Differential rate equations for 0, 1st, 2nd order — derivations: Question bank

This is a misconception hunt. Every line below is a trap the topic sets. Read the prompt, answer out loud before you reveal, and — most importantly — check that your reasoning matches the justification, not just the yes/no. If you only get the verdict right but the reason wrong, you have learned nothing.

New to the machinery? Rebuild it first from the parent note, and keep Rate Laws and Reaction Order, Half-Life Calculations, and Integrated Rate Laws Applications open in another tab.

True or false — justify

A zero-order reaction has a rate that never changes at all, even as reactant runs out.

False — the rate stays

k

only while reactant is present; once

[A]

hits zero the reaction stops abruptly, so the constant-rate line has a hard floor at

[A] = 0

The half-life of a first-order reaction gets shorter as the reaction proceeds.

False — first-order half-life is

t_{1/2} = 0.693/ k

, which contains no concentration term, so every successive half-life is identical regardless of how much is left.

For a second-order reaction, the time to fall from 1.0 M to 0.5 M is shorter than the time to fall from 0.5 M to 0.25 M.

True — second-order half-life

t_{1/2} = 1/ (k [A]_{0})

grows as

[A]_{0}

falls, so each successive halving takes longer; a dilute reaction crawls.

The rate constant

k

is a property of the reaction that changes as concentration drops.

False —

k

depends on temperature (and catalyst), not on concentration; it is exactly what stays fixed while concentration changes, which is why we can integrate the rate law at all.

A straight line on a plot of

[A]

t

guarantees the reaction is zero-order.

True for the reactant tested — only zero-order gives

[A]

itself linear in

t

; first- and second-order curve on that axis, so this specific linear plot is the zero-order fingerprint.

The negative sign in

- d [A] / d t = k [A]^{n}

means the rate is a negative number.

False —

d [A] / d t

is itself negative (concentration falls), so the minus sign flips it to make the rate a positive quantity, as a rate should be.

Doubling

[A]

doubles the rate for a second-order reaction.

False — rate

\propto [A]^{2}

, so doubling concentration quadruples the rate; doubling only holds for first-order.

The units of

k

are the same for all three orders.

False — dimensional consistency forces

k = M^{1 - n} s^{- 1}

, so zero-order is

M s^{- 1}

, first-order is

s^{- 1}

, second-order is

M^{- 1} s^{- 1}

Spot the error

"For first-order I integrated

\int d [A] / [A] = ln [A]

, so the integrated law is

[A]_{t} = [A]_{0} - k t

The integral is correct but the final form is wrong — logs subtract, giving

ln [A]_{t} - ln [A]_{0} = - k t

, i.e.

[A]_{t} = [A]_{0} e^{- k t}

; the linear-difference form belongs to zero-order.

"Second-order integrates to

\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} - k t

because there's a minus sign in the rate law."

Wrong sign — carrying the antiderivative

- [A]^{- 1}

through the limits flips things so the correct law is

\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t

; the reciprocal grows with time because concentration shrinks.

"I plotted

ln [A]

t

, got a straight line, so the reaction is second-order."

A straight

ln [A]

plot signals first-order; second-order linearizes only when you plot

1/ [A]

t

— the left side of the integrated law is always your y-axis.

"The reaction

2 A \to P

must be second-order because of the coefficient 2."

The stoichiometric coefficient does not set the order — order is experimentally determined; a reaction written

2 A \to P

can be zero-, first-, or second-order depending on its mechanism.

k

for the ammonia-on-platinum decomposition is

2.5 \times 1 0^{- 4} s^{- 1}

The exponent is fine but the units are wrong — a zero-order

k

carries

M s^{- 1}

, so it should read

2.5 \times 1 0^{- 4} M s^{- 1}

; the units themselves encode the order.

"Since rate =

- \frac{1}{a} \frac{d [ A ]}{d t}

, I can drop the

\frac{1}{a}

factor whenever I like."

