastest: three H₂ are used per reaction event, so H₂ vanishes three times as fast as N₂.
Common error to name: claiming all three species change at the same 6.0×10−4 M/s. They do not — only the normalized rate r is shared.
Answer:r=3.0×10−4; N₂ at 3.0×10−4; H₂ at 9.0×10−4 M/s.
Recall Solution 4.2
Idea: average rate × time interval = concentration change (rearranged from r=−Δ[A]/Δt, coefficient 1 so r equals the consumption rate).
First interval (0→100 s):Δ[A]=−rΔt=−(4.0×10−3)(100)=−0.40M.[A]100=1.00−0.40=0.60M.Second interval (100→300 s):Δ[A]=−(1.5×10−3)(200)=−0.30M.[A]300=0.60−0.30=0.30M.Speeding up or slowing? The average rate dropped from 4.0×10−3 to 1.5×10−3 M/s: the reaction is slowing down as A is used up — the standard behaviour explained by Collision theory.
Answer:[A]100=0.60 M, [A]300=0.30 M; slowing down.
Master relation for 2A+B→3C:
r=−21dtd[A]=−11dtd[B]=31dtd[C].(a) From C (coefficient 3):
r=31(9.0×10−4)=3.0×10−4M/s.(b) Solve each species, keeping signs (A and B are reactants → negative):
dtd[A]=−2r=−2(3.0×10−4)=−6.0×10−4M/s,dtd[B]=−1⋅r=−3.0×10−4M/s.(c) The student's error: they dropped the negative sign. The reactant A is being consumed, so d[A]/dt must be negative. The magnitude 6.0×10−4 is correct, but the sign is not.
Correct value:dtd[A]=−6.0×10−4M/s.
Answer: (a) r=3.0×10−4; (b) d[A]/dt=−6.0×10−4, d[B]/dt=−3.0×10−4; (c) missing minus sign, correct is −6.0×10−4 M/s.
Recall Solution 5.2
(a) Instantaneous rate at t=0. Differentiate [A]=[A]0(1+k[A]0t)−1. Using the chain rule (derivative of u−1 is −u−2u′ with u=1+k[A]0t, u′=k[A]0):
dtd[A]=−[A]0⋅(1+k[A]0t)2k[A]0=−(1+k[A]0t)2k[A]02.
At t=0 the denominator is 1:
rinst(0)=−dtd[A]0=k[A]02=0.20×(0.50)2=0.050M/s.(b) Average rate over 0–10 s. Endpoint concentrations:
[A]0=0.50 M.
1+k[A]0t=1+0.20×0.50×10=1+1.0=2.0, so [A]10=0.50/2.0=0.25 M.
ravg=−100.25−0.50=100.25=0.025M/s.(c) Why initial rates are trustworthy: the instantaneous rate at the very start (0.050 M/s) is exactly k[A]02 — a clean expression in the known starting concentration, before any product has built up or reverse reactions kick in. The average (0.025 M/s) is already blurred by 10 s of slowing. Measuring the rate at t=0 (initial-rate method) isolates the true starting speed and is the basis of finding the rate law.
Answer: (a) 0.050 M/s; (b) 0.025 M/s; (c) initial rate =k[A]02, unblurred by later slowdown.
Recall Quick self-check questions
What is the reaction rate for 2A→B if [A] drops at 0.04 M/s? ::: r=21(0.04)=0.02 M/s.
Average or instantaneous: which uses a tangent line? ::: Instantaneous.
Average or instantaneous: which uses a chord (two endpoints)? ::: Average.
For a reactant, why does the definition carry a minus sign? ::: Because Δ[A] is negative and rate is reported positive; the minus flips it.
In N2+3H2→2NH3, which species changes concentration fastest? ::: H2 (coefficient 3).
