Factors That Affect the Rate of Reactions

Lessons

Senior 6 · Chemistry

Factors That Affect the Rate of Reactions

3 lessons

Lesson 2 of 3

Factors That Change Reaction Rates

25 min

Racing the bacteria

Fresh milk leaves the cow at about body temperature, and in that warm milk bacteria multiply and their enzymes turn lactose into lactic acid within hours. A milk collection centre therefore pours the milk into a refrigerated stainless-steel tank and chills it within a few hours. Nothing is added and nothing is removed: lowering the temperature alone slows every one of those reactions.

In Senior 3 you met the factors that change a rate. In this lesson you explain them quantitatively with collision theory and the Maxwell–Boltzmann distribution, which is the level the national examination expects at A-level.

Definition

Collision theory: particles react only when they collide (1) with at least a minimum energy and (2) in a suitable orientation. Collisions that meet both conditions are effective (successful) collisions.

Activation energy, $E_\text{a}$ : the minimum energy that colliding particles must have for a reaction to occur; unit $\mathrm{kJ\,mol^{-1}}$ . Example: the decomposition of hydrogen iodide, $\mathrm{2HI \rightarrow H_2 + I_2}$ , has $E_\text{a} \approx 184\,\mathrm{kJ\,mol^{-1}}$ , so it is very slow at room temperature.

Catalyst: a substance that increases the rate of a reaction by providing an alternative pathway of lower activation energy, and is chemically unchanged at the end. Example: iron in the Haber process.

Rate as a product of three factors

Collision theory gives the rate as

\text{rate} = Z \times P \times f

where $Z$ is the collision frequency (collisions per unit volume per second), $P$ is the steric (orientation) factor, the fraction of collisions with a suitable orientation, and $f$ is the fraction of collisions with energy at least $E_\text{a}$ . For a gas, kinetic theory gives

f = e^{-\frac{E_\text{a}}{RT}}

with $R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ and $T$ in kelvin. Every factor below works by changing $Z$ , $f$ or the pathway.

Factor	Effect on rate	Explanation
Nature of the reactants	Reactions between oppositely charged ions in solution are almost instant; reactions that break strong covalent bonds are slow	$E_\text{a}$ depends on which bonds must break. $\mathrm{Ag^{+} + Cl^{-}}$ needs no bond breaking; $\mathrm{H_2 + I_2}$ must break $\mathrm{H{-}H}$ and $\mathrm{I{-}I}$
Concentration (solutions)	Usually increases rate	More particles per unit volume, so $Z$ rises; the exact dependence is the rate law (Unit 13)
Pressure (gases)	Usually increases rate	$c = \frac{p}{RT}$ : raising the pressure raises the concentration of each gas, so $Z$ rises
Surface area (solids)	Increases rate	Only surface particles can collide; a powder exposes far more surface than a lump
Temperature	Increases rate, often sharply	$f$ rises steeply; $Z$ rises only slightly (worked example 2)
Catalyst	Increases rate	A new pathway with lower $E_\text{a}$ , so $f$ rises at the same temperature
Light	Increases the rate of photochemical reactions	Photons break bonds to form reactive radicals: $\mathrm{H_2 + Cl_2}$ explodes in sunlight but hardly reacts in the dark

Worked example 1: how much does crushing increase the surface?

Problem. A cube of calcium carbonate of side 1.0 cm is ground into cubes of side 1.0 mm. Calculate the total surface area before and after, and state the factor by which the initial rate of reaction with excess dilute hydrochloric acid should increase.

Solution. Before: $6 \times (1.0\,\mathrm{cm})^2 = 6.0\,\mathrm{cm^2}$ .

Number of small cubes: $\left(\frac{10\,\mathrm{mm}}{1.0\,\mathrm{mm}}\right)^3 = 1000$ . Area of each: $6 \times (0.10\,\mathrm{cm})^2 = 0.060\,\mathrm{cm^2}$ . Total: $1000 \times 0.060 = 60\,\mathrm{cm^2}$ .

The surface area increases ten times, so the initial rate should increase by about a factor of 10, because the rate of a solid–liquid reaction is proportional to the area in contact with the acid (other conditions fixed). Dividing the side by $n$ multiplies the area by $n$ .

