Pseudo-first-order kinetics: Worked examples — Chemistry deep dive

This page is the drill ground for pseudo-first-order kinetics. The parent note built the idea; here we make sure that no matter what an exam throws at you — a strange fraction remaining, a half-life question, a "is it even pseudo-first-order?" trap, a units trick, or a limiting case where the whole approximation breaks — you have already seen a fully worked twin of it.

Everything rests on one equation from the parent, so let us re-earn every symbol in it before we start.

The scenario matrix

Every pseudo-first-order problem you will ever meet lives in one of these cells. The examples below are labelled with the cell they cover.

#	Cell class	What makes it tricky	Example
C1	Given $t$ → find $[A]$	pick exponential form	Ex 1
C2	Given fraction/percent → find $t$	pick log form, convert percent right	Ex 2
C3	Recover true $k$ from measured $k^{'}$	divide by $[B]_{0}$ , watch units	Ex 3
C4	Half-life & multiples ( $t_{1/2}$ , $t_{3/4}$ …)	half-life independent of $[A]_{0}$	Ex 4
C5	Graphical / slope reading	slope $= - k^{'}$ , curved vs straight	Ex 5
C6	Degenerate: excess too small	approximation breaks, quantify error	Ex 6
C7	Limiting: $t \to 0$ and $t \to \infty$	sanity endpoints, $[A] \to [A]_{0}$ and $\to 0$	Ex 7
C8	Real-world word problem + exam twist	translate words, two-excess trap	Ex 8

Cells that are "signs/quadrants" in a trig topic become, in kinetics, the direction of the inequality ( $[B]_{0} ≫ [A]_{0}$ vs comparable) and the endpoints of time ( $0$ , finite, $\infty$ ). We cover all of them.

Worked examples

Worked example Example 1 — Cell C1: given time, find concentration

Sucrose inversion has $k^{'} = 6.0 \times 1 0^{- 5} s^{- 1}$ , starting at $[A]_{0} = 0.50 M$ . What is $[A]$ after 2 hours?

Forecast: Two hours is short compared to how slowly $k^{'}$ acts here. Guess: still most of it left — somewhere near $0.3$ – $0.35 M$ .

Convert time to seconds: $t = 2 \times 3600 = 7200 s$ . Why this step? $k^{'}$ carries units $s^{- 1}$ , so the exponent $k^{'} t$ is only dimensionless if $t$ is in seconds.
We are given $t$ and want $[A]$ , so use the exponential form: $[A] = [A]_{0} e^{- k^{'} t}$ . Why this step? The unknown $[A]$ is already alone on the left — no algebra needed.
Exponent: $k^{'} t = (6.0 \times 1 0^{- 5}) (7200) = 0.432$ . Why this step? This number is "how many decay-lengths have passed"; $0.432$ means we are well under one, so little has reacted — matching our forecast.
$[A] = 0.50 \times e^{- 0.432} = 0.50 \times 0.6492 = 0.325 M$ .

Verify: Fraction left $= 0.325/0.50 = 0.65$ , so $35%$ reacted in 2 h — consistent with the forecast that "most" remains. Units: $M \times (dimensionless) = M$ . ✓

Worked example Example 2 — Cell C2: given percent completion, find time

Ethyl acetate hydrolysis, $k^{'} = 2.5 \times 1 0^{- 4} s^{- 1}$ . How long for 75% completion?

Forecast: 75% gone is between one half-life (50%) and two (75% is exactly two half-lives — since $1 - \frac{1}{4} = \frac{3}{4}$ ). So expect roughly $2 \times t_{1/2}$ .

"75% completion" means 75% consumed, so 25% remains: $[A] = 0.25 [A]_{0}$ , i.e. $[A] / [A]_{0} = 0.25$ . Why this step? The equation only cares about the ratio remaining, never the absolute start value — so we never even need $[A]_{0}$ .
Given a fraction, want a time → use the log form: $ln \frac{[ A ]}{[ A ] _{0}} = - k^{'} t$ . Why this step? The logarithm undoes the exponential and pulls $t$ out of the exponent into a place we can solve.
$ln (0.25) = - 1.3863$ , so $- 1.3863 = - k^{'} t$ .
$t = \frac{1.3863}{2.5 \times 1 0 ^{- 4}} = 5545 s \approx 92.4 min$ . Why this step? Divide by $k^{'}$ to isolate $t$ ; converting to minutes ( $/60$ ) makes the scale human.

Verify: $t_{1/2} = ln 2/ k^{'} = 0.6931/2.5 \times 1 0^{- 4} = 2773 s$ . Two half-lives $= 5545 s$ — exactly our answer, confirming the forecast that 75% = two half-lives. ✓

Worked example Example 4 — Cell C4: half-life and its multiples

For a pseudo-first-order reaction with $k^{'} = 0.0231 s^{- 1}$ : (a) find $t_{1/2}$ ; (b) find the time for the concentration to drop to one-eighth of its start.

