2.2 Rate of chemical change

$rate = change in concentration / time$

An instantaneous rate is the gradient of a tangent at the stated point on a concentration–time, volume–time or mass–time graph. Keep the units consistent.

Choose two well-separated points on the tangent, not on the curve, to calculate its gradient. A reactant concentration has a negative gradient, so report its disappearance rate as a positive magnitude unless a signed change is requested.

Requested rate	Graph operation	Evidence check
mean over an interval	secant gradient between interval endpoints	quote the interval and units
initial	tangent gradient at t = 0	choose well-separated points on the tangent
instantaneous at time t	tangent gradient at that time	do not use two points on the curved trace

A measured mass, pressure or gas volume is a rate proxy only when its change is tied to reaction progress under the stated conditions. Preserve reactant/product slope sign or report a positive disappearance/formation magnitude as requested.

Representative question

[Maximum number: 3]

Determine the instantaneous rate of reaction to two significant figures when $\left[\mathrm{Br}_{2}\right]=0.0080 \mathrm{~mol} \mathrm{dm}^{-3}$ .

tangent drawn on curve at $0.0080 \mathrm{~mol} \mathrm{dm}^{-3}$
$\text { «Rate }=\frac{\Delta\left[\mathrm{Br}_{2}\right]}{\Delta \mathrm{t}}=\text { » } 2.8 \times 10^{-5} \ll \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} \gg \text { }$
two significant figures in final answer

Marking guidance:

Accept range of $2.6 \times 10^{-5}$ to $3.1 \times 10^{-5}$ « $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ ».
Award [2 max] for $3.3 \times 10^{-5} \| \mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1} »$ .
Award [1 max] for $3.33 \times 10^{-5}$ « $\mathrm{mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}$ ».
Ignore negative sign for M2.
Award M3 for any numerical result with 2 significant figures.

A successful collision needs sufficient kinetic energy to overcome activation energy and a suitable orientation of the reacting particles.

Temperature raises average kinetic energy and changes collision frequency and the fraction of particles with energy at least Ea. Collision frequency alone does not guarantee reaction.

At one temperature, only the fraction of collisions above Ea and with a productive orientation can react. Raising temperature increases that fraction, not the energy of every particle by the same amount. Use collision frequency to explain concentration or pressure effects and the energy distribution to explain the stronger temperature effect.

Representative question

[Maximum number: 3]

Explain, using collision theory, how an increase in temperature increases the reaction rate.

Any 3 of the following:

temperature increases kinetic energy/speed of molecules more frequent collisions

more molecules have $\mathrm{E} \geq \mathrm{Ea}$ at higher temperature

larger ratio/percentage of collisions are successful

M2 requires time reference for probability, chance, or number of
collisions.

3 max

Change	Main collision consequence
concentration or pressure up	more frequent collisions
surface area up	more collisions at a solid surface
temperature up	more frequent and more energetic collisions
catalyst	alternative lower-Ea pathway

Explain a predicted rate change through collision frequency or the fraction of effective collisions, not just by saying particles move faster.

Powdered CaCO₃ reacts faster than equal-mass chips because more surface sites are exposed, not because its particles have higher kinetic energy. For each changed condition, identify exactly what changes—collision frequency, energy distribution or pathway—and hold other variables constant in a fair comparison.

Representative question

[Maximum number: 2]

The student then carried out the experiment at other acid concentrations with all other conditions remaining unchanged.

\begin{tabular}{|c|c|}
\hline $\left[\mathbf{H}^{+}\right] / \mathbf{~ m o l ~ d m}^{-\mathbf{3}}$ & Relative rate of reaction \\
\hline 0.05 & 0.0025 \\
\hline 0.10 & 0.0051 \\
\hline 0.20 & 0.0100 \\
\hline
\end{tabular}

State and explain the relationship between the rate of reaction and the concentration of acid.

Relationship:

rate of reaction is «directly» proportional to [ $\mathrm{H}^{+}$ ]

OR

rate of reaction $\alpha\left[\mathrm{H}^{+}\right]$

Explanation: more frequent collisions/more collisions per unit of time «at greater concentration»

Marking guidance:

Accept "doubling the concentration doubles the rate".

Do not accept "rate increases as concentration increases".

Do not accept collisions more likely.

$Ea = minimum kinetic energy for an effective collision$

A Maxwell–Boltzmann curve shows the distribution of particle kinetic energies. At higher temperature the peak is lower and shifts right; the area beyond Ea is larger, so more particles can react.

Two Maxwell–Boltzmann curves for the same number of particles have equal total area. At higher T the curve is broader with a lower peak and a larger area to the right of a fixed Ea line; the peak does not move to Ea and no particle count is lost.

Representative question

[Maximum number: 3]

Explain why the reaction rate increases with temperature, adding annotations to the following Maxwell-Boltzmann graph to assist your explanation.

Explanation:

A catalyst provides an alternative reaction pathway with lower activation energy. It does not change the energy levels of the reactants or products.

