The rate constant of a reaction has units s⁻¹. The order of the reaction is:

Cannot be determined from the units

For the reaction 2NO + O₂ → 2NO₂, the experimentally found rate law is rate = k[NO][O₂]. The order of the reaction is:

3, from the stoichiometric coefficients

Adding a catalyst to a reaction:

Lowers Ea for both directions equally, leaving ΔH unchanged

Lowers Ea for the forward reaction only

Lowers Ea for both directions equally, leaving ΔH unchanged

Makes the reaction more exothermic

The hydrolysis of an ester in a large excess of water is found to be first order. This is because:

The concentration of water is essentially constant, so it drops out of the rate law

Water does not take part in the reaction

The concentration of water is essentially constant, so it drops out of the rate law

Raising the temperature by 10 K can double a reaction rate mainly because:

The fraction of collisions with energy above Ea rises sharply

Molecules move about twice as fast

The fraction of collisions with energy above Ea rises sharply

The activation energy falls with temperature

The collision frequency doubles

Chemical Kinetics — Rate Laws, Order and Activation Energy

Kinetics supplies the answer thermodynamics refused to give: how fast. The whole chapter rests on one distinction that must become automatic — order is experimental, molecularity is theoretical. You cannot read the order of a reaction off its balanced equation, however tempting the coefficients look, because the equation describes the overall stoichiometry while the rate is set by the slowest single step. A reaction written with four reactant molecules can turn out to be first order, and one written with a single molecule can turn out to be second order. Every exam question that looks like a trap in this chapter is built on candidates forgetting that sentence.

Rate, Rate Law and Order

For aA + bB → cC + dD the rate is defined so that all four expressions agree: rate = −(1/a)d[A]/dt = −(1/b)d[B]/dt = +(1/c)d[C]/dt. The negative signs make reactant rates positive, and dividing by the coefficient makes the single number unambiguous. So in N₂ + 3H₂ → 2NH₃, hydrogen disappears three times as fast as nitrogen and ammonia appears twice as fast, yet the rate of reaction is one value.

The rate law rate = k[A]ˣ[B]ʸ must be found by experiment. The exponents x and y are the orders with respect to each reactant and their sum is the overall order. They can be zero, fractional or even negative; they need bear no relation to the stoichiometric coefficients. Molecularity, by contrast, is the number of species colliding in a single elementary step, so it is always a small positive whole number and is defined only for an elementary reaction.

🎯 Why the rate-determining step controls the rate law

Think of a highway with one narrow bridge. The number of cars reaching the far side per hour is set entirely by the bridge, no matter how wide the road is on either side of it. In a multi-step reaction the slow step is the bridge: species that react only after it do not appear in the rate law, because they are never the bottleneck. This explains an otherwise baffling observation — the acid-catalysed hydrolysis of an ester is written with two reactants but is often first order overall, because water is present in such vast excess that its concentration barely changes and it drops out of the rate expression. This is called a pseudo first-order reaction, and it is the standard example.

Zero, first and second order compared
Order	Integrated form	Straight-line plot and half-life
Zero	[A] = [A]₀ − kt	[A] vs t; t½ = [A]₀/2k, so it depends on [A]₀
First	ln[A] = ln[A]₀ − kt	ln[A] vs t; t½ = 0.693/k, independent of [A]₀
Second	1/[A] = 1/[A]₀ + kt	1/[A] vs t; t½ = 1/k[A]₀
Units of k	mol L⁻¹s⁻¹ (zero), s⁻¹ (first), L mol⁻¹s⁻¹ (second)	k has units mol^(1−n)L^(n−1)s⁻¹ for order n

🧠 Read the order off the units of k

Because k must make the rate come out in mol L⁻¹s⁻¹, its own units are fixed by the order and nothing else. So if a question gives k = 4.5 × 10⁻³ s⁻¹ without saying the order, the reaction is first order — and if k is quoted in L mol⁻¹s⁻¹ it is second order. This is the fastest possible route through several exam questions and it works in reverse too: an order of ½ demands units of mol^½L^−½s⁻¹, which is a useful sanity check on any fractional-order answer you calculate.

