Pseudo-first-order kinetics: Question bank — Chemistry deep dive

Before we start, three pieces of notation must be earned before any question uses them:

Keep order and molecularity apart and half these traps disappear; keep the units of $k^{'}$ vs $k$ apart and the other half do.

Picture the whole page in one figure first. The two panels below are the mental image every question secretly refers to — a genuinely second-order run bends on a $ln [A]$ -vs- $t$ plot, while the same reaction flooded with $B$ straightens out.

And the reason $B$ can be treated as constant is purely visual: the flooded reservoir barely dips while $A$ empties. The next figure shows exactly how big the "excess" must be before that claim is safe.

Two short derivations we will lean on repeatedly:

True or false — justify

Pseudo-first-order kinetics changes the true molecularity of the reaction.

False — molecularity counts colliding particles in a step and is fixed by the mechanism; flooding one reactant only changes what our measured rate law looks like, not the physical collision.

A pseudo-first-order plot of

ln [A]

t

being straight proves the reaction is genuinely first-order overall.

False — it only proves first-order in A under those conditions. Straightness is engineered by holding

[B]

constant; the true overall order is still second (see Method of Isolation).

The pseudo-first-order rate constant

k^{'}

and the true rate constant

k

have the same units.

False —

k^{'}

s^{- 1}

(first-order) while

k

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

(second-order); they differ by the units of the excess concentration

[B]_{0}

, since

k^{'} = k [B]_{0}

(see Integrated Rate Laws).

If you double

[B]_{0}

(still in large excess), the measured

k^{'}

doubles — assuming the reaction is first-order in

B

True — with

Rate = k [A] [B]

we have

k^{'} = k [B]_{0}^{1}

, so

k^{'}

scales linearly; but if it were order

n

B

, doubling

[B]_{0}

would multiply

k^{'}

2^{n}

If you double

[B]_{0}

, the half-life of A changes.

True — the Half-life is

t_{1/2} = ln 2/ k^{'} = ln 2/ (k [B]_{0})

, so doubling

[B]_{0}

halves the half-life even though "first-order half-life is constant" still holds for a fixed run.

For a fixed pseudo-first-order run, the half-life of A is independent of

[A]_{0}

True — that is the signature of first-order behaviour; within one run

k^{'}

is constant, so

t_{1/2} = ln 2/ k^{'}

does not depend on how much A you started with.

Water in an aqueous hydrolysis is a spectator and does not appear in the true rate law.

False — water is a genuine reactant; it is simply at

\approx 55.5 M

and barely changes, so it hides inside

k^{'}

rather than vanishing.

A reaction can be pseudo-second-order.

True — for a third-order reaction

k [A] [B] [C]

, flooding both

B

and

C

collapses it to

k^{''} [A]

with

k^{''} = k [B]_{0} [C]_{0}

; but flooding only one of the three leaves a genuine second-order dependence, so "pseudo-lower-order" is a general idea.

Doubling

[A]_{0}

(the limiting reactant) while keeping the same excess of B changes

k^{'}

False —

k^{'} = k [B]_{0}

depends only on the excess reactant; changing the limiting reactant's start amount changes how long the run takes but not

k^{'}

A reaction that is order

2

in the flooded reactant

B

still gives a straight

ln [A]

-vs-

t

line when

A

is first-order.

True — flooding fixes

[B]_{0}^{2}

as a constant inside

k^{'} = k [B]_{0}^{2}

, so the

A

-dependence is unchanged; the plot is straight, but

k^{'}

now scales as the square of

[B]_{0}

Spot the error

A student measures

k^{'} = 0.05 s^{- 1}

with

[B]_{0} = 1.0 M

and reports the reaction's true rate constant as

0.05 s^{- 1}

They forgot to divide out the excess:

k = k^{'} / [B]_{0} = 0.05/1.0 = 0.05 M^{- 1} s^{- 1}

— same number, but the units change to second-order, which is the whole point.

A student uses a 10× excess of B, consumes 50% of A, and claims

[B]

was constant.

When half of A reacts, B has dropped from

10 [A]_{0}

9.5 [A]_{0}

— a 5% change — which is large enough to bend the "straight" line; 10× is too small for reliable pseudo-first-order.

Given the ester hydrolysis run, a student writes

k = k^{'} \times [H_{2} O] = 2.5 \times 1 0^{- 4} \times 55.5

The relation is

k = k^{'} / [B]_{0}

, not

k^{'} \times [B]_{0}

; you remove the flooded concentration to recover the true

k

, so

k = 2.5 \times 1 0^{- 4} /55.5 = 4.5 \times 1 0^{- 6} M^{- 1} s^{- 1}

A student sees a curved

ln [A]

-vs-

t

plot and concludes the reaction is first-order but with a changing

k

A rate "constant" cannot change during a run at fixed temperature; curvature means the assumed order is wrong — here both concentrations vary, so it is genuinely second-order, not first with a drifting

k

A student says: "Since

k^{'} = k [B]_{0}

, raising

[B]_{0}

speeds the reaction, so the true rate constant

k

increased."

k

is untouched — it depends only on temperature and Activation Energy (via Collision Theory); only the apparent constant

k^{'}

grew because it carries

[B]_{0}

A student floods a third-order reaction with only

B

and writes

k^{''} = k [B]_{0} [C]_{0}

You cannot fold in

[C]_{0}

— you did not flood

C

, so

[C]

still varies; the correct collapse is

Rate = k [B]_{0} [A] [C]

, a pseudo-second-order law, not pseudo-first.

