Foundations — Pseudo-first-order kinetics

Before you can trust the parent note, you need every letter, bracket, and squiggle it throws at you to mean something concrete. Let us build each one from nothing, in an order where every idea leans on the one before it.

1. Concentration and the square-bracket notation $[A]$

The picture. Imagine a jar of water with red dots (molecules of A) floating in it. $[A]$ is how crowded the red dots are — dots per litre, not the total number of dots. Two jars can hold the same crowding while one is a test tube and the other a swimming pool.

Why the topic needs it. Every rate law is written in these brackets. When we say "B barely changes", we mean the crowding $[B]$ barely changes — that is the whole trick.

The subscript $0$ is a time-stamp: $[A{]}_0$ means "the crowding of A at the start, when the clock reads $t=0$". Later, plain $[A]$ means "right now".

2. Rate, and the derivative $-\frac{d[A]}{dt}$

Unpacking the symbols one at a time.

• $t$ = time, measured in seconds ($\text{s}$). The clock.
• $d[A]$ = a tiny change in the concentration of A (in $\text{M}$).
• $dt$ = a tiny slice of time (in $\text{s}$).
• $\frac{d[A]}{dt}$ = (tiny change in $[A]$) ÷ (tiny slice of time) = the slope of the curve when you plot $[A]$ against $t$, in $\text{M}{\text{s}}^{-1}$.

Why the tool "derivative" and not just "change ÷ time"? Because the reaction slows down as A runs out — the curve bends. A single big "change ÷ time" would only give the average speed over a long stretch. The derivative asks a sharper question: "How fast is it going right at this instant?" — the steepness of the tangent line touching the curve at one point.

Why the minus sign? As time moves forward, a reactant decreases, so $d[A]$ is negative and the raw slope $\frac{d[A]}{dt}$ is a negative number. A "rate" is a speed and should be a positive number, so we flip the sign with a minus to make it positive. The minus is not extra physics — it is bookkeeping so "rate" reads as a comfortable positive value.

The mirror convention for products. A product is being created, so its concentration rises and its slope is already positive. There we keep a plus sign: $\text{Rate}=+\frac{d[\text{Product}]}{dt}$ So the same reaction has one rate written two ways — a minus in front of a shrinking reactant, a plus in front of a growing product — chosen so the number always comes out positive.

3. The rate constant $k$ and the rate law

The picture. Think of $k$ as the "eagerness" of the collision: even if two crowds ($[A]$ and $[B]$) are present, $k$ decides how likely a meeting actually turns into a product. This eagerness comes from Collision Theory and depends on temperature through the Activation Energy.

Why multiply $[A]$ times $[B]$? For A and B to react they must meet. The chance of a red dot meeting a blue dot goes up if there are more red dots and if there are more blue dots — the two chances multiply, just like the chance of rolling two sixes is $\frac{1}{6}\times \frac{1}{6}$. That multiplication is the heart of a second-order rate law.

4. "Much greater than" $\gg$ and the excess idea

The picture — the bucket and the cupful. Picture a huge bucket of blue water (B) and one small cup of red dye (A). Pour the whole cup in; the bucket's blue level does not visibly drop. Even after all the red is used up, B has fallen by maybe $1\%$. That "did-not-move" is what lets us treat $[B]$ as a constant.

Why the topic needs it. This single inequality is the permission slip for the whole trick. Without $\gg$, both concentrations change and the reaction stays genuinely second-order.

5. The pseudo constant $k^{\prime }$ (k-prime)

The picture. Take the second-order rate law $\text{Rate}=k[A][B{]}_0$ and mentally box up the two constants $k$ and $[B{]}_0$ into one grey box labelled $k^{\prime }$. What is left outside the box is a single concentration $[A]$ — so the law now looks first-order: $\text{Rate}=k^{\prime }[A]$.

Why introduce a new symbol at all? Because $k^{\prime }$ is exactly what your experiment measures directly (the slope of a straight line, see §7), whereas the true $k$ is hidden inside it. Keeping them as separate symbols stops you confusing an ${\text{s}}^{-1}$ number with an ${\text{M}}^{-1}{\text{s}}^{-1}$ number.

6. The natural log $\ln$ and Euler's number $e$

Why does $e$ show up here at all? Look at the pseudo rate law $-\frac{d[A]}{dt}=k^{\prime }[A]$. In words: the speed of falling is proportional to how much is left. The bigger the pile, the faster it shrinks; as it shrinks, it slows. We do not have to guess the answer — we can derive it in one honest step.

The one-line integration. Start from the pseudo rate law and gather every $[A]$ on the left, every $t$ on the right (this move is called separating the variables — it lets each side be added up on its own): $\frac{d[A]}{[A]}=-k^{\prime }dt$ Now add up (integrate) both sides — the left from the start value $[A{]}_0$ to the current $[A]$, the right from time $0$ to time $t$: ${\int }_{[A{]}_0}^{[A]}\frac{d[A]}{[A]}=-k^{\prime }{\int }_0^{t}dt$ The left integral is exactly the one that produces a natural log (that is what $\ln$ is for — it is the running total of $1/x$), and the right is just $t$: $\ln [A]-\ln [A{]}_0=-k^{\prime }t$ Rearranging, then undoing the log with $e$, gives the two twin forms: $\ln [A]=\ln [A{]}_0-k^{\prime }t⟺[A]=[A{]}_0{e}^{-k^{\prime }t}$ So $e$ is not a random choice — it is forced on us by the integration; it is the unique function whose rate of change is proportional to itself.

Two log facts the worked examples use.

• $\ln (0.25)=-1.386$ (used to find the time for $75\%$ completion).
• $e^{-0.432}=0.649$ (used to find sucrose left after 2 hours).

7. Reading a $\ln [A]$-versus-$t$ graph

Read the two panels above directly. In the left panel the vertical axis is $[A]$ in $\text{M}$ and the horizontal axis is $t$ in $\text{s}$; the cyan line is the curved exponential $[A]=[A{]}_0{e}^{-k^{\prime }t}$ — hard to read a slope from. In the right panel the vertical axis is now $\ln [A]$ (a pure number, no units) and the horizontal axis is still $t$ in $\text{s}$; the same data becomes a perfectly straight amber line. The amber arrow marks its downward tilt: that tilt is $-k^{\prime }$. Reading $k^{\prime }$ off the right panel is easy exactly because a straight line has one single slope everywhere.

Compare with $y=mx+c$ from school: match $y=\ln [A]$, $x=t$, slope $m=-k^{\prime }$, intercept $c=\ln [A{]}_0$. The straight-line test is the experimental heart of the Method of Isolation and connects directly to the Integrated Rate Laws.

8. Molecularity vs order (a word-trap)

Pseudo-first-order changes the order we observe (from 2 to 1) but never touches the molecularity — two molecules still collide. That is why "pseudo" (Latin-ish for "false/apparent") is the perfect word.

Prerequisite map

Each box is a symbol you now own; the arrows show which feeds which, ending at the parent topic Pseudo-first-order kinetics. From here you can branch onward to Half-life, Enzyme Kinetics, Second-Order Reactions, Buffer Solutions and Activation Energy.

Equipment checklist