⚠️ A constant half-life proves first order — and only first order

First order is the unique case where t½ = 0.693/k contains no concentration term, so successive half-lives are equal: 100 → 50 → 25 → 12.5 in three identical intervals. For zero order the half-life shortens as the reaction proceeds, because it is proportional to the remaining concentration; for second order it lengthens. So a data table showing equal successive half-lives is telling you the order without any calculation, and a table showing them changing rules first order out immediately. This is also why radioactive decay, which is strictly first order, has a half-life that is a property of the isotope alone.

Temperature, Arrhenius and Activation Energy

The Arrhenius equation is k = A e^(−Ea/RT), or in the form used for graphs, ln k = ln A − Ea/RT. A plot of ln k against 1/T is therefore a straight line of slope −Ea/R, which is how activation energy is measured. For two temperatures, ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), the working equation for almost every numerical in this section.

🎯 Why 10 degrees can double a rate when the molecules speed up only 2 %

Going from 300 K to 310 K raises the average molecular speed by about 1.6 % — nowhere near enough to explain a doubled rate. The answer is that reaction needs not average energy but energy above Ea, and that fraction is governed by the exponential e^(−Ea/RT), which sits in the steep tail of the Maxwell distribution. A small rightward shift of the whole distribution moves a large proportion of the tail past the threshold. So temperature acts not by making collisions faster but by making a far greater fraction of them effective. This is also why a reaction with a large Ea is much more temperature-sensitive than one with a small Ea — the tail it depends on is further out.

⚠️ A catalyst lowers Ea for both directions equally

A catalyst provides an alternative path with a lower activation barrier, and because the barrier is lowered from the same transition state, the forward and reverse activation energies fall by the same amount. Two consequences are examined. First, ΔH of the reaction is untouched, since ΔH = Ea(forward) − Ea(reverse) and both dropped equally — the catalyst cannot make a reaction more exothermic. Second, the equilibrium constant is unchanged, so a catalyst can never increase the yield at equilibrium, only the speed of reaching it. A question offering "the catalyst increases the equilibrium yield" is offering a thermodynamic impossibility.

Two theories underpin the equation. Collision theory says the rate is the collision frequency multiplied by the fraction with enough energy and the fraction correctly oriented, which is where the pre-exponential factor A comes from — A is essentially collision frequency times an orientation probability. Transition state theory describes reactants climbing to an activated complex at the top of the barrier, a species so short-lived it can never be isolated, from which the system falls either forward to products or back to reactants. Both make the same qualitative prediction, and the paper tests the vocabulary rather than the derivations.

Key takeaways

Rate = −(1/a)d[A]/dt; divide by the coefficient so all reactants and products give one number.
Order is experimental and can be zero, fractional or negative; molecularity is theoretical and always a small whole number.
The slowest step sets the rate law; a vast excess reactant drops out, giving pseudo first order.
First order: ln[A] vs t is linear and t½ = 0.693/k is independent of the starting concentration.
Equal successive half-lives prove first order; they shorten for zero order and lengthen for second.
The units of k reveal the order: s⁻¹ is first order, L mol⁻¹s⁻¹ is second, mol L⁻¹s⁻¹ is zero.
ln k vs 1/T is a straight line of slope −Ea/R — this is how activation energy is measured.
A catalyst lowers Ea equally in both directions, so ΔH and K are untouched — only the speed changes.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

