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

This is your self-test for the parent derivations. Work each problem on paper FIRST, then open the collapsible solution. The exercises climb a ladder: L1 Recognition (spot the order) → L2 Application (plug into a law) → L3 Analysis (extract $k$ from data) → L4 Synthesis (combine ideas) → L5 Mastery (design/prove).

Before we start, one figure fixes every symbol you'll use — read it once so nothing below is a surprise.

The three laws we lean on, all derived in the parent:

Level 1 — Recognition

Exercise 1.1 (L1)

A reaction has rate constant with units $s^{- 1}$ . What is its order? What plot gives a straight line?

Recall Solution

WHAT we ask: the units of $k$ act as a fingerprint. Rate always has units $M s^{- 1}$ , and $k [A]^{n} = M s^{- 1}$ , so $k = \frac{M s ^{- 1}}{M ^{n}} = M^{1 - n} s^{- 1} .$ WHY: matching $M^{1 - n}$ to the given $s^{- 1}$ (i.e. $M^{0}$ ) forces $1 - n = 0$ , so $n = 1$ . Answer: first-order. The straight-line plot is $ln [A]$ vs $t$ (slope $- k$ ).

Exercise 1.2 (L1)

Below are three sketches of concentration data. Which order does each match? See the figure below.

Recall Solution

The trick: each order goes straight only after the right transform of the y-axis.

Left panel — $[A]$ vs $t$ is a straight falling line → zero-order.
Middle panel — $ln [A]$ vs $t$ is straight → first-order.
Right panel — $\frac{1}{[ A ]}$ vs $t$ is straight → second-order. WHY: the integrated law's left side IS your y-axis. Zero uses $[A]$ , first uses $ln [A]$ , second uses $1/ [A]$ .

Level 2 — Application

Exercise 2.1 (L2)

A zero-order reaction has $[A]_{0} = 0.500 M$ and $k = 0.010 M s^{- 1}$ . Find $[A]$ after $20 s$ .

Recall Solution

Use $[A]_{t} = [A]_{0} - k t$ . $[A]_{t} = 0.500 - (0.010) (20) = 0.500 - 0.200 = 0.300 M .$ WHY the subtraction: in zero-order the reactant is consumed at a fixed pace $k$ , so each second removes the same chunk — a straight line down.

Exercise 2.2 (L2)

A first-order reaction has $k = 0.0693 s^{- 1}$ . Starting from $[A]_{0} = 1.00 M$ , find $[A]$ after $30 s$ .

Recall Solution

Use $[A]_{t} = [A]_{0} e^{- k t}$ . $k t = (0.0693) (30) = 2.079.$ $[A]_{t} = 1.00 e^{- 2.079} = 0.125 M .$ WHY the exponential (not subtraction): here the pace is proportional to how much is left, so it slows as it goes — mathematically that "self-shrinking" decay is exactly $e^{- k t}$ . Note $2.079 = 3 \times 0.693$ , so this is exactly 3 half-lives: $1 \to \frac{1}{2} \to \frac{1}{4} \to \frac{1}{8} = 0.125 M$ . See Half-Life Calculations.

Exercise 2.3 (L2)

A second-order reaction has $k = 0.50 M^{- 1} s^{- 1}$ and $[A]_{0} = 0.20 M$ . Find $[A]$ after $10 s$ .

Recall Solution

Use $\frac{1}{[ A ] _{t}} = \frac{1}{[ A ] _{0}} + k t$ . $\frac{1}{[ A ] _{t}} = \frac{1}{0.20} + (0.50) (10) = 5.0 + 5.0 = 10.0 M^{- 1} .$ $[A]_{t} = \frac{1}{10.0} = 0.10 M .$ WHY reciprocals: second-order linearises in $1/ [A]$ , so you add into the reciprocal, then flip back at the end.

Level 3 — Analysis

Exercise 3.1 (L3)

For a first-order reaction, $[A]$ drops from $0.80 M$ to $0.20 M$ in $60 s$ . Find $k$ and $t_{1/2}$ .

Recall Solution

Step 1 — find $k$ . Use $ln \frac{[ A ] _{t}}{[ A ] _{0}} = - k t$ . $ln \frac{0.20}{0.80} = ln (0.25) = - 1.386 = - k (60) .$ $k = \frac{1.386}{60} = 0.0231 s^{- 1} .$ Step 2 — half-life. $t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0231} = 30 s .$ Sanity check (the elegant way): $0.80 \to 0.40 \to 0.20$ is exactly 2 half-lives in $60 s$ , so $t_{1/2} = 30 s$ — matches. See Integrated Rate Laws Applications.

Exercise 3.2 (L3)

A zero-order reaction starts at $[A]_{0} = 0.60 M$ and reaches $0.00 M$ in $120 s$ . Find $k$ and $t_{1/2}$ .

