2.8.1 · D5Chemical Kinetics
Question bank — Rate of reaction — average vs instantaneous
This is a question bank built to break your misconceptions before an exam does. Each line is a prompt; the part after ::: is the reasoning. Cover the answer, think, then reveal. These are conceptual traps — no heavy arithmetic (that lives in the calculation decks). Parent: Rate of Reaction — Average vs Instantaneous.
True or false — justify
The average rate always equals the instantaneous rate at the midpoint of the interval.
False. That only holds if the concentration–time curve is a straight line over the interval; real reactant curves bend (they slow down), so the midpoint tangent is not the interval average.
For a reactant, and have the same units.
True. Both are (mol/L)/s = M/s; and describe a change in the same quantity over time, so the units are identical — only the size of the time gap differs.
The instantaneous rate is just an average rate over a very small time interval.
True in spirit. It is the limit of the average rate as ; make the interval vanishingly small and the average over it becomes the exact rate at that instant.
A reaction rate can legitimately be reported as a negative number.
False. Rate is defined as positive; the minus sign in exists precisely to flip the negative slope of a disappearing reactant into a positive rate.
For , the rate of disappearance of A equals the rate of appearance of B.
False. Two A vanish per one B formed, so A disappears twice as fast as B appears; only after dividing by the coefficients do the normalized rates match.
The average rate over a long interval is generally smaller than the initial instantaneous rate.
True for typical reactions. Rate falls as reactant is used up, so the early fast period is diluted by later slow periods, pulling the average below the starting rate.
Doubling the length of the time interval always halves the average rate.
False. Average rate is , and over the longer interval is not fixed — the numerator changes too, usually less than proportionally because the reaction slows.
The reaction rate is the same number whether you track A, B, C or D.
True — that is the entire point of dividing by stoichiometric coefficients. Each species has its own , but collapses them to one shared value.
Spot the error
"Rate M/s, so the rate is M/s."
The negative sign of the definition was dropped. Correct rate M/s; a reactant's rate must come out positive.
"For , since NO₂ forms at M/s, the reaction rate is M/s."
NO₂ has coefficient 4, so M/s. Forgetting to divide by 4 inflates the reaction rate fourfold.
"I found the tangent slope at s to be M/s, so the instantaneous rate is M/s."
Same sign trap: M/s. The tangent slope is negative; the rate is its positive counterpart.
"[A] fell from 0.5 to 0.2 M over 0–100 s, so the rate at s is M/s."
That value is the average over the whole 0–100 s window, not the instantaneous rate at 50 s; the latter needs the tangent slope exactly at 50 s.
"Because O₂ has coefficient 1 in , the O₂ formation rate equals the reaction rate."
True here only because the coefficient is 1: . The statement is right but for the wrong stated reason if you thought coefficients never matter — they do, this one just happens to be 1.
"The units of rate are mol, since we track how many moles react."
Wrong — rate tracks concentration per time, so units are M/s = mol·L⁻¹·s⁻¹. Moles alone forget both the volume (concentration) and the time.
Why questions
Why do we need both an average and an instantaneous definition instead of just one?
Because reaction speed changes continuously; average captures the overall throughput of an interval, while instantaneous captures the exact speed at one moment (the "speedometer reading").
Why does the minus sign appear only in front of reactant terms, not product terms?
Reactant concentration decreases, giving a negative slope; the minus flips it positive. Product concentration increases, giving a positive slope already, so no sign is needed.
Why do we divide each species' rate by its stoichiometric coefficient?
To get one single "reaction rate" independent of which molecule you watch; coefficients say how many of each are consumed/made per reaction event, so dividing normalizes to "per event."
Why does the instantaneous rate need a tangent line rather than a chord (straight line between two points)?
A chord gives the average slope between two separated points (an average rate); the tangent touches the curve at one instant, giving the exact slope — the derivative — there.
Why is the initial instantaneous rate often the largest rate the reaction ever has?
At the reactant concentration is highest, so collisions between reactant molecules are most frequent (see Collision theory); as reactant is used up the rate falls.
Why can two students tracking N₂O₅ and O₂ in the same flask report different raw values yet agree on the reaction rate?
Their species vanish/appear at different molecular speeds (set by coefficients 2 and 1), but divides those differences away, leaving one shared .
Why isn't the average rate over 0–1000 s a good estimate of the rate at s for a decaying reaction?
The average is dominated by the fast early period; by s the reactant is nearly depleted and the true instantaneous rate is far smaller than the whole-interval average.
Edge cases
What is the instantaneous rate at the exact moment a reactant runs out (concentration hits zero)?
For an irreversible reaction it approaches zero — with no reactant left the tangent to the flat curve is horizontal, so .
If a concentration–time graph is a perfectly straight line, what is true about average vs instantaneous rate?
They are equal everywhere — a straight line has constant slope, so every tangent matches every chord; this is the rare zero-order-like case.
For a reaction at equilibrium, what does the net rate of change of concentration equal, and does that mean nothing is happening?
Net so concentrations are constant, but forward and reverse reactions still run at equal speed — dynamic balance, not a dead stop.
Can the instantaneous rate ever exceed every average rate that contains that instant?
Yes. If the reaction speeds up momentarily (e.g. an autocatalytic burst), the tangent at that point can be steeper than any surrounding chord, so its rate beats the enclosing averages.
What happens to the average rate as you shrink the interval toward zero around a fixed time ?
It converges to the instantaneous rate at — this limit is the definition of the derivative and hence of instantaneous rate.
Is "rate of reaction" defined when the balanced equation is unknown or unbalanced?
No single can be assigned — you need the stoichiometric coefficients to divide by; without a balanced equation you can only report species-specific values.
For a zero-order reaction, how does the instantaneous rate compare across different times?
It stays constant regardless of concentration (rate is independent of [A]), so instantaneous rate equals average rate throughout — a link explored further in Rate law and order of reaction and Integrated rate laws.
Recall One-line self-test
Say the two-word rule that fixes the sign and the two-word rule that fixes the coefficient. Sign :::: "add minus (for reactants)"; Coefficient :::: "divide by (stoichiometric number)".