The rate constant of a reaction has units s⁻¹. The order of the reaction is:
1. Zero
2. First
3. Second
4. Cannot be determined from the units
Show answer
Answer: B — First
Since rate has units mol L⁻¹s⁻¹ and rate = k[A]ⁿ, the units of k are mol^(1−n)L^(n−1)s⁻¹. Setting that equal to s⁻¹ requires n = 1. The units of k are therefore a complete fingerprint of the order — zero order would give mol L⁻¹s⁻¹ and second order L mol⁻¹s⁻¹ — so the last option is wrong for a reason worth using as a shortcut in the exam.
A reaction shows successive half-lives of 20, 20 and 20 minutes. The order is:
1. Zero
2. First
3. Second
4. Third
Show answer
Answer: B — First
Only for first order does t½ = 0.693/k contain no concentration term, so only first order gives equal successive half-lives. Zero order has t½ = [A]₀/2k, which shrinks as the concentration falls, and second order has t½ = 1/k[A]₀, which grows. Constant half-life is therefore a complete diagnostic and needs no further calculation.
For the reaction 2NO + O₂ → 2NO₂, the experimentally found rate law is rate = k[NO][O₂]. The order of the reaction is:
1. 3, from the stoichiometric coefficients
2. 2
3. 1
4. 2 with respect to NO
Show answer
Answer: B — 2
Order is read only from the experimental rate law, where the exponents are 1 and 1, giving an overall order of 2. The coefficient of 2 in the balanced equation would suggest third order and second order in NO, and both of those options are offered precisely to catch candidates who read the order off the equation instead of the rate law. Only for an elementary step do the two happen to coincide.
Adding a catalyst to a reaction:
1. Lowers Ea for the forward reaction only
2. Lowers Ea for both directions equally, leaving ΔH unchanged
3. Makes the reaction more exothermic
4. Increases the equilibrium yield
Show answer
Answer: B — Lowers Ea for both directions equally, leaving ΔH unchanged
The catalyst provides a lower-barrier path through a new transition state, and since ΔH = Ea(forward) − Ea(reverse), lowering both by the same amount leaves the enthalpy change and hence K untouched. That rules out the last two options as thermodynamic impossibilities: a catalyst can change how fast equilibrium arrives, never where it lies.
A plot of ln k against 1/T gives a straight line whose slope equals:
1. Ea/R
2. −Ea/R
3. ln A
4. −Ea/2.303R
Show answer
Answer: B — −Ea/R
Taking natural logarithms of k = Ae^(−Ea/RT) gives ln k = ln A − (Ea/R)(1/T), so the slope is −Ea/R and the intercept is ln A. The option −Ea/2.303R is the slope you get when plotting log₁₀ k rather than ln k, which is why the base of the logarithm on the axis must be checked before reading a slope.
The hydrolysis of an ester in a large excess of water is found to be first order. This is because:
1. Water does not take part in the reaction
2. The concentration of water is essentially constant, so it drops out of the rate law
3. The reaction is elementary
4. Esters are always first order
Show answer
Answer: B — The concentration of water is essentially constant, so it drops out of the rate law
Water is a genuine reactant, but in vast excess its concentration barely changes over the course of the reaction, so the term [H₂O] is effectively a constant that gets absorbed into k. The reaction is truly second order but behaves as first order — hence the name pseudo first-order. The first option misstates the chemistry: water is consumed, it simply does not appear in the observed rate law.
For a first-order reaction with k = 0.0693 min⁻¹, the time needed for 75 % completion is:
1. 10 min
2. 15 min
3. 20 min
4. 30 min
Show answer
Answer: C — 20 min
The half-life is 0.693/0.0693 = 10 min, and 75 % completion means one quarter remains, which is two successive halvings — so 20 min. The answer 15 min comes from assuming the second half takes only half as long as the first, but for first order every halving takes exactly the same time regardless of how much is left.
Raising the temperature by 10 K can double a reaction rate mainly because:
1. Molecules move about twice as fast
2. The fraction of collisions with energy above Ea rises sharply
3. The activation energy falls with temperature
4. The collision frequency doubles
Show answer
Answer: B — The fraction of collisions with energy above Ea rises sharply
From 300 K to 310 K the average speed rises only about 1.6 % and collision frequency similarly little, so neither can account for a doubling. What changes sharply is the exponential factor e^(−Ea/RT): a small shift of the Maxwell distribution moves a large proportion of its high-energy tail past the activation threshold. Ea itself is a property of the reaction path and does not depend on temperature, which rules out the third option.