Chemical Kinetics · Lesson 10 of 10
Chapter Summary and Practice
“Connect measured rates, kinetic laws, mechanisms and temperature dependence through a complete chapter synthesis and mixed answered problems.”
• Choose a consistent rate definition and apply stoichiometric conversions. • Infer and test kinetic order using experiments, units and integrated laws. • Connect mechanisms, excess-reactant conditions and half-life behaviour. • Combine Arrhenius calculations with energy and catalyst reasoning while stating assumptions.
Start with the observable and the conditions
Chemical kinetics connects what changes in an experiment to how rapidly molecular events take place. The observed quantity might be a concentration, gas pressure, optical signal or amount of a parent nuclide. Before using a formula, identify which species the measurement follows and whether volume, temperature, catalyst, solvent and other conditions remain fixed. An apparent change in concentration can include dilution, and a total gas pressure can include products rather than only the reactant.
The next step is to define rate. A species disappearance rate and a stoichiometrically normalised rate differ by the species coefficient. Consistently changing that convention rescales a rate constant but does not change the concentration exponents. Balanced equations control material relationships; experimental laws describe how composition influences speed. Keeping these two roles distinct prevents many numerical and mechanistic errors.
Let experiments determine the concentration dependence
| Question | Evidence or equation | Main caution |
|---|---|---|
| What is the average rate? | Finite concentration change / elapsed time | Use the requested interval and sign convention. |
| What is the instantaneous rate? | Tangent slope or derivative | A secant average is not automatically an instantaneous value. |
| What are the partial orders? | Controlled initial-rate comparisons | Hold other conditions fixed; do not read them from overall coefficients. |
| What is overall order? | Sum of empirical exponents | Fractional, zero and negative partial orders can occur. |
| What are k units? | Concentration^(1−n) / time | Specify the rate unit and whether concentrations or pressures are used. |
| Is a mechanism established? | Consistency with stoichiometry and kinetic evidence | A matching law supports but does not uniquely prove a mechanism. |
Molecularity applies to one elementary event, while observed order applies to a macroscopic concentration dependence. A catalyst is regenerated; an intermediate is formed and consumed; a transition state lies at an activated region between states. A rate-controlling step can contain an intermediate whose concentration depends on earlier steps. Therefore the molecularity of that step need not equal the overall empirical order.
Integrated laws connect concentration and time
| Feature | Zero-order disappearance | First-order disappearance |
|---|---|---|
| Differential law | −d[R]/dt=k₀ | −d[R]/dt=k₁[R] |
| Integrated law | [R]t=[R]₀−k₀t | ln([R]₀/[R]t)=k₁t |
| Remaining concentration | Linear fall in a valid regime | [R]₀exp(−k₁t) |
| Straight-line plot | [R] against t; slope −k₀ | ln([R]/Cref) against t; slope −k₁ |
| Half-life | [R]₀/(2k₀) | ln2/k₁ |
| Constant units for M and s | M s⁻¹ | s⁻¹ |
| Important limit | Do not extend to negative concentration | No exact zero at finite time in the exponential model |
Zero order often reflects a saturated process with limited throughput; first order removes equal fractions in equal times. Under a pseudo-first-order condition, a nearly constant excess reactant or fixed solvent activity can be included in an observed coefficient. A first-order time trace then does not prove that the full mechanism has only one reacting entity or that varying the excess component would have no effect.
Problem
For 2A→B, define A disappearance by −d[A]/dt=k_d[A], with k_d=0.0100 s⁻¹ and [A]₀=0.200 M. Find remaining A, its instantaneous disappearance and B appearance after one A half-life.
- 1.The A half-life is ln2/0.0100=69.3147 s. At that time [A]=0.100 M by definition.
- 2.A disappearance then equals k_d[A]=0.0100×0.100=0.00100 M s⁻¹.
- 3.Because two A make one B, B appears at half that value, 0.000500 M s⁻¹.
- 4.The integrated constant was defined for species disappearance. Calling both species rates simply “the rate” without their definitions would conceal the factor of two.
Gas-pressure measurements need a stoichiometric model
For initially pure A undergoing A→B+C at constant T and V, the remaining partial pressure is 2p₀−p_t. For a stipulated 2A→2B+C gas conversion it is 3p₀−2p_t. Both follow by tracking how many moles of each gas appear per amount consumed. The integrated first-order law then uses the remaining A pressure, not total pressure. If product association, additional reactions, changing temperature or an initial inert-gas contribution matters, the bookkeeping must be modified.
Temperature changes kinetic access to a pathway
| Representation | Slope or interpretation | Conditions |
|---|---|---|
| ln k against 1/T | Slope=−E_a/R | Fixed numerical unit for k. |
| log₁₀ k against 1/T | Slope=−E_a/[(ln10)R] | Decimal logarithm requires its conversion factor. |
| ln k against 1000/T | Slope=−E_a/(1000R) | Restore the horizontal-axis scale. |
| Energy profile | Barrier top versus endpoint difference | State the vertical energy type; reaction coordinate is not time. |
| Energy distribution | Area above a threshold grows on warming | Total probability area remains one. |
An exothermic endpoint does not guarantee a rapid process. A catalyst can provide another route with easier kinetic access while leaving the net equilibrium condition unchanged. Effective collisions need suitable energy and orientation; collision count alone is not a rate constant. A molecular drawing must preserve atoms, charge and valence, and an activated substitution arrangement must use partial rather than five full bonds to carbon.
