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Loading home…Chemical kinetics studies how fast a reaction proceeds and the factors that control its rate.
Chemical equilibrium studies how far a reversible reaction goes and the final balance of reactants and products.
Kinetics deals with rate laws, order, activation energy, and mechanisms.
Equilibrium deals with Kc, Kp, and Le Chatelier’s principle.
A reaction may be thermodynamically favourable yet kinetically slow if Ea is high.
Kinetics answers “how quickly?” while equilibrium answers “how completely?” under given conditions.
Increasing concentration of reactants usually increases collision frequency, so the rate rises for reactions with positive order.
Raising temperature increases the fraction of molecules with energy ≥ Ea and also raises collision frequency.
For solids, greater surface area exposes more particles, increasing successful collisions with the other reactant.
Powdered reactants therefore react faster than large lumps of the same mass.
These factors change the rate, not the thermodynamic ΔG of the reaction.
Industrial processes carefully control all three to achieve a useful rate safely.
According to collision theory, reactant particles must collide to react.
Collision frequency is how often particles collide per unit time and volume.
Only collisions with enough energy to reach the activation energy can form products.
Proper orientation is also needed so bonds rearrange correctly.
Higher concentration or temperature increases collision frequency.
Higher temperature mainly increases the energetic fraction of collisions, which often has a larger effect on rate.
A catalyst provides an alternative pathway with a lower activation energy.
More collisions then become successful at the same temperature, so the rate increases.
The catalyst is regenerated and does not appear as a net reactant or product.
It speeds both forward and reverse reactions equally in a reversible system.
Therefore it shortens the time to reach equilibrium but does not change Kc or Kp.
Enzymes are biological catalysts that are highly specific for their substrates.
If the reaction occurs in a single elementary step, the rate law follows the stoichiometry of that step.
Rate = k[R–X][H₂O].
The reaction is first order in R–X and first order in H₂O.
Overall order is 2 (second-order reaction).
If water is the solvent and present in large excess, [H₂O] is nearly constant.
Then the observed rate may appear first order in R–X only (pseudo-first-order).
The rate law is Rate = k[A]².
Second order in A means doubling [A] multiplies the rate by four.
Overall second order means the sum of exponents in the rate law is 2.
No other reactant appears in the rate law, so the reaction is zero order in other species present.
The units of k for a second-order reaction are typically mol⁻¹ dm³ s⁻¹.
This rate law must be determined experimentally unless the step is known to be elementary.
(a) A catalyst is regenerated after each catalytic cycle, so a small amount can process many reactant molecules.
It lowers Ea, greatly increasing the fraction of successful collisions.
(b) As a reaction proceeds, reactant concentrations fall.
For reactions with positive order, lower concentration means a lower instantaneous rate.
Products may also inhibit the reaction in some systems.
Hence measured rate is usually highest at the start and decreases with time.
Reaction rate is the change in concentration of a reactant or product per unit time.
For a → products, average rate ≈ −Δ[A]/Δt (or +Δ[product]/Δt).
Instantaneous rate is the slope of the concentration–time curve at a chosen time.
The rate law expresses rate in terms of concentrations: Rate = k[A]m[B]n …
Here k is the rate constant at a fixed temperature.
k depends on temperature and catalyst but not on reactant concentrations.
The exponents m, n, … are orders with respect to each reactant.
Overall order is the sum of these exponents.
Order is experimental; it need not equal stoichiometric coefficients.
The method of initial rates measures rate at the very start for several runs.
Concentrations are known accurately before products build up.
One reactant concentration is changed while others are held constant.
The ratio of initial rates gives the order in the varied reactant.
Example: if doubling [A] doubles initial rate, order in A is 1.
If doubling [A] quadruples rate, order in A is 2.
Repeating for each reactant builds the full rate law.
k is then found by substituting one set of data into the rate law.
Zero-order, first-order, and second-order reactions have characteristic units of k.
Graphical methods (plots of [A], ln[A], 1/[A] vs t) can confirm order.
Half-life behaviour also distinguishes common orders.
Together these tools link measured kinetics to a usable rate equation.
Most reactions speed up when temperature rises.
Collision theory says particles need energy at least equal to Ea.
The Maxwell–Boltzmann distribution shows molecular kinetic energies at a given T.
Only the high-energy tail has E ≥ Ea.
A small increase in T shifts the curve and greatly enlarges that tail.
Collision frequency also rises slightly with T, but the energetic factor dominates.
The Arrhenius equation is k = A e(−Ea/RT).
A is the frequency (pre-exponential) factor.
Ea is the activation energy and R is the gas constant.
As T increases, e^(−Ea/RT) increases, so k increases.
Taking logs: ln k = ln A − Ea/(RT).
