Rate equations

In a nutshell

A rate equation is the experimentally found link between the rate of a reaction and the concentrations of its reactants. It tells you how much each reactant actually controls the rate, and, because you cannot read it off the balanced equation, it gives a window into the reaction mechanism.

This subtopic is about writing and using that equation: finding the orders and the rate constant from data, working out how the rate constant changes with temperature (the Arrhenius equation), and using the orders to point at the rate-determining step.

Assumed knowledge: Kinetics for collision theory, activation energy and the Maxwell-Boltzmann distribution.

Core content

The rate equation

For a reaction between A and B, the rate equation has the form:

$rate = k[A]^m[B]^n$

• $[A]$ and $[B]$ are the concentrations of the reactants (mol dm^-3).
• $k$ is the rate constant.
• $m$ and $n$ are the orders of reaction with respect to A and B. AQA restricts them to 0, 1 or 2.
• The overall order is the sum of the individual orders, $m + n$.

The single most important fact about a rate equation is that it is determined experimentally. The orders are not the balancing numbers in the chemical equation, and you cannot predict them by looking at the equation.

The order tells you how the rate responds when you change a concentration:

Plotting rate against the concentration of one reactant shows each order as a distinct shape:

The order of reaction with respect to a reactant is the power to which that reactant's concentration is raised in the rate equation. The rate constant $k$ is the proportionality constant in the rate equation; it is fixed at a given temperature but changes when the temperature changes.

Still don't get it? · what "order" really means

Forget chemistry for a moment. Imagine a recipe where the baking time depends on how much dough you use. You do not get to guess the rule by reading the ingredients list, you have to run the oven and watch.

Run one: you double the dough and the time doubles. So time is "first order" in dough, it tracks the amount one-for-one. Run two: you double the dough and the time goes up four times. Now it is "second order", doubling the input quadruples the output ($2^2 = 4$). Run three: you double the dough and nothing changes. That ingredient is "zero order", it has no say in the timing at all.

Back to the exam: the order is exactly this, the power you have to raise a reactant's concentration to so that the rate equation matches the measured rates. You find it by experiment, one reactant at a time, by seeing what happens to the rate when you change that concentration. Crucially, it is not the big number in front of the reactant in the balanced equation. A species can even be zero order, which means it appears in the overall equation but does not appear in the rate equation at all.

The units of the rate constant

The units of $k$ are not fixed, they depend on the overall order. Work them out every time by rearranging the rate equation to make $k$ the subject and cancelling the concentration units.

Rate always carries the units mol dm^-3 s^-1, and each concentration carries mol dm^-3, so for a second-order overall reaction:

$\text{units of } k = \frac{\text{mol dm}^{-3}\,\text{s}^{-1}}{(\text{mol dm}^{-3})^2} = \text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$

The units are worth their own mark, and they are one of the commonest marks lost, so always quote them.

The rate constant and temperature: the Arrhenius equation

Raising the temperature increases the rate constant $k$ (and so increases the rate). The relationship is the Arrhenius equation:

$k = Ae^{-E_a/RT}$

• $A$ is the Arrhenius constant (same units as $k$).
• $E_a$ is the activation energy (must be in J mol^-1 to match $R$).
• $R$ is the gas constant, 8.31 J K^-1 mol^-1.
• $T$ is the temperature in kelvin (K).

Why $k$ rises with temperature (the explanation that earns the mark): at a higher temperature a greater proportion of molecules have energy greater than or equal to the activation energy $E_a$, so a greater proportion of collisions are successful. Saying only "there are more successful collisions" is not enough, the mark needs the idea of energy relative to $E_a$. This links straight back to the Maxwell-Boltzmann distribution.

The equation is usually rearranged by taking natural logs into the equation of a straight line:

$\ln k = -\frac{E_a}{R}\cdot\frac{1}{T} + \ln A$

Comparing this with $y = mx + c$: a plot of $\ln k$ (y) against $\dfrac{1}{T}$ (x) is a straight line with gradient $-\dfrac{E_a}{R}$ and y-intercept $\ln A$. That is how activation energy is found from experimental data (worked below).

