Rates of reaction

The rate of reaction is how quickly reactants are converted into products. Formally it is the change in concentration of a reactant or product per unit time:

$$\text{rate} = \frac{\Delta[\text{product}]}{\Delta t} = -\frac{\Delta[\text{reactant}]}{\Delta t}$$

The minus sign keeps the value positive because reactant concentration falls as time passes. Typical units are $\text{mol dm}^{-3}\,\text{s}^{-1}$.

On a concentration-time graph the rate at any moment is the gradient of the tangent. The reaction is fastest at the start (steepest slope) and slows as reactants run out, levelling off to zero when the reaction finishes.

We rarely measure concentration directly. Instead we track any property that changes with it:

• Gas volume or pressure (collect gas in a syringe).
• Mass loss (gas escaping from an open flask on a balance).
• Colour / light absorbed (colorimetry).
• Electrical conductivity if ion numbers change.
• Time for a fixed change - e.g. the "disappearing cross" in the thiosulfate-acid reaction, where rate $\propto 1/t$.

Collision theory explains why reactions happen and what controls their speed. For a reaction to occur, particles must:

1. Collide with one another, and

2. Collide with energy equal to or greater than the activation energy $E_a$, and

3. Collide with the correct orientation (an effective collision).

Most collisions do not lead to reaction - either the particles are too slow (too little energy) or they hit at the wrong angle. Only the small fraction that satisfy all three conditions are successful collisions.

The rate of reaction therefore depends on the frequency of successful collisions per second. Anything that increases either:

• the number of collisions per second, or
• the proportion that have enough energy,

will increase the rate.

This single idea is the key to explaining every rate factor. For example, more concentrated solutions pack particles closer together, so they collide more often; hotter particles move faster and carry more energy, so a larger share of collisions clear the energy barrier.

Each factor that changes rate does so through collision theory.

• Concentration (solutions): higher concentration means more particles per unit volume, so more frequent collisions and a faster rate.
• Pressure (gases): increasing pressure squeezes gas particles closer together (same effect as concentration), increasing collision frequency.
• Surface area (solids): breaking a solid into smaller pieces or a powder exposes more particles to collision, so a powdered solid reacts faster than a lump.
• Temperature: raising temperature has a large effect. Particles move faster (slightly more frequent collisions) but, far more importantly, a much greater fraction of particles now have energy $\geq E_a$, so many more collisions succeed.
• Catalyst: provides a lower-$E_a$ pathway, raising the fraction of successful collisions.

A useful rule of thumb often quoted is that rate roughly doubles for every 10 degree C rise - this is mainly due to the energy effect, not the small change in collision frequency.

When explaining any factor in an exam, always link it back to collision frequency and/or the fraction of collisions with $E \geq E_a$.

The Maxwell-Boltzmann distribution shows how kinetic energies are spread among the particles in a sample at a fixed temperature.

Key features of the curve (energy on the x-axis, number of particles on the y-axis):

• It starts at the origin (the number of particles with exactly zero energy is zero) and the high-energy tail never touches the x-axis (a few particles always have very high energy).
• It has a peak at the most probable energy, then a long tail.
• The total area under the curve = total number of particles.
• The area beyond $E_a$ represents the fraction of particles able to react.

Raising the temperature keeps the curve starting at the origin, but the peak moves to higher energy and becomes lower and broader (the total area is unchanged). Crucially, the area to the right of $E_a$ increases significantly, so many more particles can react - this is why temperature has such a strong effect on rate.

Adding a catalyst does not change the distribution, but it lowers $E_a$, moving the $E_a$ line to the left, so a larger area (more particles) lies beyond it. Always draw the $E_a$ line well into the tail.

A catalyst speeds up a reaction by providing an alternative reaction pathway with a lower activation energy. More colliding particles now have energy $\geq E_a$, so a greater fraction of collisions succeed and the rate rises.

Important facts:

• A catalyst is not consumed - it is chemically unchanged at the end (it may be involved in intermediate steps but is regenerated).
• It does not change the enthalpy change $\Delta H$ of the reaction; reactants and products keep the same energies.
• It lowers $E_a$ for both the forward and reverse reactions equally, so it does not shift the position of equilibrium - it only helps equilibrium arrive faster.
• Only a small amount is needed.

On an energy profile, the catalysed pathway has a lower hump (lower peak) than the uncatalysed one, but the same start and end levels.

Types include heterogeneous catalysts (different phase, e.g. solid iron in the Haber process) which work by adsorbing reactants onto a surface, and homogeneous catalysts (same phase as the reactants). Enzymes are highly efficient biological catalysts.

Most reactions do not happen in a single collision. A reaction mechanism breaks the overall change into a sequence of simple elementary steps, each involving only one or two (rarely three) particles.

The steps must add up to the overall balanced equation, and any species made in one step and used up in a later step is a reaction intermediate (it does not appear in the overall equation).