Temperature and the Maxwell–Boltzmann distribution

The molecules in a gas (or a solution) do not all have the same energy. Figure 2.1 shows the Maxwell–Boltzmann distribution of kinetic energies at two temperatures. Note three features:

The curve starts at the origin (no molecule has zero energy) and never meets the energy axis at high energy.
The area under the curve is the total number of molecules, so it is the same at both temperatures.
At the higher temperature $T_{2}$ the peak is lower and moves to higher energy, and the tail beyond $E_\text{a}$ is much larger.

Figure 2.1: The Maxwell–Boltzmann distribution at two temperatures. The shaded areas are the molecules with enough energy to react.

How a catalyst works

A catalyst does not change the distribution of energies; it moves the activation energy to the left. The reaction goes by a different route, usually through an intermediate formed with the catalyst, with a lower energy barrier (Figure 2.2). The enthalpy change $\Delta H$ is unchanged because the reactants and products are the same, and the catalyst speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium; it only brings the system to equilibrium sooner.

Figure 2.2: Energy profile of an exothermic reaction with and without a catalyst.

Type	Meaning	Examples
Homogeneous	Catalyst in the same phase as the reactants	$\mathrm{H^{+}(aq)}$ in esterification; $\mathrm{Fe^{2+}(aq)}$ in $\mathrm{S_2O_8^{2-} + 2I^{-} \rightarrow 2SO_4^{2-} + I_2}$
Heterogeneous	Catalyst in a different phase; reaction on its surface (adsorption, reaction, desorption)	Fe in the Haber process; $\mathrm{V_2O_5}$ in the Contact process; Ni in hydrogenating vegetable oils; Pt, Pd and Rh in car catalytic converters
Biological (enzymes)	Protein catalysts, very specific, with an optimum temperature and pH	Amylase (starch to maltose); catalase ( $\mathrm{2H_2O_2 \rightarrow 2H_2O + O_2}$ )
Autocatalysis	A product of the reaction catalyses it	$\mathrm{Mn^{2+}}$ in the reaction of manganate(VII) with ethanedioate ions

The surfaces of heterogeneous catalysts can be poisoned: a substance such as lead or sulfur adsorbs strongly and blocks the active sites. Leaded petrol destroys a catalytic converter for this reason.

Worked example 2: why a small temperature rise matters so much

Problem. A reaction has $E_\text{a} = 50\,\mathrm{kJ\,mol^{-1}}$ . Calculate the fraction of collisions with enough energy at 298 K and at 308 K. Compare the increase with the increase in collision frequency, which is proportional to $\sqrt{T}$ .

Solution.

f_{298} = e^{-\frac{50\,000}{8.314 \times 298}} = 1.72 \times 10^{-9}, \qquad f_{308} = e^{-\frac{50\,000}{8.314 \times 308}} = 3.31 \times 10^{-9}

Ratio: $\frac{3.31 \times 10^{-9}}{1.72 \times 10^{-9}} = 1.93$ . The fraction of effective collisions almost doubles.

Collision frequency: $\sqrt{\frac{308}{298}} = 1.017$ , an increase of only 1.7 %.

Conclusion. A 10 K rise roughly doubles the rate of a reaction with $E_\text{a}$ near $50\,\mathrm{kJ\,mol^{-1}}$ , and almost all of the increase comes from more molecules having $E \ge E_\text{a}$ , not from more frequent collisions.

Worked example 3: autocatalysis in a titration

Problem. When acidified potassium manganate(VII) is run into warm ethanedioic acid, the first few drops take many seconds to lose their purple colour; later drops are decolourised almost at once. The equation is $\mathrm{2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O}$ . Explain the observation and sketch the shape of a $[\mathrm{MnO_4^{-}}]$ –time graph for one portion of manganate(VII) added at the start.

Solution. The product $\mathrm{Mn^{2+}}$ catalyses the reaction. Two negative ions ( $\mathrm{MnO_4^{-}}$ and $\mathrm{C_2O_4^{2-}}$ ) repel, so the direct reaction has a high activation energy. Once a little $\mathrm{Mn^{2+}}$ has formed, it provides a lower-energy route (it is oxidised to $\mathrm{Mn^{3+}}$ , which then oxidises ethanedioate). The rate therefore increases as the reaction proceeds, until the reactants run low. The graph is S-shaped: a slow start, a steep fall in the middle, then levelling off. Check of the equation: charges on the left $- 2 - 10 + 16 = + 4$ ; on the right $2 \times (+2) = +4$ .