Still don't get it? · turning Arrhenius into a straight line

A curved graph is hard to read a number off. The trick chemists use everywhere is to bend the maths until the graph goes straight, because a straight line hands you two easy numbers: its slope and where it crosses the axis.

Start with $k = Ae^{-E_a/RT}$, a curve. Take the natural log of both sides. Logs turn a product into a sum and pull the power down in front, so the right-hand side becomes $\ln A - \dfrac{E_a}{RT}$.

Now tidy it to look like $y = mx + c$. Put $\ln k$ as your "y". Put $\dfrac{1}{T}$ as your "x". Then the slope "m" is $-\dfrac{E_a}{R}$ and the intercept "c" is $\ln A$.

So when the exam says "plot $\ln k$ against $1/T$", it is asking you to straighten the Arrhenius curve. You measure the gradient of that line, and because the gradient equals $-E_a/R$, the activation energy is $E_a = -\text{gradient} \times R$. The gradient is negative, so the minus sign makes $E_a$ come out positive.

Deducing orders from initial-rate data

The standard method: run several experiments, changing the concentration of one reactant at a time and measuring the initial rate each time. Compare two experiments in which only one concentration changes:

• concentration doubled, rate unchanged → zero order
• concentration doubled, rate doubled → first order
• concentration doubled, rate quadrupled (x4) → second order

In general, if a concentration is multiplied by a factor $x$, the rate is multiplied by $x^{(\text{order})}$. Once you have each order, write the rate equation and calculate $k$ by substituting one experiment's data into the rearranged equation.

Watch the second-order case carefully: if tripling a concentration (x3) multiplies the rate by nine, the order is 2, because $3^2 = 9$, not 3.

Concentration-time graphs

When you follow a single reaction over time (a continuous monitoring method), you get a concentration-time graph. The rate at any instant is the gradient of the graph at that point, found by drawing a tangent to the curve.

• The reaction is fastest at the start, where the curve is steepest.
• The initial rate is the gradient of the tangent drawn at time = 0. Initial rate is preferred because you know the concentrations exactly at the start.

A zero-order reactant is a special case: its rate does not depend on its concentration, so as it is used up the rate stays constant. Its concentration-time graph is a straight line with a constant gradient, and because $rate = k[A]^0 = k$, the magnitude of that gradient is the rate constant $k$.

Orders and the reaction mechanism

A mechanism is the sequence of steps a reaction actually goes through. The steps have different speeds, and the slowest step controls the overall rate. This slowest step is the rate-determining step (RDS).

The orders in the rate equation match how many molecules of each species appear in (or before) the rate-determining step:

• a species that appears in the slow step contributes to the rate equation (its order matches the number of its molecules in that step)
• a species that appears only after the slow step is zero order, it does not appear in the rate equation

Worked reasoning, for the overall reaction NO₂(g) + CO(g) → NO(g) + CO₂(g), where experiment gives $rate = k[NO_2]^2$:

1. Step 1 (slow): NO₂ + NO₂ → NO + NO₃
2. Step 2 (fast): NO₃ + CO → NO₂ + CO₂

The slow step contains two NO₂ molecules and no CO, which is exactly why the reaction is second order in NO₂ and zero order in CO. The orders match the slow step, not the overall equation.

This also explains catalysis evidence: if a species appears in the rate equation but not in the overall balanced equation (for example H⁺), it must be taking part in or before the rate-determining step, which is evidence that it acts as a catalyst.

Required practical 7: measuring the rate of a reaction

The spec attaches Required practical 7 here: measuring a rate by an initial rate method and by a continuous monitoring method.

Continuous monitoring method follows one reaction's progress over time to build a concentration-time (or amount-time) graph.

• A common technique is collecting a gas in a gas syringe and recording its volume at regular time intervals, for a reaction such as Mg + 2HCl → MgCl₂ + H₂.
• A colorimeter can monitor a reaction that changes colour, taking readings at timed intervals.
• The rate at any time is the gradient of a tangent; the initial rate is the tangent at $t = 0$.