Problem
Two runs at 300 and 310 K show a doubled observed rate. Can activation energy always be found by treating the rate ratio as k₂/k₁?
- 1.First check that reactant concentrations, catalyst, solvent and rate definition are comparable. If these changed, their effects can contribute to the rate ratio.
- 2.If the same law and composition apply and the Arrhenius parameters are approximately constant, the rate ratio can represent k₂/k₁.
- 3.Only then use E_a=Rln2/(1/300−1/310). The result is conditional on those controls, not implied by the rate numbers alone.
- 4.An experiment with several temperatures allows a fuller check for curvature or scatter than a single pair.
Check the answer as well as the arithmetic
A reliable kinetic calculation ends by substituting the result into the original relationship. Check that concentration is non-negative, remaining fraction lies between zero and one, the sign of a predicted temperature effect matches the model and every logarithm has a valid positive argument. Keep guard digits for small differences and fitted slopes, but report final precision honestly. A correctly executed formula cannot repair an unbalanced equation, ambiguous rate definition or missing temperature.
Quick Quiz
Quiz
Which statement correctly separates roles?
Explanation: A balanced equation provides amount relationships, whereas a measured law supplies empirical exponents.
Which law has a concentration-independent first half-life?
Explanation: The first-order half-life is ln2/k; the zero-order half-life contains the initial concentration.
Why does a pressure-derived M s⁻¹ rate need temperature?
Explanation: Under ideal gas behaviour [A]=p_A/(RT), so pressure and concentration rates differ by RT at constant temperature.
What does a catalyst preserve for the same net reaction at fixed T?
Explanation: The pathway and kinetics can change while the net thermodynamic endpoint difference remains the same.
A measured Arrhenius dataset has scatter. Which report is most defensible?
Explanation: A fitted model should use a stated method and preserve both unit conventions and uncertainty.
Practice Problems
- For 2A→P, A drops from 0.60 to 0.40 M in 50 s. Find A disappearance and normalised reaction rates.
- A law r=k[A]¹ᐟ²[B]³ᐟ² is measured in M s⁻¹. Find overall order, k units and the effect of doubling A while quadrupling B.
- A zero-order disappearance process starts at 0.500 M with k₀=0.00500 M s⁻¹. Find the time to reach 0.200 M and its first half-life.
- A first-order process starts at 0.400 M and has half-life 20.0 min. Find k and concentration after 50.0 min.
- First-order A→B+C begins with pure A at 0.800 bar. After 200 s, total pressure is 1.20 bar. Find k and the pressure-based A disappearance rate then.
- Explain why a third-order overall law can arise from a bimolecular slow step and identify the required assumption in a simple example.
- An excess-B process has v=k[A][B], k=0.0200 M⁻¹s⁻¹ and [B]≈5.00 M. Find k_obs and its observed half-life in A.
- At 300 K a rate constant is 0.0100 s⁻¹. Predict its value at 330 K for E_a=50.0 kJ mol⁻¹ and constant A.
- An ln k versus 1/T fit has slope −8000 K. Find E_a and explain how the calculation differs for a decimal-log plot with the same numerical slope.
- A profile has reactants at 0, products at −20 and a transition-state peak at 50 kJ mol⁻¹. A catalytic route lowers its effective highest barrier but keeps endpoints. What changes and what does not?
- A parent nuclide has half-life 10.0 years. What fraction remains after 25.0 years in a closed decay-only model, and why is this not a general biological retention prediction?
- Audit the statements “ten half-lives is exactly 99.9% conversion,” “all encounters react,” and “a third-order overall equation is necessarily termolecular.”
The species disappearance rate is (0.60−0.40)/50=0.00400 M s⁻¹. Divide by the stoichiometric coefficient two to obtain r=0.00200 M s⁻¹. Product P appears at r because its coefficient is one. This is an average across the 50 s interval.
Key Takeaways
• Identify the measured species, rate convention and controlled conditions before calculating. • Use stoichiometry for amount ratios and experimental laws for concentration dependence. • Order, molecularity, intermediates and transition states describe different aspects of a pathway. • Zero- and first-order integrated laws have different graphs and half-life behaviour. • Pseudo-first-order coefficients can hide nearly constant medium or excess-reactant factors. • Derive reactant partial pressure before applying an integrated gas-phase law. • Arrhenius calculations require kelvin, matching energy units and consistent logarithms. • Catalysts change kinetic pathways while preserving the same net equilibrium condition. • Check physical bounds, missing information, experimental scatter and numerical precision in every answer.
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Catalysts and Effective Collisions
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