Warning

Common mistakes

Saying a catalyst "gives the particles more energy". It lowers the activation energy; the energy distribution is unchanged.
Drawing the higher-temperature Maxwell–Boltzmann curve with a higher peak, or letting the curves enclose different areas. The number of molecules does not change.
Saying a catalyst increases the yield. It changes the rate only; the position of equilibrium is unchanged.
Explaining a temperature rise only by "more collisions". The main reason is that a larger fraction of collisions have $E \ge E_\text{a}$ .

Safety in the manganate(VII) experiment: wear eye protection; ethanedioic acid is harmful if swallowed and manganate(VII) is an oxidising agent that stains skin. Warm the acid in a water bath to about 60 °C, never over a flame to boiling.

Tip

For any "explain" question on rate, write two sentences: (1) what happens to the frequency of collisions, and (2) what happens to the fraction of collisions with $E \ge E_\text{a}$ . Concentration, pressure and surface area change (1); temperature changes mainly (2); a catalyst changes (2) by moving $E_\text{a}$ .

Siebert Science: Effect of Temperature and Catalysts on Rate and Maxwell-Boltzmann Distributions. Watch how the area to the right of the activation-energy line grows when the curve flattens (higher temperature) and when the line moves left (catalyst).

✦

Practice

State the two conditions a collision must meet to be effective.
Use collision theory to explain why magnesium ribbon reacts faster with $2.0\,\mathrm{mol\,dm^{-3}}$ hydrochloric acid than with $0.50\,\mathrm{mol\,dm^{-3}}$ acid.
A 1.0 cm cube of zinc is cut into cubes of side 0.50 cm. How many cubes are formed, and what is the new total surface area?
Sketch a Maxwell–Boltzmann distribution, mark $E_\text{a}$ , and show on the same sketch how a catalyst increases the number of molecules able to react.
Distinguish homogeneous from heterogeneous catalysis, with one example of each.
For $E_\text{a} = 40\,\mathrm{kJ\,mol^{-1}}$ , calculate the fraction of collisions with enough energy at 300 K and at 320 K, and the ratio of the two.
A student says: "A catalyst increases the yield of ammonia in the Haber process." Evaluate the statement.

Show answersHide answers

Energy at least equal to $E_\text{a}$ , and a suitable orientation.
The more concentrated acid has more $\mathrm{H^{+}}$ ions per unit volume, so they collide with the magnesium surface more often; the fraction of effective collisions is the same, so more effective collisions occur per second.
$2^{3} = 8$ cubes; area $8 \times 6 \times (0.50)^2 = 12\,\mathrm{cm^2}$ , double the original $6.0\,\mathrm{cm^2}$ .
One curve from the origin with a tail; the catalysed activation energy $E_\text{a}(\text{cat})$ is a line to the left of $E_\text{a}$ , and the larger area to its right represents more molecules that can react. The curve itself does not change.
Homogeneous: same phase, e.g. $\mathrm{H^{+}(aq)}$ in esterification. Heterogeneous: different phase, e.g. solid iron with gaseous $\mathrm{N_2}$ and $\mathrm{H_2}$ .
$f_{300} = e^{-\frac{40\,000}{8.314 \times 300}} = 1.08 \times 10^{-7}$ ; $f_{320} = 2.95 \times 10^{-7}$ ; ratio 2.72.
Wrong. Iron speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but its position, and so the equilibrium yield, is unchanged. The catalyst lets the plant work at a temperature low enough for a reasonable yield while still reacting fast enough.

Sources and credits

Image: CoolingTank.jpg by Alexandre Soares, via Wikimedia Commons, licensed under CC BY 3.0.
Image: Assam CTC Hanamizuki 2009.JPG by USAGI-WRP, via Wikimedia Commons, licensed under CC BY 3.0.
Video: Effect of Temperature and Catalysts on Rate and Maxwell-Boltzmann Distributions, Siebert Science, embedded from YouTube.
The text, tables, diagrams, worked examples and Practice items were written and drawn by Ganzaa.

1Racing the bacteria

2Rate as a product of three factors

3Temperature and the Maxwell–Boltzmann distribution

4How a catalyst works

Racing the bacteria

Rate as a product of three factors

Temperature and the Maxwell–Boltzmann distribution

How a catalyst works