Forecast: One-eighth is $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$ = three half-lives. So (b) should be $3 \times$ (a).

Half-life: set $[A] / [A]_{0} = \frac{1}{2}$ in $ln ([A] / [A]_{0}) = - k^{'} t$ : $ln \frac{1}{2} = - k^{'} t_{1/2}$ . Why this step? Half-life is defined as the time for the ratio to hit $\frac{1}{2}$ .
$t_{1/2} = \frac{ln 2}{k ^{'}} = \frac{0.6931}{0.0231} = 30.0 s$ . Why this step? $ln 2$ appears because halving is the same fixed fraction every time — this is why first-order half-life is independent of $[A]_{0}$ .
For one-eighth: $[A] / [A]_{0} = \frac{1}{8}$ , so $ln \frac{1}{8} = - 3 ln 2$ , giving $t = 3 ln 2/ k^{'} = 3 t_{1/2}$ .
$t = 3 \times 30.0 = 90.0 s$ .

Verify: Plug $t = 90$ into $[A] / [A]_{0} = e^{- k^{'} t} = e^{- 0.0231 \times 90} = e^{- 2.079} = 0.125 = \frac{1}{8}$ . ✓ Matches the forecast (three half-lives).

Worked example Example 5 — Cell C5: reading the graph (geometric)

Two experiments track $ln [A]$ against $t$ . Experiment 2 (B in $100 \times$ excess) gives a straight line through $(0, - 4.605)$ and $(100, - 9.605)$ . Find $k^{'}$ , and the true $k$ if $[B]_{0} = 1.0 M$ .

Forecast: A straight downward line means pseudo-first-order confirmed. The steeper the fall, the bigger $k^{'}$ . Over 100 s the value dropped by 5 units → slope of magnitude $5/100 = 0.05$ .

Slope of $ln [A]$ vs $t$ is $\frac{Δ ln [ A ]}{Δ t} = \frac{- 9.605 - ( - 4.605 )}{100 - 0} = \frac{- 5.000}{100} = - 0.0500 s^{- 1}$ . Why this step? The integrated law $ln [A] = ln [A]_{0} - k^{'} t$ is $y = c - k^{'} t$ : its slope is literally $- k^{'}$ .
So $k^{'} = 0.0500 s^{- 1}$ (drop the minus — $k^{'}$ is positive; the minus lives in the "decay" direction).
Recover true $k$ : $k = k^{'} / [B]_{0} = 0.0500/1.0 = 0.0500 M^{- 1} s^{- 1}$ . Why this step? Same division as Ex 3 — strip the excess concentration back out.

Verify: Intercept $ln [A]_{0} = - 4.605 \Rightarrow [A]_{0} = e^{- 4.605} = 0.010 M$ — the stated start. In the figure, the red curved line (Experiment 1, comparable concentrations) is not straight, so it is NOT pseudo-first-order — exactly why we vary the excess. ✓

Worked example Example 6 — Cell C6 (degenerate): the approximation breaks

A student uses only a $10 \times$ excess: $[A]_{0} = 0.10 M$ , $[B]_{0} = 1.0 M$ . When half of A is consumed, by what percent has $[B]$ actually changed, and is calling it "constant" honest?

Forecast: Only $10 \times$ excess is the parent note's warning zone. Consuming half of A = $0.05 M$ , and B loses the same amount. $0.05$ out of $1.0$ is $5%$ — noticeable.

Amount of A consumed at 50%: $Δ = 0.50 \times 0.10 = 0.050 M$ . Why this step? Stoichiometry $A + B \to$ products means one B vanishes per A consumed.
New $[B] = 1.0 - 0.050 = 0.950 M$ .
Percent change in B: $\frac{0.050}{1.0} \times 100 = 5.0%$ . Why this step? This is the error you inject by pretending $[B]$ is fixed at $[B]_{0}$ .

Verify: Compare to a proper $100 \times$ excess ( $[A]_{0} = 0.10$ , $[B]_{0} = 10$ ): B would change by $0.05/10 = 0.5%$ . So going from $10 \times$ to $100 \times$ cuts the error tenfold — this is why the parent demands $[B]_{0} / [A]_{0} \geq 50$ – $100$ . At $5%$ , the "line" in Ex 5 would visibly bend. ✓

Worked example Example 7 — Cell C7 (limiting): the two endpoints of time

Using $[A] = [A]_{0} e^{- k^{'} t}$ with any positive $k^{'}$ , evaluate the physical endpoints: (a) $t \to 0$ ; (b) $t \to \infty$ . Confirm the model never misbehaves.

