GENERAL CHEMISTRY · CHEMICAL KINETICS

Measure the change.
Defend the model.

Learn all three official topics through rate, energy, and time ledgers, worked examples, closed-note retrieval, and twelve original five-choice questions.

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3guided lessons

12practice questions

9study objectives

5choices per item

A repeatable kinetics routine

Observable. Order. Model. Verify.

01Name the observable
Identify the species, concentration change, elapsed time, slope sign, coefficient, temperature, or energy profile being measured.
02Infer the order
Use controlled rate changes, a linearized plot, or successive half-life intervals—never an unsupported net-equation shortcut.
03Select the model
Choose a rate law, integrated law, Arrhenius relationship, profile difference, or order-specific half-life equation.
04Verify the boundary
Check coefficients, units, kelvins, positive barriers, nonnegative concentrations, and rate-versus-equilibrium claims.

Three prerequisite-aware lessons

Let the evidence choose the law.

Each lesson pairs two semantic ledgers with a fully worked example and closed-note retrieval before multiple-choice practice.

LESSON 1 · 23 MIN

Study + retrieve

Rate laws from measured change

Normalize concentration–time change to one reaction rate, infer empirical orders and rate-constant units, and select the integrated law supported by a linearized plot.

ESSENTIAL QUESTIONWhat was measured, what concentration dependence does the experiment support, and which time model preserves the units and graph?

Rate-law evidence ledger

Normalize a species slope to one reaction rate
Measured species	Signed slope	Reaction-rate operation
Reactant A	Δ[A]/Δt < 0	−(1/a)Δ[A]/Δt
Product B	Δ[B]/Δt > 0	+(1/b)Δ[B]/Δt
Equation rate	Nonnegative forward value	Divide by stoichiometric coefficient
Instantaneous rate	Tangent slope	Normalize at the stated time

Order diagnostics for one reactant
Order	Linear plot and slope	k units · half-life
Zero	[A] vs t · −k	M·s⁻¹ · [A]₀/(2k)
First	ln[A] vs t · −k	s⁻¹ · ln2/k
Second	1/[A] vs t · +k	M⁻¹·s⁻¹ · 1/(k[A]₀)
Overall order	Sum empirical exponents	Determines the units of k

RATE / ENERGY / TIME LEDGER · CAPTIONS INCLUDED

Define the measured rate

A reactant has a negative concentration slope and a product a positive slope. Divide each signed species slope by its signed stoichiometric number—or apply the equivalent reactant minus sign—to report one reaction rate for the equation as written.

Species slope first
Normalize by the coefficient

Infer order instead of copying coefficients

In rate = k[A]^m[B]^n, compare trials where one concentration changes and the others stay fixed. The exponents are empirical for an overall reaction; their sum sets overall order and therefore the units required for k.

Hold other variables fixed
Units close the rate law

Choose the time model from linear evidence

Zero order makes [A] versus t linear with slope −k, first order makes ln[A] versus t linear with slope −k, and second order makes 1/[A] versus t linear with slope +k. Apply only the model supported over the stated conditions.

Transform before assigning order
Slope sign identifies k

Worked example

For a reaction whose ln[A] versus time plot is linear with slope −0.035 min⁻¹, [A]₀ = 0.800 M. What is [A] after 20.0 min?

1
A linear ln[A] plot identifies first-order behavior, so ln[A]t = ln[A]₀ − kt.
2
The slope is −k, giving k = 0.035 min⁻¹; use the same minute unit as the elapsed time.
3
Calculate [A]t = (0.800 M)e^(−0.035×20.0) = 0.397 M.

ConclusionThe graph chooses the law before arithmetic begins. Using raw [A] as a straight line would apply the zero-order model instead.

Close the notes first

Retrieve the model.

01For 2A → B, how does −Δ[A]/Δt compare with the reaction rate?

It is twice the reaction rate.

The A disappearance slope must be divided by its coefficient of 2.

02Can the exponents of an overall rate law be copied from the balanced equation?

No—not without experimental evidence or an explicitly elementary step.

Overall stoichiometry does not uniquely reveal the mechanism or empirical concentration dependence.

03Which transformed concentration is linear for a second-order single-reactant model?

1/[A] versus time.

Its slope is +k and its intercept is 1/[A]₀.

LESSON 2 · 22 MIN

Study + retrieve

Activation barriers and catalytic paths

Use effective-collision conditions, Arrhenius temperature dependence, and reaction-coordinate profiles to explain speed without changing thermodynamic endpoints or equilibrium position.

ESSENTIAL QUESTIONDoes the evidence change collision frequency, the energetic fraction, the pathway barrier, or only a thermodynamic endpoint?

Activation-energy evidence ledger

Collision and Arrhenius evidence
Rate influence	What changes	Boundary
Collision frequency	Encounter count	Not every collision reacts
Energetic fraction	Particles at or above E_a	Temperature must be in kelvins
Orientation	Bond-forming geometry	Energy alone is insufficient
ln k vs 1/T	Slope = −E_a/R	E_a and R need compatible units

Reaction profile and catalyst boundaries
Quantity	Energy difference	Interpretation
Forward E_a	Transition state − reactants	Forward barrier
Reverse E_a	Transition state − products	Reverse barrier
ΔH	Products − reactants	Endpoint change, not barrier
Catalyst	Lower alternative barrier	Same endpoints and K at fixed T

RATE / ENERGY / TIME LEDGER · CAPTIONS INCLUDED

Require energy and orientation

A collision is productive only when the particles meet with sufficient energy to cross the activation barrier and a geometry that can rearrange the necessary bonds. Concentration, pressure, surface area, and temperature influence rate through different parts of that collision picture.

