5.1.1 Rates of Reaction

For a concentration measurement, rate of reaction is the change in concentration of a reactant or product divided by the time taken. If you follow a reactant, its concentration falls; if you follow a product, it rises.

I would keep the definition tied to what you measure. A common concentration-rate unit is mol dm⁻³ s⁻¹, although the time unit can change with the experiment.

The order with respect to a reactant is the power of its concentration in the rate equation. It tells you how strongly changing that concentration changes the rate, with the other relevant conditions held constant.

If you double the concentration: zero order gives no rate change, first order doubles the rate, and second order makes the rate four times larger. That quick check is often the fastest way into an initial-rates question.

• zero order: ×2 concentration → rate ×1
• first order: ×2 concentration → rate ×2
• second order: ×2 concentration → rate ×4

Once you have the rate equation, add the individual powers to get the overall order. For rate = k[A]²[B], the reaction is second order in A, first order in B and third order overall.

Do not use the coefficients in the balanced equation for this. Reaction orders come from experimental evidence unless a mechanism gives you a justified route to them.

The rate constant, k, is the proportionality constant that links the concentrations in the rate equation to the reaction rate. For a particular reaction at a fixed temperature, changing a reactant concentration changes the rate but not k.

Temperature does change k, and that becomes important when you reach the Arrhenius equation. The units of k depend on the overall order, so never assume they are always s⁻¹.

The half-life, t₁/₂, is the time taken for the concentration of a reactant to fall to half its value. You can measure successive half-lives from a concentration–time graph by halving the concentration again each time.

Those half-lives stay constant for a first-order reaction. That constancy is evidence you can use later to identify first-order behaviour.

A multi-step mechanism can contain several elementary steps, but one step controls how fast the overall reaction can proceed. We call that the rate-determining step.

The experimentally measured rate equation has to be consistent with that step and any earlier fast steps that supply species to it. This is why the rate equation can tell you something about a mechanism that the overall equation cannot.

Compare two experiments where one reactant concentration changes while the others stay the same. Then compare the concentration factor with the rate factor. If doubling [A] leaves the rate unchanged, A is zero order; if the rate doubles, A is first order; if it quadruples, A is second order.

When more than one concentration changes, choose a better pair of experiments if you can. If you cannot, use the rate ratios carefully and account for the order you already know.

Once you have the order for each reactant, put those concentrations into the rate equation with the orders as powers. A zero-order reactant disappears from the written equation because [A]⁰ = 1.

For example, first order in A, zero order in B and second order in C gives rate = k[A][C]². I would write the rate equation before calculating k; it keeps the rest of the maths much cleaner.

Rearrange the rate equation for the quantity you need, then substitute concentrations in mol dm⁻³ and the rate in the units given. Keep the unrounded calculator value until the end.

For rate = k[A][B]², if rate = 4.00 × 10⁻⁴ mol dm⁻³ s⁻¹, [A] = 0.200 mol dm⁻³ and [B] = 0.100 mol dm⁻³, then k = 0.200 in the appropriate units.

Do not memorise one unit for k. Start from k = rate divided by the concentration terms in your rate equation, then cancel the units algebraically.

A useful pattern falls out of that maths: zero order gives mol dm⁻³ s⁻¹, first order gives s⁻¹, second order gives dm³ mol⁻¹ s⁻¹, and third order gives dm⁶ mol⁻² s⁻¹.

• overall order 0 → mol dm⁻³ s⁻¹
• overall order 1 → s⁻¹
• overall order 2 → dm³ mol⁻¹ s⁻¹
• overall order 3 → dm⁶ mol⁻² s⁻¹

On a concentration–time graph, a zero-order reactant falls in a straight line because its concentration decreases by the same amount in equal time intervals. A first-order reactant gives a curve because the rate slows as the concentration falls.

For the first-order curve, equal fractions disappear in equal times: the concentration keeps halving after the same time interval. That constant half-life is the stronger check if the graph makes the shape hard to judge.

If the question asks for the rate at one particular time, draw a tangent to the curve at that point and work out its gradient using two well-separated points on the tangent. Do not use two points from the curve itself.

For a reactant concentration the gradient is negative because the concentration is falling. Reaction rate is normally quoted as positive, so reverse the sign unless the question specifically asks for the gradient.

A first-order reaction has a constant half-life. So if 0.80 mol dm⁻³ falls to 0.40 in the same time that 0.40 falls to 0.20, that is first-order evidence.