Recall Solution

WHAT is special about zero-order: it actually hits zero at a finite time $t_{end} = [A]_{0} / k$ (unlike first/second order, which only approach zero). Set $[A]_{t} = 0$ : $0 = 0.60 - k (120) \Rightarrow k = \frac{0.60}{120} = 0.0050 M s^{- 1} .$ Half-life: $t_{1/2} = \frac{[ A ] _{0}}{2 k} = \frac{0.60}{2 ( 0.0050 )} = 60 s .$ WHY it's half of $t_{end}$ : the line falls at constant speed, so it reaches the halfway concentration in exactly half the total time. Clean geometry.

Level 4 — Synthesis

Exercise 4.1 (L4)

A first-order reaction is $75%$ complete in $50 min$ . How long until it is $99%$ complete?

Recall Solution

WHAT "75% complete" means: $25%$ remains, so $[A]_{t} / [A]_{0} = 0.25$ . Step 1 — find $k$ : $ln (0.25) = - k (50) \Rightarrow k = \frac{1.386}{50} = 0.02773 min^{- 1} .$ Step 2 — time for 99% complete ( $1%$ remains, ratio $= 0.01$ ): $ln (0.01) = - k t \Rightarrow t = \frac{4.605}{0.02773} = 166 min .$ Elegant check: $75%$ complete = 2 half-lives (since $0.25 = (1/2)^{2}$ ), so $t_{1/2} = 25 min$ . $99%$ complete means about $0.01 = (1/2)^{6.64}$ , i.e. $6.64$ half-lives $= 6.64 \times 25 \approx 166 min$ . Both routes agree.

Exercise 4.2 (L4)

Two reactions start at the same $[A]_{0} = 0.10 M$ . Reaction P is second-order with $k_{P} = 2.0 M^{- 1} s^{- 1}$ . Reaction Q is first-order with $k_{Q} = 0.20 s^{- 1}$ . Which has the shorter first half-life, and by how much?

Recall Solution

Reaction P (second-order): $t_{1/2} = \frac{1}{k _{P} [ A ] _{0}} = \frac{1}{( 2.0 ) ( 0.10 )} = 5.0 s .$ Reaction Q (first-order): $t_{1/2} = \frac{0.693}{k _{Q}} = \frac{0.693}{0.20} = 3.465 s .$ Answer: Q is shorter by $5.0 - 3.465 = 1.535 s \approx 1.54 s$ . WHY the comparison matters: second-order half-life depends on $[A]_{0}$ while first-order does not — so the winner can flip if you change the starting concentration. That concentration-sensitivity comes from the collision picture in Collision Theory.

Level 5 — Mastery

Exercise 5.1 (L5)

Prove that for a first-order reaction the half-life is independent of starting concentration, but for a second-order reaction the half-life doubles each time you halve $[A]_{0}$ .

Recall Solution

First-order. Set $[A]_{t} = \frac{1}{2} [A]_{0}$ in $ln ([A]_{t} / [A]_{0}) = - k t$ : $ln (\frac{1}{2}) = - k t_{1/2} \Rightarrow t_{1/2} = \frac{l n 2}{k} .$ WHY it's concentration-free: the $[A]_{0}$ cancelled inside the ratio, leaving only constants. Halving anything takes the same time. Second-order. Set $[A]_{t} = \frac{1}{2} [A]_{0}$ in $\frac{1}{[ A ] _{t}} - \frac{1}{[ A ] _{0}} = k t$ : $\frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2} \Rightarrow \frac{1}{[ A ] _{0}} = k t_{1/2} \Rightarrow t_{1/2} = \frac{1}{k [ A ] _{0}} .$ WHY doubling: $t_{1/2} \propto 1/ [A]_{0}$ . Halve $[A]_{0}$ and $t_{1/2}$ doubles. Physically, fewer molecules means fewer collisions per second (rate $\propto [A]^{2}$ ), so each successive half takes longer.

Exercise 5.2 (L5)

A student measures a zero-order reaction: $[A]_{0} = 0.40 M$ , $k = 0.020 M s^{- 1}$ . (a) Give the exact time the reaction finishes. (b) What is $[A]$ at $t = 30 s$ ? (c) At $t = 25 s$ ? Explain why one answer must be reported carefully.

Recall Solution

(a) Finish time. Reaction ends when $[A] = 0$ : $t_{end} = \frac{[ A ] _{0}}{k} = \frac{0.40}{0.020} = 20 s .$ (b) At $t = 30 s$ . The naive formula gives $0.40 - (0.020) (30) = 0.40 - 0.60 = - 0.20 M$ — impossible. Concentration cannot be negative. Once $t > t_{end}$ , all A is gone, so $[A] = 0 M$ . (c) At $t = 25 s$ . Same reasoning: $25 > 20$ , so $[A] = 0 M$ . WHY report carefully: the zero-order line $[A]_{0} - k t$ is only valid while reactant remains ( $t \leq t_{end}$ ). Beyond that the physics floors it at zero — a degenerate case the blind formula ignores.

Recall Self-Check Scoreboard

How many did you nail on the first try, without peeking? Level 1 correct means you can identify order. Level 5 correct means you can prove and defend the laws. Weakest level ::: revisit that section of the parent note, then re-attempt only that level's problems.