A student determines

k^{'}

from

[A] = [A]_{0} e^{- k^{'} t}

but plugs in

t

in minutes while

k^{'}

is in

s^{- 1}

Units must match: convert time to seconds (or

k^{'}

min^{- 1}

) before exponentiating, or the exponent is meaningless.

Why questions

Why does flooding one reactant turn a curved second-order plot into a straight first-order line?

Holding

[B] \approx [B]_{0}

removes the second varying concentration, so

Rate = k [B]_{0} [A] = k^{'} [A]

— the mathematical form of first-order — and only then does

ln [A]

fall linearly (see the left→right panels above).

Why do we require roughly 50–100× excess rather than just "more"?

The fractional change in B over the whole run is about

[A]_{0} / [B]_{0}

; at 100× that is ~1% error, acceptable, whereas at 10× it can reach ~5–10%, which visibly distorts

k^{'}

Why does the exponent

n

on the flooded reactant not appear in the apparent order?

Because

Rate = k [A]^{m} [B]_{0}^{n} = k^{'} [A]^{m}

hides all of

[B]_{0}^{n}

inside

k^{'} = k [B]_{0}^{n}

; the slope still reports

m

, while

n

only shows up if you vary

[B]_{0}

and watch how

k^{'}

scales.

Why is pseudo-first-order the natural regime for reactions in water?

Pure water sits at

\approx 55.5 M

, dwarfing typical solute concentrations by thousands of times, so any reaction that consumes water is automatically flooded and behaves first-order in the solute.

Why do we bother "hiding"

[B]

inside

k^{'}

instead of solving the full second-order equation?

The full second-order integrated law mixes both concentrations and is awkward to fit; the pseudo trick isolates one variable, giving a clean straight line and one easily measured slope.

Why does enzyme catalysis at low substrate become pseudo-first-order in substrate?

When

[S] ≪ K_{M}

, the Michaelis–Menten rate simplifies to

(V_{ma x} / K_{M}) [S]

, a constant times

[S]

— exactly the pseudo-first-order form, because the enzyme is effectively in relative excess.

Why does the Method of Isolation use deliberate excess for each reactant in turn?

By flooding all reactants but one, the rate depends only on that one; repeating for each reactant reveals its individual order, letting you rebuild the full rate law piece by piece.

Why does raising temperature change

k^{'}

even at fixed

[B]_{0}

k^{'} = k [B]_{0}

and the true

k

climbs with temperature via the Arrhenius/Activation Energy dependence, so

k^{'}

rises even though the flooded concentration is unchanged.

Edge cases

What happens to the pseudo-first-order approximation as

[A]_{0}

approaches

[B]_{0}

The excess ratio collapses toward 1,

[B]

can no longer be treated as constant, and the behaviour reverts to true second-order — the straight line curves (right panel of the figure loses its straightness).

What if the reaction is appreciably reversible,

A + B ⇌ P

The reverse rate

k_{- 1} [P]

can no longer be ignored; even with

B

flooded the integrated law becomes

ln ([A] - [A]_{e q}) = ln ([A]_{0} - [A]_{e q}) - (k^{'} + k_{- 1}) t

, so the plot is straight against

[A] - [A]_{e q}

and its slope is the sum

k^{'} + k_{- 1}

, not

k^{'}

alone.

Why does the reversible correction subtract

[A]_{e q}

instead of using

[A]

directly?

Because at equilibrium the net rate is zero, not

[A] = 0

; the driving quantity is the distance from equilibrium

[A] - [A]_{e q}

, and that combination decays as a clean single exponential where a plain

ln [A]

would curve.

What if

[B]_{0}

is enormous but B is also a product?

If B is regenerated or barely consumed it stays constant regardless, so the pseudo-first-order form still holds — the key is constancy of

[B]

, not whether it is technically consumed.

What is the "order" of a pseudo-first-order reaction if you refuse to flood any reactant?

There is no pseudo behaviour at all — you measure the genuine order (e.g. second overall), because pseudo-first-order is a condition you impose, not an intrinsic property.

In the limit

t \to \infty

for an irreversible pseudo-first-order run, what does

[A]

do?

[A] = [A]_{0} e^{- k^{'} t} \to 0

; A is fully consumed exponentially, while the flooded B settles at

[B]_{0} - [A]_{0} \approx [B]_{0}

, still practically unchanged. (For a reversible run it instead levels off at

[A]_{e q} > 0

What happens to

k^{'}

[B]_{0} \to 0

(excess removed entirely)?

k^{'} = k [B]_{0} \to 0

, meaning the apparent rate vanishes — correctly reflecting that with no B present the reaction cannot proceed; the pseudo picture degenerates.

Does a pseudo-first-order reaction have a pH dependence when acid-catalyzed (like sucrose inversion)?

Yes — the catalyst

[H^{+}]

folds into the effective constant too, so

k^{'}

depends on pH even though the plot stays first-order in sucrose; changing buffer pH shifts

k^{'}

If you accidentally start a "pseudo" run with B in only 3× excess, will the

ln [A]

plot be perfectly straight, perfectly curved, or slightly bent?

Slightly bent — B changes appreciably as A is consumed, so the line has gentle curvature rather than a clean slope, and any

k^{'}

you extract is only a rough average.

Recall One-line self-test

"Pseudo" means ::: apparent/false — the reaction is truly higher-order; we only made it look first-order by holding one concentration constant. To recover true $k$ from $k^{'}$ you ::: divide by the excess concentration, $k = k^{'} / [B]_{0}^{n}$ (with $n$ the order in the flooded reactant), which also fixes the units back to those of the genuine rate constant.