Measure more than one half-life if the graph gives you enough room. One approximate halving could be coincidence; repeated equal half-lives make the pattern much clearer.

Use k = ln 2 / t₁/₂ for a first-order reaction. Because ln 2 = 0.693, you can go directly between k and half-life once the time units are consistent.

If t₁/₂ is in seconds, k comes out in s⁻¹; if it is in minutes, k comes out in min⁻¹. OCR gives the relationship, so you only need to use it accurately.

A rate–concentration graph gives a different set of shapes from a concentration–time graph. Zero order is horizontal because rate does not depend on concentration. First order is a straight line through the origin because rate ∝ [A]. Second order curves upwards because rate ∝ [A]².

I would check the axes before doing anything else. OCR has explicitly seen students mix up these two graph types.

• zero order: horizontal line
• first order: straight line through the origin
• second order: upward curve

If rate = k[A], a graph of rate against [A] is a straight line through the origin. Compare that with y = mx: the gradient is k.

Use two well-separated points on the best-fit line to calculate the gradient, and keep the axis scales in the calculation. Because this is first order, the gradient has units of s⁻¹ if rate is in mol dm⁻³ s⁻¹ and concentration is in mol dm⁻³.

Run a series of experiments with known starting concentrations and change one reactant concentration at a time where possible. Measure the initial rate, then compare experiments to work out the order with respect to each reactant.

A clock method can be used when the same small, fixed amount of reaction produces an observable change each time. If that change happens early enough, 1/t is proportional to the initial rate, so you can compare 1/t values between experiments.

With continuous monitoring, you follow a quantity throughout the reaction instead of recording only an initial rate. Choose something that changes with reaction progress, such as gas volume, mass, concentration or absorbance, and record it at regular intervals.

Plot the measurement against time. You can then find rates from tangent gradients and, where suitable, use the graph shape or half-life to get kinetic information.

Colorimetry is useful when the concentration of a coloured reactant or product changes during the reaction. Choose a suitable wavelength or filter, measure absorbance as the reaction proceeds, and relate the absorbance to concentration.

If you need actual concentrations, use calibration data from known concentrations. Keep the same cuvette setup and wavelength throughout so that changes in absorbance are telling you about concentration, not a change in the measurement conditions.

Look at the rate-determining step and ask which reactant particles have to be available for that step to happen. If an intermediate is needed, include the earlier fast step that makes it. Together, those steps tell you which concentration terms and powers you expect in the rate equation.

Do not simply copy the overall equation into a rate equation. A mechanism can have intermediates that cancel from the overall equation, and the observed order can depend on species used before the rate-determining step.

Start with the experimental rate equation: your proposed rate-determining step has to explain the observed concentration powers. Then add any faster steps needed to make intermediates or complete the reaction.

Finally, add the mechanism steps together. Intermediates should cancel and you should recover the balanced overall equation. If the steps do not sum to the overall reaction, the mechanism is not finished.

When temperature rises, the Maxwell–Boltzmann distribution shifts so that a much larger fraction of particles has energy at least equal to Eₐ. That gives a larger fraction of successful collisions.

The important Year 13 link is that k increases as temperature increases. The activation energy has not fallen; you have changed the energy distribution of the particles, not the reaction pathway.

The Arrhenius equation links the rate constant to temperature: k = Ae^(−Eₐ/RT). Here k is the rate constant, A is the pre-exponential factor, Eₐ is activation energy, R is the gas constant and T is absolute temperature in kelvin.

OCR supplies the equation and R = 8.314 J K⁻¹ mol⁻¹. If you use that R value, put Eₐ into J mol⁻¹ before substituting. You need to use A correctly, but OCR does not require an explanation of what A represents.

For an Arrhenius plot of ln k against 1/T, compare ln k = −Eₐ/R(1/T) + ln A with y = mx + c. The gradient is −Eₐ/R, so Eₐ = −gradient × R.

The gradient should be negative. Check the scale on the 1/T axis before using it: if the graph labels the axis with a ×10⁻³ factor, that factor belongs in the numbers you use for the gradient.

On the same graph, the y-intercept is ln A, not A. Once you have the intercept, use A = e^(intercept) to get A.

If the graph does not extend to 1/T = 0, do not treat the visible left-hand edge as the intercept. The intercept is the value of ln k when 1/T = 0. A has the same units as k.