How fast? — OCR A-Level Chemistry

This row is a study guide for the quantitative core of reaction kinetics, not an assessed answer target. Work through it in three linked stages. First, order: for a reactant, order is the power to which its concentration is raised in the experimentally determined rate equation, and it is found from data, not from the stoichiometric equation. Second, rate equation: assemble rate = k[A]^m[B]^n, where m and n are the orders with respect to A and B and the sum m + n is the overall order. Third, rate constant: k is the constant linking rate to concentrations at a fixed temperature, and its units depend on the overall order, so derive them from the equation rather than memorising a list. Use a table of initial rates to deduce each order by comparing experiments where only one concentration changes, then calculate k and its units.

This criterion requires you to explain and use six linked terms. Rate of reaction is the change in concentration of a reactant or product per unit time. Order is the power to which a reactant's concentration is raised in the experimentally determined rate equation. Overall order is the sum of the individual orders. The rate constant k links rate to concentrations at a fixed temperature and carries units set by the overall order. Half-life is the time for the concentration of a reactant to fall to half its value; for a first-order reaction it is constant and independent of initial concentration. The rate-determining step is the slowest step of a mechanism and controls the overall rate. In a multiple-choice question, match each option against these precise definitions and against data such as a constant half-life or a rate equation.

This row is a lead-in to the sub-clause on deducing orders from experimental data, so read it as a heading that frames the skill rather than as a standalone assessed statement. In OCR A Level Chemistry (H432) section 5.1.1 How fast?, the deduction work means taking a table of initial rate measurements at varied concentrations and working out how the rate depends on each reactant. The learner should practise the method: pick two experiments where only one reactant's concentration changes, compare how the rate changes, and decide whether the order is zero, first or second. Treat the wording as a prompt to study the whole sub-clause, not to memorise a definition.

This sub-clause requires the learner to deduce the order with respect to each reactant from experimental initial-rate data. The method: find two experiments where only one reactant's concentration changes, then compare the ratio of rates with the ratio of concentrations. If doubling the concentration leaves the rate unchanged the order is zero; if the rate doubles the order is first; if the rate quadruples the order is second. Repeat for each reactant, then combine the individual orders into the overall rate equation. The MCQ tests whether the learner can apply this comparison correctly and read off the order for a named reactant.

A rate equation links the rate of reaction to the concentrations of reactants raised to their orders. For rate = k[A]ᵐ[B]ⁿ, the symbol k is the rate constant, [A] and [B] are concentrations in mol dm⁻³, and the superscripts m and n are the orders with respect to A and B. Each order is 0, 1 or 2 and is found by experiment, not from the balanced equation. An order of 0 means concentration has no effect on rate; order 1 means rate is directly proportional to concentration; order 2 means rate is proportional to concentration squared. The overall order is m + n. For example, if doubling [A] doubles the rate, m = 1; if doubling [B] leaves the rate unchanged, n = 0.

The rate constant k is found by rearranging the rate equation. For rate = k[A]ᵐ[B]ⁿ, rearrange to k = rate ÷ ([A]ᵐ[B]ⁿ). Substitute the rate (in mol dm⁻³ s⁻¹) and the concentrations (in mol dm⁻³) raised to their orders, then evaluate. The units of k depend on the overall order m + n. Since rate has units mol dm⁻³ s⁻¹ and the concentration product has units (mol dm⁻³)^(m+n), the units of k are mol^(1−(m+n)) dm^(−3(1−(m+n))) s⁻¹. For overall order 0, k has units mol dm⁻³ s⁻¹; for order 1, s⁻¹; for order 2, dm³ mol⁻¹ s⁻¹; for order 3, dm⁶ mol⁻² s⁻¹. Related quantities such as rate or a concentration can be found by substituting known values into the rearranged equation.

This reading item introduces how the shape of a rate–concentration graph reveals the order with respect to a reactant. You plot initial rate on the y-axis against the concentration of one reactant on the x-axis, keeping all other reactant concentrations constant. A straight line through the origin shows that rate is proportional to concentration, so the reaction is first order in that reactant. A horizontal line shows that changing the concentration has no effect on rate, so the order is zero. A curve that steepens as concentration rises indicates that the order is greater than one, typically second order. To read the order from a graph, compare how the rate changes when the concentration doubles: doubling the rate means first order, no change means zero order, and quadrupling the rate means second order.

This reading item covers what a concentration–time graph shows and how to extract rate information from it. You plot the concentration of a reactant on the y-axis against time on the x-axis. The gradient of the curve at any point equals the rate of reaction at that moment, so a steeper gradient means a faster rate. To find the rate at a chosen time, draw a tangent to the curve at that time and calculate its gradient as the change in concentration divided by the change in time, giving units such as mol dm⁻³ s⁻¹. As the reaction proceeds the curve flattens because reactant concentration falls and the rate decreases. The half-life is the time taken for the concentration to halve; a constant half-life indicates a first-order reaction. Compare successive half-lives to judge the order.