Both conditions are required
Collision count is not reaction count

Read the Arrhenius relationship

In k = Ae^(−Ea/RT), temperature is absolute and Ea must share energy units with R. A plot of ln k versus 1/T has slope −Ea/R, so a steeper negative slope corresponds to a larger positive activation energy.

Kelvins and compatible energy units
Slope = −Ea/R

Separate barrier from endpoint

Forward Ea runs from reactants to the transition state, reverse Ea from products to that state, and ΔH compares product and reactant endpoints. A catalyst provides a lower-barrier path but does not change ΔG°, ΔH, K, or the equilibrium composition at fixed temperature.

Peak gap is Ea
Catalyst changes path, not endpoints

Worked example

An Arrhenius plot of ln k versus 1/T has slope −9.00 × 10³ K. Use R = 8.314 J/(mol·K) to find Ea.

1
Use slope = −Ea/R, so Ea = −(slope)R.
2
Multiply (9.00 × 10³ K)(8.314 J/(mol·K)) = 7.48 × 10⁴ J/mol.
3
Convert to 74.8 kJ/mol and keep the positive barrier magnitude.

ConclusionThe plotted slope is negative, but activation energy is positive. Neither that barrier nor the rate constant alone states whether the reaction is thermodynamically favored.

Close the notes first

Retrieve the model.

01What two conditions make a collision productive?

Adequate energy and proper orientation.

Either condition can fail even when particles collide.

02What is the slope of ln k versus 1/T?

−Ea/R.

The linear form is ln k = ln A − (Ea/R)(1/T).

03Does a catalyst change K at a fixed temperature?

No.

It changes accessible barriers and approach time, not the thermodynamic endpoint ratio.

LESSON 3 · 19 MIN

Study + retrieve

Half-life as an order diagnostic

Read half-life from concentration–time evidence, use the zero-, first-, or second-order equation, and restrict repeated equal-time halving to a supported first-order model.

ESSENTIAL QUESTIONWhich amount is being halved, does the interval stay constant, and what reaction order makes that pattern valid?

Half-life decision ledger

Half-life formula and concentration dependence
Order	Half-life	If [A]₀ doubles
Zero	[A]₀/(2k)	t₁/₂ doubles
First	ln2/k	t₁/₂ is unchanged
Second	1/(k[A]₀)	t₁/₂ halves
Definition	Time to halve the current amount	Does not guarantee equal intervals

Successive intervals and repeated fractions
Evidence	Supported model	Safe use
Intervals shorten	Zero order	Use the integrated law
Intervals stay constant	First order	Remaining fraction = (1/2)ⁿ
Intervals lengthen	Second order	Use changing starting concentration
Nuclear half-life	First-order decay model	Nuclear applications belong to Nuclear Reactions

RATE / ENERGY / TIME LEDGER · CAPTIONS INCLUDED

Halve the amount present now

A half-life is the time for the current amount or concentration to fall to one-half of its value at the start of that interval. The definition does not promise that successive intervals have equal duration.

Start of the interval matters
Half of what remains

Match half-life to reaction order

For one reactant, zero order gives t₁/₂ = [A]₀/(2k), first order gives t₁/₂ = ln 2/k, and second order gives t₁/₂ = 1/(k[A]₀). Only first order removes the initial concentration from the expression.

Zero grows with [A]₀
Second shrinks with [A]₀

Use interval trends as evidence

As concentration falls, successive zero-order half-lives shorten, first-order half-lives stay constant, and second-order half-lives lengthen. Use (1/2)^n after n equal half-lives only after first-order behavior is established.

Shorter · constant · longer
Repeated halves need first order

Worked example

A first-order reactant starts at 0.640 M and has a half-life of 8.0 min. What concentration remains after 24.0 min?

1
The elapsed time is 24.0/8.0 = 3.00 half-lives.
2
For established first-order behavior, the remaining fraction is (1/2)³ = 1/8.
3
Multiply 0.640 M by 1/8 to obtain 0.0800 M.

ConclusionEqual successive intervals are justified here because the model is first order. Applying the same shortcut to a zero- or second-order reaction would impose the wrong time pattern.

Close the notes first

Retrieve the model.

01Which common order has a half-life independent of initial concentration?

First order.

Its expression is ln 2/k.

02What happens to successive half-life intervals for a second-order reaction?

They become longer.

t₁/₂ is inversely proportional to the concentration present at the start of the interval.

03After four first-order half-lives, what fraction remains?

1/16.

The fraction is (1/2)⁴.

All twelve Chemical Kinetics problems

Choose the evidence before the equation.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

12 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions are draft and have not been calibrated to the official score scale. Use each explanation to repair the exact rate normalization, empirical order, plot, barrier, catalyst, or half-life decision.

DAT TRAIN does not claim topic quotas because the official manual does not publish them. Thermodynamics owns ΔG and spontaneity; Chemical Equilibria owns Q, K, and equilibrium composition; Nuclear Reactions owns nuclear-decay applications.