Because the catalysed Ea is lower, a larger fraction of the same distribution lies beyond the threshold. The reaction therefore has more effective collisions at the same temperature.

At fixed temperature a catalyst does not change the Maxwell–Boltzmann distribution; it moves the threshold to a lower Ea, increasing the area beyond it. On an energy profile it changes the pathway and peak, not ΔH, reactant/product energies or the equilibrium constant.

Representative question

[Maximum number: 2]

Sketch the Maxwell-Boltzmann energy distribution curve for this reaction. Label the activation energy with and without a catalyst on the diagram.

Fraction of particles
$\xrightarrow[\text { Kinetic energy }]{E_{a} \text { with catalyst }}$
correct shape curve starting at the origin, without touching the x axis at high energy.
( $\mathrm{E}_{\mathrm{a}}$ ) catalysed $<\left(\mathrm{E}_{\mathrm{a}}\right)$ uncatalysed on x axis.

A mechanism is a sequence of elementary steps. The slowest step is the rate-determining step; an intermediate is formed in one step and consumed in a later step, whereas a transition state is the high-energy configuration at a barrier.

A proposed mechanism must be compared with the experimental rate equation and stoichiometry. Matching a rate equation can support a mechanism but does not prove it, because different mechanisms may give the same expression.

Add all elementary steps and cancel intermediates to recover the overall equation. Then derive the rate dependence expected from the slow step, eliminating an intermediate when necessary, and compare with experiment. A catalyst consumed early and regenerated later cancels from the overall equation but is not an intermediate.

Representative question

[Maximum number: 1]

Suggest why experimental confirmation of the rate equation would not prove that the mechanism is correct.

In a multistep profile, peaks are transition states and valleys between peaks are intermediates. Each step has its own Ea; the largest barrier from its preceding intermediate identifies the rate-determining step.

$ΔH = energy(products) − energy(reactants)$

For each step, measure Ea upward from its own preceding reactant or intermediate valley to the next peak. The rate-determining barrier is the largest of those step barriers, which need not be the peak with the greatest absolute height above the page baseline.

Representative question

[Maximum number: 4]

Sketch an energy profile for the two-step reaction, labelling reactants, intermediate and products, activation energies, $E_{\mathrm{a}}$ , and overall enthalpy change, $\Delta H$ . Assume that the reaction is exothermic.

reaction coordinate

products lower than reactants

intermediate at an energy minimum

$E_{\mathrm{a}}$ of both steps correctly marked

$\Delta H$ from reactants down to products

Accept any combination of peak heights.

Accept double-headed arrows or lines with no heads.

M1 and M4 may be awarded for a curve with a single maximum.

Ignore the energy position of the intermediate.

Molecularity is the number of reacting particles in one elementary step: one is unimolecular, two is bimolecular and three is termolecular.

Count particles in the individual elementary step, not coefficients in the overall reaction equation.

A step A + 2B → products is termolecular because three reacting particles meet in that elementary event. Molecularity is always a positive whole-number description of one proposed step; reaction order is experimental and can be zero, fractional or unrelated to overall stoichiometric coefficients.

Representative question

[Maximum number: 1]

Identify the molecularity of the rate-determining step in this reaction.

$rate = k[A]^x[B]^y$

Use experimental trials that vary one concentration independently. Compare the rate factor with the concentration factor to infer each exponent; the balanced overall equation alone cannot determine the rate equation.

If doubling [A] while holding [B] constant quadruples rate, the order in A is 2; if doubling [B] leaves rate unchanged, the order in B is 0, giving rate = k[A]². Choose trial pairs with only one changed concentration before combining exponents.

Representative question

[Maximum number: 2]

Two more trials ( 2 and 3 ) were carried out. The results are given below.

Trials	Volume of $\mathbf{0 . 2 0 ~ m o l ~ d m}{ }^{\mathbf{- 3}} \mathbf{K I} \boldsymbol{(} \mathbf{a q} \boldsymbol{)} \boldsymbol{/} \mathbf{c m}^{\mathbf{3}}$ Volume of $\mathbf{0 . 2 0 ~ m o l ~ d m}{ }^{\mathbf{- 3}}$ $\mathbf{K I} \boldsymbol{(} \mathbf{a q} \boldsymbol{)} \boldsymbol{/} \mathbf{c m}^{\mathbf{3}}$	Volume of deionised water / $\mathbf{c m}^{\mathbf{3}}$	Volume of 3\% $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) / \mathbf{c m}^{3}$ Volume of 3\% $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})$ $/ \mathbf{c m}^{3}$	Average rate of reaction $/ \mathrm{cm}^{3} \mathbf{O}_{\mathbf{2}}(\mathrm{g}) \mathbf{s}^{-1}$
1	10.0	15.0	5.0
2	10.0	10.0	10.0	0.0429
3	20.0	5.0	5.0	0.0451

Determine the rate equation for the reaction and its overall order, using your answer from (b)(i).