You deduce order from a concentration–time graph, not a rate–concentration graph. Plot concentration of the reactant against time. For zero order the concentration falls at a constant rate, so the graph is a straight line with a constant negative gradient; the half-life shortens as concentration drops. For first order the concentration falls exponentially, giving a curve whose gradient decreases steadily, and crucially the half-life is constant: each successive halving of concentration takes the same time. To test for first order, measure two or three successive half-lives from the graph; equal values confirm first order. A straight line confirms zero order. Read the shape and the half-life pattern together, and remember the order is with respect to that specific reactant.

The rate of reaction at a particular moment equals the gradient of the concentration–time graph at that moment. Draw a tangent to the curve at the chosen time, then pick two points far apart on the tangent and calculate gradient = change in concentration ÷ change in time, giving units such as mol dm⁻³ s⁻¹. For a straight-line portion the gradient is constant, so the rate is constant; for a curve the tangent must be redrawn at each time because the rate changes. Take care with signs: concentration falls with time, so the gradient is negative, and the rate is reported as a positive magnitude. Use the tangent, not a chord between two data points, when the rate at that instant is required.

A first order reaction has a rate proportional to the concentration of one reactant, so its concentration–time curve falls steeply at first and then flattens. Its defining feature is a constant half-life: the time for the concentration to halve is the same at every starting concentration. To measure t₁/₂ from a graph, read the concentration at a chosen start point, halve that value, find the time at which the curve reaches it, and subtract the start time. Repeat from a second, lower concentration. If the two intervals are equal within reading error, the reaction is first order and that interval is t₁/₂.

For a first order reaction the half-life is constant: the time for the concentration of the reactant to fall from any value to half that value stays the same as the reaction proceeds. Because t₁/₂ does not depend on concentration, it can be measured once and used to find the rate constant with k = ln 2 / t₁/₂. Here ln 2 ≈ 0.693. If t₁/₂ is in seconds, k comes out in s⁻¹; if t₁/₂ is in minutes, convert to seconds first, or report k in min⁻¹ consistently. Example: t₁/₂ = 120 s gives k = 0.693 / 120 s = 5.78 × 10⁻³ s⁻¹. A larger k means a shorter half-life and a faster reaction.

A rate–concentration graph plots the initial rate of a reaction on the vertical axis against the concentration of one reactant on the horizontal axis, with all other reactant concentrations held constant. The shape of the line reveals the order with respect to that reactant. A horizontal line (rate unchanged as concentration rises) means zero order. A straight line through the origin, where rate is directly proportional to concentration, means first order. An upward curve whose steepness increases, where rate is proportional to the square of the concentration, means second order. To read such a graph, identify the axes, note whether the line is flat, straight or curved, and state the order with a reason. For example, doubling the concentration and finding the rate quadruples indicates second order. Always keep other variables fixed so the graph isolates one reactant.

Order with respect to a reactant is deduced from how the initial rate responds to changes in that reactant's concentration, with all other concentrations kept constant. Zero order gives a horizontal rate–concentration line because rate is unaffected by concentration. First order gives a straight line through the origin because rate is directly proportional to concentration, so doubling concentration doubles rate. Second order gives a curve because rate is proportional to the square of concentration, so doubling concentration quadruples rate. To deduce order, compare rates at two concentrations: if rate is unchanged the order is 0, if rate doubles when concentration doubles the order is 1, and if rate quadruples the order is 2. State the order and support it with the observed change in rate.

For a first order reaction the rate equation is rate = k[A], so a graph of rate against [A] is a straight line through the origin whose gradient equals k. Alternatively, plotting ln(rate) against ln[A] gives a line of gradient 1 for first order and intercept ln k. In practice you obtain rates at several concentrations (for example from initial-rate experiments or from gradients of concentration–time tangents), then plot rate on the y-axis against [A] on the x-axis. The gradient, found from the coordinates of two well-separated points on the line, is k. Because rate has units mol dm⁻³ s⁻¹ and [A] has units mol dm⁻³, k carries units s⁻¹ for a first order reaction. Read the gradient from the line of best fit, not from a single data point, and quote k with its unit.

Two experimental approaches reveal how reaction rate depends on concentration. The initial rates method runs several separate experiments, varying the concentration of one reactant while holding others constant, and measures the rate at the very start of each reaction (for example from the gradient of the first part of a concentration–time curve or from a clock reaction time). Continuous monitoring follows a single reaction as it proceeds, recording a physical property that changes with concentration, such as gas volume, mass loss, pH or absorbance. Colorimetry exploits the absorption of light by a coloured species: absorbance is proportional to concentration (Beer–Lambert), so tracking absorbance over time gives a concentration–time curve from which rate can be found. Control variables such as temperature and concentrations of other reactants must be kept constant for valid comparisons.

A multi-step reaction proceeds through a sequence of steps, and the rate-determining step is the slowest step in that sequence. Because it is the slowest, it limits the overall rate, so the experimental rate equation reflects the species involved up to and including this step. Species that appear in the rate equation must be involved in the rate-determining step or in a fast equilibrium before it; a species that only reacts in a later fast step does not appear in the rate equation. For example, if the rate equation is rate = k[A][B], then A and B (or species in equilibrium with them) take part in or before the slow step. A proposed mechanism is consistent with the rate equation only if this condition holds.