A plot of ln k versus 1T is a straight line of slope −Ea/R.
This plot is used to find Ea from experimental rate constants.
Reactions with high Ea are more temperature-sensitive.
A rule of thumb is that rate often roughly doubles for a 10 °C rise near room temperature, but this is not universal.
Catalysts lower Ea, which also increases k at the same T.
Temperature does not change the meaning of order, but it changes k strongly.
Industrial reactors balance faster rates at high T against cost, safety, and equilibrium yield.
Understanding both the distribution curve and Arrhenius maths is essential for kinetic control.
Industrial kinetic control often means choosing T for a target rate safely.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
A mechanism is the sequence of elementary steps that convert reactants into products.
An elementary step occurs in a single molecular event.
The molecularity of a step equals the number of species colliding in that step.
The slowest step is the rate-determining step (RDS).
The overall rate cannot exceed the rate of the RDS.
The experimental rate law often matches the stoichiometry of the RDS (and preceding equilibria).
Fast steps after the RDS do not usually appear in the rate law.
Example 1: slow step 2A + B → products gives Rate = k[A]²[B].
Example 2: if A + B ⇌ C (fast equilibrium) then C → products (slow), rate may depend on [A][B].
Intermediates are formed and consumed; they cancel in the overall equation.
Catalysts may appear in early steps and be regenerated later.
A valid mechanism must reproduce the overall balanced equation.
It must also predict the observed rate law and any known intermediates.
Changing conditions can change which step is slow, altering the rate law.
Mechanism proposals are tested, not assumed from stoichiometry alone.
Spectroscopy and trapping experiments help detect intermediates.
School kinetics emphasises linking RDS ideas to simple rate laws.
Always state that order is experimental until a mechanism is established.
Clear RDS reasoning explains why coefficients and orders may differ.
Writing both the overall equation and the proposed steps prevents stoichiometry mistakes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Activation energy Ea is the minimum energy barrier that must be overcome for reactants to form products.
It is the energy difference between reactants and the transition state.
The activated complex (transition state) is an unstable arrangement of atoms at the top of the barrier.
Bonds are partially broken and partially formed in this fleeting state.
On a reaction profile, enthalpy (or potential energy) is plotted against reaction coordinate.
Reactants start at one energy level; products end at another.
The peak height from reactants to the transition state is Ea(forward).
From products back to the same peak is Ea(reverse).
ΔH ≈ Ea(forward) − Ea(reverse) for the elementary process on the profile.
An exothermic reaction has products lower than reactants; endothermic has products higher.
A catalyst provides a new pathway with a lower peak.
Both forward and reverse Ea values are reduced on the catalysed path.
More molecules can cross the barrier at the same temperature.
ΔH (or ΔG) between reactants and products is unchanged by the catalyst.
Homogeneous catalysts share the phase of reactants; heterogeneous ones provide a surface.
Enzymes bind substrates in an active site that stabilises the transition state.
Lowering Ea increases rate exponentially via Arrhenius behaviour.
Profiles must not confuse kinetics (barrier height) with thermodynamics (reactant–product gap).
Exam answers should label reactants, products, Ea, activated complex, and catalysed path.
Label ΔH and Ea separately so thermodynamics is not confused with kinetics.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Gibbs free energy change ΔG tells whether a process is thermodynamically spontaneous under given conditions.
If ΔG < 0, the forward reaction is spontaneous (product-favoured in the thermodynamic sense).
If ΔG > 0, the reverse is favoured; if ΔG = 0, the system is at equilibrium.
Spontaneity does not mean the reaction is fast.
Kinetics depends on the activation energy pathway, not only on ΔG.
Diamond converting to graphite is spontaneous (ΔG < 0) yet extremely slow at room temperature.
A high Ea keeps the rate negligible even when products are more stable.
ΔG° is related to the equilibrium constant: ΔG° = −RT ln K.
A large K means products dominate at equilibrium, but arrival at equilibrium can still be slow.
Catalysts change rate by lowering Ea; they do not change ΔG or K.
Temperature affects both ΔG (through ΔH − TΔS) and rate (through Arrhenius k).
Exothermic reactions can still be slow if Ea is large.
Endothermic reactions need continuous energy supply and may also be slow.
In industry, chemists seek conditions where thermodynamics and kinetics are both acceptable.
School answers should clearly separate “will it go?” (ΔG/K) from “how fast?” (rate/Ea).
Energy profiles show ΔH (thermodynamics) and Ea (kinetics) on the same diagram.
Students must not assume negative ΔG implies a rapid reaction.
Likewise, a fast reaction is not automatically product-favoured at equilibrium.
Correct use of both ideas prevents common exam mistakes.
Exam answers should give one slow-but-spontaneous example and one fast-but-equilibrium-limited idea.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.