The rate-determining step (RDS) is the slowest step - it acts as a bottleneck that controls the overall rate, just as the slowest worker sets the pace of a production line.

The crucial consequence: the experimentally determined rate equation reflects the species involved up to and including the RDS. Species that react only after the RDS do not appear in the rate equation.

• If a reactant appears in the rate equation, it is involved in (or before) the RDS.
• If a reactant in the overall equation is absent from the rate equation, it must enter after the RDS.

This is why a proposed mechanism can be tested: a valid mechanism must be consistent with the observed rate equation.

Molecularity is the number of particles (atoms, ions or molecules) that take part in an elementary step.

• Unimolecular: one particle (e.g. a molecule breaking apart). Molecularity = 1.
• Bimolecular: two particles collide. Molecularity = 2 (the most common type).
• Termolecular: three particles collide simultaneously - molecularity = 3, but this is rare because a successful three-body collision is very unlikely.

For an elementary step, the rate is proportional to the concentration of each reacting particle, so the orders do match the molecularity of that step. For example, an elementary step $\text{A} + \text{B} \rightarrow \text{products}$ has rate $= k[\text{A}][\text{B}]$.

Be careful: molecularity applies only to a single elementary step, not to the overall reaction. The overall order is determined by experiment and reflects the rate-determining step. Do not confuse:

• Molecularity - a theoretical count for one elementary step.
• Order - an experimental quantity for the overall reaction.

They coincide only for the elementary steps themselves.

The rate equation (rate law) links rate to reactant concentrations:

$$\text{rate} = k[\text{A}]^m[\text{B}]^n$$

The powers $m$ and $n$ are the orders with respect to A and B; the overall order is $m + n$. These orders are found by experiment and generally cannot be read from the balanced equation.

Order from how rate responds to concentration:

• Zero order ($[\text{A}]^0$): changing $[\text{A}]$ has no effect on rate.
• First order ($[\text{A}]^1$): doubling $[\text{A}]$ doubles the rate.
• Second order ($[\text{A}]^2$): doubling $[\text{A}]$ makes the rate 4 times larger ($2^2$).

Use initial-rates data: compare experiments where only one reactant changes, and see how the rate responds.

Graphical recognition (concentration-time, $[\text{A}]$ vs $t$):

• Zero order: a straight line with constant (negative) gradient.
• First order: a curve with a constant half-life (concentration takes the same time to halve repeatedly).
• Second order: a steeper-then-shallower curve whose half-life increases as concentration falls.

A graph of rate vs concentration is horizontal (zero order), a straight line through the origin (first order), or an upward curve (second order).

The rate constant $k$ is the proportionality constant in the rate equation $\text{rate} = k[\text{A}]^m[\text{B}]^n$. It is constant for a given reaction at a given temperature.

Properties of $k$:

• A larger $k$ means a faster reaction (at given concentrations).
• $k$ does not depend on concentration; it does depend strongly on temperature (and on the presence of a catalyst).
• Raising the temperature increases $k$, which is why rate increases.

Units of $k$ depend on the overall order, because rate always has units $\text{mol dm}^{-3}\,\text{s}^{-1}$:

• Zero order: $\text{mol dm}^{-3}\,\text{s}^{-1}$.
• First order: $\text{s}^{-1}$.
• Second order: $\text{mol}^{-1}\,\text{dm}^{3}\,\text{s}^{-1}$.

Work units out by rearranging: $k = \dfrac{\text{rate}}{[\,]^{\text{overall order}}}$, then cancel.

To find $k$ from data, substitute one set of rate and concentration values into the rate equation and solve, including the correct units.

The Arrhenius equation quantifies how the rate constant depends on temperature. The data booklet gives both forms:

$$k = A e^{-E_a / RT}$$

$$\ln k = \frac{-E_a}{RT} + \ln A$$

Here $A$ is the frequency (pre-exponential) factor (related to collision frequency and orientation), $E_a$ is the activation energy, $R = 8.31\ \text{J K}^{-1}\text{mol}^{-1}$, and $T$ is in kelvin.

The exponential term $e^{-E_a/RT}$ is the fraction of collisions with enough energy to react. Raising $T$ makes this fraction larger, so $k$ - and hence the rate - increases.

The Arrhenius plot: the linear form has the shape $y = mx + c$ with

• $y = \ln k$, $x = \dfrac{1}{T}$,
• gradient $= -\dfrac{E_a}{R}$ (always negative),
• intercept $= \ln A$.

So $E_a = -R \times \text{gradient}$, giving $E_a$ in $\text{J mol}^{-1}$ (divide by 1000 for $\text{kJ mol}^{-1}$). A steeper (more negative) line means a larger $E_a$. Always work in kelvin and keep $E_a$ in joules when using $R = 8.31$.