For a multi-step reaction, prediction of the rate equation from a proposed mechanism depends on identifying the rate-determining step. The slowest step controls the overall rate, so the species that collide in that step, and their numbers, determine the orders in the rate equation. If the rate-determining step follows a fast equilibrium, the concentration of an intermediate may need to be substituted using the equilibrium expression, which can raise the order of a reactant. For example, a fast pre-equilibrium producing an intermediate followed by a slow step can give an order of 2 in a reactant even though only one molecule appears in the slow step. A reliable method is to write the slow step, substitute any intermediate from earlier equilibria, and compare the resulting expression with the experimental rate equation.

The rate-determining step is the slowest step of a mechanism and controls the overall rate. Any species that appears in this slow step as a reactant before the rate-determining step contributes to the rate equation, with its order equal to its number of particles in that step. Species that only enter in a later fast step do not appear in the rate equation. For example, if the slow step is A + B → X, the rate equation is rate = k[A][B], first order in both. If the slow step involves two A particles, the order in A is 2. A species formed in a fast equilibrium before the slow step can also appear, but its order reflects the step. To check consistency, compare the species and their powers in the proposed rate equation with the reactants of the slow step.

To propose steps in a mechanism, start from the rate equation: the species it names, and their orders, must be reactants in the rate-determining step. The overall balanced equation shows the net change, so any species in the overall equation but absent from the rate equation must enter in a fast step after the slow step. For example, if the overall equation is A + B → C and the rate equation is rate = k[A], then A alone is in the slow step, and B enters in a later fast step. Build the slow step first, then add fast steps that consume remaining reactants and produce the products, ensuring the steps sum to the overall equation and conserve atoms and charge.

This row is a study pointer, not an assessed answer target: it tells you to read the temperature treatment in 5.1.1 How fast? alongside the collision material in 3.2.2 f–g. As you read, connect three ideas. First, raising the temperature increases the proportion of collisions whose energy is at least the activation energy, so the rate of reaction rises. Second, the rate constant k is the quantity that carries that change in the rate equation, so k increases with temperature. Third, the increase is steep because the fraction of molecules with enough energy grows roughly exponentially with temperature. Build a short chain of reasoning you can reproduce from memory, and check each link against the specification wording rather than memorising a phrase.

Write a qualitative explanation, not a calculation. Start from collision theory: particles react only when they collide with energy at least equal to the activation energy. Raising the temperature gives particles greater average kinetic energy, so a larger proportion of collisions has energy at least the activation energy; the frequency of successful collisions rises and the rate of reaction increases. Because the rate equation expresses rate through the rate constant k, the same change means k increases with temperature. The rise in k is steep rather than proportional, since the fraction of sufficiently energetic collisions grows roughly exponentially with temperature. Lowering the temperature reverses each step. Link every stage explicitly so the chain from temperature to rate to k is complete.

This row is a course-guidance heading in 5.1.1 How fast?, introducing the Arrhenius equation as the model linking the rate constant k to temperature T. Read it as a signpost, not as a separate assessed statement: the examinable content sits in the surrounding sub-statements, which give the exponential form k = Ae^(–Ea/RT), define A and Ea, and require use of the equation and its logarithmic form. When you study this section, learn to identify k, A, Ea, R and T, keep T in kelvin, and use R = 8.314 J K⁻¹ mol⁻¹ so that Ea comes out in J mol⁻¹. Practise converting Ea to kJ mol⁻¹ when a question asks for it, and check that a graph of ln k against 1/T has gradient –Ea/R. Treat the heading as a checklist prompt: can you state the equation, rearrange it, and explain what each term means physically?

The Arrhenius equation k = Ae^(–Ea/RT) shows that k rises exponentially as T increases, because T appears in the exponent –Ea/RT. Here A is the pre-exponential factor, Ea the activation energy in J mol⁻¹, R the gas constant 8.314 J K⁻¹ mol⁻¹ and T the temperature in kelvin. A small rise in T makes the exponent less negative, so the exponential term grows and k increases; a larger Ea makes the exponent more negative and k smaller at a given T. Taking natural logs gives ln k = ln A – Ea/(RT), so a graph of ln k against 1/T is a straight line of gradient –Ea/R and intercept ln A. In a multiple-choice item, check the exponent sign, the kelvin conversion and the units before choosing an option.

The Arrhenius equation k = A e^(–Ea/RT) links the rate constant k to temperature T in kelvin, the activation energy Ea and the pre-exponential factor A. Taking natural logs gives ln k = –Ea/RT + ln A, which has the form y = mx + c with y = ln k, x = 1/T, gradient m = –Ea/R and intercept c = ln A. To find Ea and A graphically, measure k at several temperatures, convert each T to kelvin, plot ln k against 1/T, draw a line of best fit, then calculate Ea = –gradient × R and A = e^(intercept). A steeper negative gradient means a larger Ea; a higher intercept means a larger A.