Initial rate method measures just the rate at the very start, often using a clock reaction: you time how long ($t$) it takes to reach a fixed, visible change (such as a colour appearing), and take the initial rate as proportional to $1/t$.

• The classic example is the iodine clock: iodine formed from the reaction of hydrogen peroxide with iodide is mopped up by a small, fixed amount of thiosulfate; once the thiosulfate runs out, free iodine turns starch blue-black.
• Repeat with the concentration of one reactant varied and the others held constant, and $1/t$ gives the order with respect to that reactant.
• A reactant held in large excess stays at an effectively constant concentration, so it appears not to affect the rate (a pseudo-order technique).

Worked examples

Example 1: deduce the rate equation and calculate the rate constant.

Order in A (experiments 1 and 2, [B] constant): [A] doubles, rate doubles, so first order in A.

Order in B (experiments 2 and 3, [A] constant): [B] doubles, rate goes x4, so second order in B.

$rate = k[A][B]^2 \quad (\text{overall order } 3)$

Rearrange and substitute experiment 1:

$k = \frac{rate}{[A][B]^2} = \frac{2.0\times10^{-3}}{(0.10)(0.10)^2}$

$k = \frac{2.0\times10^{-3}}{1.0\times10^{-3}} = 2.0$

Units: $\dfrac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}$

$k = 2.0\ \text{mol}^{-2}\,\text{dm}^{6}\,\text{s}^{-1}$

The working line (substitution) earns one mark; the answer with its units earns the next. Using any other experiment gives the same $k$, which is a good check.

Example 2: use the rate equation to find a new rate.

Using $rate = k[A][B]^2$ from Example 1, the initial rate was $2.0\times10^{-3}$ mol dm^-3 s^-1. Both [A] and [B] are now doubled. Find the new initial rate.

Multiply the rate by the factor for each reactant: A is first order so x2¹, B is second order so x2².

$\text{factor} = 2^1 \times 2^2 = 8$

$rate_{new} = 8 \times (2.0\times10^{-3}) = 1.6\times10^{-2}\ \text{mol dm}^{-3}\,\text{s}^{-1}$

You do not need to recalculate from scratch. The quick route is to combine the factors; students lose this mark by halving or doubling once instead of applying every order.

Example 3: find $k$ for a zero-order reactant from a concentration-time graph.

A reactant is zero order. Its concentration falls linearly from 1.00 to 0.40 mol dm^-3 over 300 s.

For a zero-order reactant, $rate = k[A]^0 = k$, so $k$ equals the magnitude of the gradient:

$\text{gradient} = \frac{0.40 - 1.00}{300 - 0} = \frac{-0.60}{300} = -2.0\times10^{-3}$

$k = 2.0\times10^{-3}\ \text{mol dm}^{-3}\,\text{s}^{-1}$

The units are mol dm^-3 s^-1 (the units of $k$ for a zero-order overall reaction are the units of rate).

Example 4: Arrhenius calculation (find the activation energy).

A reaction has $k = 3.3\times10^{-3}$ mol^-1 dm³ s^-1 at 25 °C. Given $\ln A = 17.1$ and $R = 8.31$ J K^-1 mol^-1, find $E_a$ in kJ mol^-1.

Convert the temperature: $T = 25 + 273 = 298$ K.

Start from the log form and rearrange for $E_a$:

$\ln k = \ln A - \frac{E_a}{RT} \implies E_a = (\ln A - \ln k)\,RT$

$\ln k = \ln(3.3\times10^{-3}) = -5.71$

$E_a = (17.1 - (-5.71)) \times 8.31 \times 298$

$E_a = 22.81 \times 8.31 \times 298 = 56\,500\ \text{J mol}^{-1}$

$E_a = +56.5\ \text{kJ mol}^{-1}$

Two traps live here: read $\ln A = 17.1$ as $\ln A$, not as $A$; and divide the joule answer by 1000 for kJ mol^-1.

Example 5: find the activation energy graphically.