Only if you don't care about matching

k

to a specific species — the

1/ a

converts the disappearance rate of A into the reaction rate, so omitting it silently rescales

k

by a factor of

a

"Half-life for zero-order is

0.693/ k

, same idea as first-order."

No — set

[A]_{t} = [A]_{0} /2

[A]_{t} = [A]_{0} - k t

to get

t_{1/2} = [A]_{0} / (2 k)

; the zero-order half-life depends on initial concentration, and

0.693

never appears.

Why questions

Why does zero-order rate refuse to depend on concentration even though there's plenty of reactant?

The rate is limited by something other than reactant abundance — usually a saturated catalyst surface or a fixed enzyme/photon supply — so extra reactant just waits in line and cannot speed things up.

Why does the natural logarithm appear specifically in the first-order law and nowhere else?

Because

\int d [A] / [A] = ln [A]

— the log is the unique antiderivative of

1/ [A]

, so it is forced to appear exactly when the rate carries the first power of concentration.

Why does the second-order rate depend on

[A]^{2}

physically, not just algebraically?

A second-order step needs two A molecules to meet; the chance of a meeting scales as (number of A)

\times

(number of A), so doubling A quadruples encounters and hence rate.

Why is a first-order half-life so useful for dating and dosing?

Because it is concentration-independent — the fraction lost per fixed time is constant, so "how much is left" depends only on how many half-lives have elapsed, giving a clean exponential clock.

Why must the units of

k

change with order at all?

Rate always carries

M s^{- 1}

, but

k

multiplies

[A]^{n}

which carries

M^{n}

; to keep the product at

M s^{- 1}

k

must absorb

M^{1 - n}

, so its units track the order automatically.

Why do we bother separating variables before integrating?

Integration needs all the

[A]

-stuff on one side and all the

t

-stuff on the other; separation is what turns a single equation linking both into two independent integrals we can evaluate.

Why does the same reaction sometimes appear first-order when it's really second-order?

If one reactant is in huge excess its concentration barely changes, folding into

k

as a pseudo-constant — this "pseudo-first-order" trick is a deliberate simplification, not a change in true mechanism (see Rate Laws and Reaction Order).

Edge cases

What does the zero-order

[A]

-vs-

t

line do at the moment reactant is exhausted?

It hits

[A] = 0

at time

t = [A]_{0} / k

and must stop — the mathematical line would go negative, but concentration cannot, so the physical model truncates there.

Can a first-order concentration ever reach exactly zero in finite time?

No —

[A]_{t} = [A]_{0} e^{- k t}

approaches zero asymptotically; the exponential never actually touches zero, so first-order reactions technically never fully "finish."

What happens to second-order half-life if you start extremely dilute, near

[A]_{0} \to 0

t_{1/2} = 1/ (k [A]_{0})

blows up toward infinity — with almost no molecules to collide, the reaction becomes vanishingly slow, matching the collision picture.

What if

k = 0

for any order — is the reaction still described by these laws?

Yes trivially — every integrated law collapses to

[A]_{t} = [A]_{0}

, meaning no reaction;

k = 0

is the degenerate "nothing happens" case common to all orders.

For a reaction that is genuinely instantaneous compared to observation time, which order model applies?

None cleanly — these differential laws assume a measurable finite rate; an essentially instantaneous step is diffusion- or mixing-limited and falls outside the concentration-driven framework entirely.

t = 0

, do all three integrated laws agree?

Yes — plug

t = 0

into each and the right-hand extra term vanishes, giving

[A]_{t} = [A]_{0}

(or

ln [A]_{0}

, or

1/ [A]_{0}

); they only diverge as time advances, which is exactly what lets us tell orders apart.

Recall One-line self-test before you leave

Say aloud: which quantity goes on the y-axis to straighten each order, and how does each half-life depend on $[A]_{0}$ ? Answer ::: y-axis is $[A]$ (0th), $ln [A]$ (1st), $1/ [A]$ (2nd); half-life $\propto [A]_{0}$ (0th), independent (1st), $\propto 1/ [A]_{0}$ (2nd).