Rate equation:

Overall order:

Rate equation:
$\text { Rate }=k\left[\mathrm{H}_{2} \mathrm{O}_{2}\right] \times[\mathrm{KI}]$

Overall order:

2

Rate constant must be included.

$overall order = x + y in rate = k[A]^x[B]^y$

The exponent gives order with respect to that reactant. Use rate factors to identify zero, first or second order, then match the order to the concentration–time or rate–concentration graph.

Use multiple representations as cross-checks: zero-order rate is independent of concentration and [A] falls linearly; first-order rate is proportional to [A] and gives exponential decay; second-order rate curves upward on a rate-versus-concentration plot. Do not infer order from one balanced equation.

Representative question

[Maximum number: 1]

Compounds P and Q were mixed together at various concentrations and the initial rate of each reaction was measured.

\begin{tabular}{|l|l|l|l|}
\hline Experiment & [P] mol dm ${ }^{-3}$ & [Q] mol dm ${ }^{-3}$ & Initial rate of reaction
(mol dm ${ }^{-3} \mathrm{~s}^{-1}$ ) \\
\hline 1 & 0.20 & 0.15 & 0.50 \\
\hline 2 & 0.10 & 0.15 & 0.25 \\
\hline 3 & 0.20 & 0.30 & 1.00 \\
\hline
\end{tabular}

What are the orders of reaction with respect to P and Q ?

Order with respect to P

Order with respect to Q

$k = rate / ([A]^x[B]^y)$

The units of k depend on the overall order and must cancel the concentration and time units in the rate equation. For a particular reaction, k changes with temperature.

Derive rather than memorize the units: first-order k has units s⁻¹, while an overall second-order law written with mol dm⁻³ and seconds gives dm³ mol⁻¹ s⁻¹. Concentration changes rate but does not change k at fixed temperature.

Representative question

[Maximum number: 2]

Calculate the value of the rate constant stating its units.

$k=\ll \frac{1.2 \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3} \mathrm{~s}^{-1}}{2.0 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3} \times 3.0 \times 10^{-3} \mathrm{~mol} \mathrm{dm}^{-3}}=>0.20$

$\mathrm{mol}^{-1} \mathrm{dm}^{3} \mathrm{~s}^{-1}$

$ln k = (−Ea/R)(1/T) + ln A$

On a plot of ln k against 1/T, gradient = −Ea/R and intercept = ln A. Use kelvin, preserve the negative gradient and convert J mol−1 to kJ mol−1 when required.

A steeper negative ln k versus 1/T gradient means a larger activation energy. When two temperatures are given, subtracting the two linear equations removes ln A and lets Ea be found without knowing the frequency factor.

$ln(k₂/k₁) = −(Eₐ/R)(1/T₂ − 1/T₁) with T in K and Eₐ in J mol⁻¹ when R = 8.31 J mol⁻¹ K⁻¹$

Representative question

[Maximum number: 2]

Determine the activation energy of the reaction, $E_{\mathrm{a}}$ , in $\mathrm{kJ} \mathrm{mol}^{-1}$ , from the Arrhenius plot. Use sections 1 and 2 of the data booklet.

gradient =-8900
$\text { « } E_{\mathrm{a}}=-(-8900 \times 8.31)=+74000 \mathrm{~J} \mathrm{~mol}^{-1}=» 74 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Award [2] for correct final answer.

Accept $E_{a}$ in the range 66 to 76 «kJ $\mathrm{mol}^{-1}{ }^{-}{ }^{-}$

Do not accept $E_{a}$ in $\mathrm{J} \mathrm{mol}^{-1}$ .

Award [1 max] for -74 « $\mathrm{kJ} \mathrm{mol}^{-1}$ ».

Accept gradient in the range: -8000 to -9100 for M1.

A is the frequency factor: it represents the frequency of collisions with proper orientation. In the linear Arrhenius form, the y-intercept is ln A.

Read the intercept, find A by exponentiating ln A, and keep A distinct from Ea: Ea describes the energy barrier, while A describes collision frequency and orientation.

From a fitted line, intercept b gives A = eᵇ, while gradient m gives Ea = −mR. Check the pair against k = Ae^(−Ea/RT) at one data temperature. A has the same units as k for the stated rate law; it is not an activation energy or a universal constant.

Representative question

[Maximum number: 2]

Calculate the numerical value of A.

$\ln \mathrm{A}=($ intercept on y-axis =) 23.2;
$\mathrm{A}=1.190 \times 10^{10} ;$

Retrieve the route: measure a tangent rate, explain effective collisions, map rate factors, read Ea and energy profiles, evaluate mechanisms, determine molecularity and orders, calculate k, then use Arrhenius gradient and intercept for Ea and A.

Check tangent versus average slope, energy versus orientation, barrier labels, intermediate versus transition state, one-variable trial comparisons, order-dependent units, kelvin temperature and the signs of gradient and Ea.

ConceptIB Chemistry HL

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