The rate constant is measured at several temperatures and $\ln k$ is plotted against $\dfrac{1}{T}$:

The gradient of the line equals $-\dfrac{E_a}{R}$. Taking two points far apart on the line (the first and last):

$\text{gradient} = \frac{-3.233 - (-5.748)}{(3.030 - 3.448)\times10^{-3}} = \frac{2.515}{-4.18\times10^{-4}} = -6.02\times10^{3}\ \text{K}$

$E_a = -\text{gradient} \times R = 6.02\times10^{3} \times 8.31 = 50\,000\ \text{J mol}^{-1}$

$E_a = +50.0\ \text{kJ mol}^{-1}$

Use points on the best-fit line and far apart, not raw data points close together, or the gradient (and $E_a$) will be inaccurate.

Common exam mistakes

• Reading the orders off the balanced equation. Orders are found by experiment only; a balancing number is not an order.
• Seeing the rate go up x9 when a concentration is tripled and calling the order 3. It is second order, because $3^2 = 9$.
• Leaving the units off the rate constant, or quoting the wrong units. The units depend on the overall order and carry their own mark, so derive them every time.
• Forgetting to square (or cube) a concentration when substituting into the rate equation to find $k$.
• Dividing by the concentration of a reactant that is zero order when finding a rate or constant. A zero-order reactant does not appear in the rate equation at all.
• For the "new rate" calculation, applying only one concentration change (halving or doubling once) instead of combining the factor for every reactant.
• Explaining the temperature effect on $k$ with "more successful collisions" alone. The mark needs a greater proportion of molecules with energy greater than or equal to $E_a$.
• In Arrhenius calculations: using °C instead of K, reading a given $\ln A$ value as $A$, or forgetting to convert J mol^-1 to kJ mol^-1.
• Drawing the $\ln k$ against $1/T$ graph too small so it fills less than half the grid, which makes the gradient (and $E_a$) inaccurate; and assuming a "best-fit line" must be straight on a concentration-time graph, where it may be a curve.
• For a rate-determining-step question, describing the rate instead of the mechanism. The mark needs the species' role in (or before) the slow step, not just "it is in the rate equation".
• Forgetting to draw a tangent to find the rate from a curved concentration-time graph, or measuring the gradient at the wrong point.

Key definitions

• Order of reaction (with respect to a reactant): the power to which the concentration of that reactant is raised in the rate equation.
• Overall order of reaction: the sum of the individual orders of all the reactants in the rate equation.
• Rate constant ($k$): the proportionality constant in the rate equation; constant for a given reaction at a given temperature.
• Rate-determining step: the slowest step in the reaction mechanism, which controls the overall rate of reaction.
• Activation energy ($E_a$): the minimum energy that colliding particles must have for a reaction to occur.
• Arrhenius constant ($A$): the constant $A$ in the Arrhenius equation $k = Ae^{-E_a/RT}$, with the same units as the rate constant.

Specification

• I can define order of reaction and rate constant.
• I can write a rate equation of the form $rate = k[A]^m[B]^n$ and state that the orders are restricted to 0, 1 or 2.
• I can perform calculations using the rate equation, including working out the units of $k$.
• I can use initial-rate data to deduce the order (0, 1 or 2) with respect to each reactant and derive the rate equation.
• I can use a concentration-time graph to deduce a rate (gradient of a tangent) and an initial rate (tangent at $t = 0$).
• I can calculate the rate constant of a zero-order reaction from the gradient of a concentration-time graph.
• I can explain the qualitative effect of temperature on the rate constant $k$.
• I can perform calculations using $k = Ae^{-E_a/RT}$.
• I can rearrange it to $\ln k = -\dfrac{E_a}{R}\cdot\dfrac{1}{T} + \ln A$ and use a plot of $\ln k$ against $1/T$ (gradient $-E_a/R$) to find $E_a$.
• I can use the orders with respect to reactants to give information about the rate-determining step.
• I can measure a rate by an initial rate method and by a continuous monitoring method (Required practical 7).

Related notes

• Kinetics
• Energetics
• Equilibrium constant Kp for homogeneous systems