Forecast: At the very start nothing has reacted → $[A]$ should equal $[A]_{0}$ . After infinite time everything is consumed → $[A] \to 0$ . A good model must give exactly these.

As $t \to 0$ : $e^{- k^{'} \cdot 0} = e^{0} = 1$ , so $[A] \to [A]_{0} \times 1 = [A]_{0}$ . Why this step? $e^{0} = 1$ is the anchor of the exponential — it guarantees the curve starts at the right height.
As $t \to \infty$ : $- k^{'} t \to - \infty$ , and $e^{- \infty} = 0$ , so $[A] \to 0$ . Why this step? A positive $k^{'}$ makes the exponent march to $- \infty$ ; the concentration decays to (but never quite reaches) zero — matching "A runs out eventually."
Degenerate case $k^{'} = 0$ (no reaction, e.g. catalyst absent): $e^{0} = 1$ for all $t$ , so $[A] = [A]_{0}$ forever — a flat line. The model correctly says "nothing happens."

Verify: Numerically with $[A]_{0} = 1, k^{'} = 0.05$ : at $t = 0$ , $[A] = 1.000$ ; at $t = 1000 s$ , $[A] = 1 \cdot e^{- 50} \approx 1.9 \times 1 0^{- 22} \approx 0$ . Both endpoints hit. ✓

Worked example Example 8 — Cell C8: real-world word problem with an exam twist

A pollutant $P$ degrades in a lake by reaction with dissolved oxygen: $P + O_{2} \to$ products, truly second-order, $k = 3.0 M^{- 1} s^{- 1}$ . The lake holds $[O_{2}] = 2.0 \times 1 0^{- 4} M$ , effectively constant (continuously resupplied from air), while $[P]_{0} = 1.0 \times 1 0^{- 8} M$ . Twist: the exam gives you $k$ (not $k^{'}$ ) and asks for the time for the pollutant to fall to 10% of its start.

Forecast: Oxygen ( $2 \times 1 0^{- 4}$ ) dwarfs pollutant ( $1 0^{- 8}$ ) by a factor $2 \times 1 0^{4}$ — massively pseudo-first-order, approximation rock-solid. We must first build $k^{'}$ ourselves, then solve for time. "10% remaining" is a bit past three half-lives ( $12.5%$ is exactly three), so expect $t$ a little over $3 t_{1/2}$ .

Confirm excess: $[O_{2}] / [P]_{0} = 2 \times 1 0^{- 4} /1 0^{- 8} = 2 \times 1 0^{4} \geq 100$ . ✓ Why this step? You must check the excess before trusting pseudo-first-order — never assume.
Build the pseudo constant: $k^{'} = k [O_{2}] = 3.0 \times 2.0 \times 1 0^{- 4} = 6.0 \times 1 0^{- 4} s^{- 1}$ . Why this step? The twist gave true $k$ ; we fold in the constant excess to get the $k^{'}$ our time-equation needs (units: $M^{- 1} s^{- 1} \times M = s^{- 1}$ ✓).
"10% remaining": $[P] / [P]_{0} = 0.10$ , use log form: $ln (0.10) = - k^{'} t$ . Why this step? Given fraction, want time → logarithm isolates $t$ .
$ln (0.10) = - 2.3026$ , so $t = 2.3026/6.0 \times 1 0^{- 4} = 3838 s \approx 64 min$ .

Verify: $t_{1/2} = ln 2/ k^{'} = 0.6931/6 \times 1 0^{- 4} = 1155 s$ ; three half-lives $= 3466 s$ gives $12.5%$ . Our $3838 s$ is a bit longer → less than $12.5%$ remains, i.e. exactly $10%$ — consistent with the forecast. Plug back: $e^{- 6 \times 1 0^{- 4} \times 3838} = e^{- 2.303} = 0.100$ . ✓

Recall Quick self-test

Given percent completion, which form do you use and why? ::: The $ln$ form, because the logarithm undoes the exponential and isolates the unknown time $t$ . Why is $t_{3/4}$ (75% done) exactly $2 t_{1/2}$ ? ::: Because 25% remaining $= \frac{1}{2} \times \frac{1}{2}$ , i.e. two successive halvings, and first-order half-life is constant. You measured $k^{'} = 0.05 s^{- 1}$ with $[B]_{0} = 2.0 M$ . True $k$ ? ::: $k = k^{'} / [B]_{0} = 0.025 M^{- 1} s^{- 1}$ . At $10 \times$ excess, how much does B change when half of A is gone? ::: About 5% — enough to bend the "straight" line, so $10 \times$ is unsafe.

Related: Integrated Rate Laws · Second-Order Reactions · Method of Isolation · Half-life · Enzyme Kinetics · back to Pseudo-first-order kinetics.