Chemistry - Chemical Kinetics

Rate of Reaction

Definition

The rate of reaction measures how fast reactants are consumed or products are formed.

$\mathrm{Rate} = \frac{\mathrm{change in amount (mol)}}{\mathrm{change in time (s)}}$

For a reaction $a\mathrm{A} + b\mathrm{B} \to c\mathrm{C} + d\mathrm{D}$:

$\mathrm{Rate} = -\frac{1}{a}\frac{d[\mathrm{A}]}{dt} = -\frac{1}{b}\frac{d[\mathrm{B}]}{dt} = \frac{1}{c}\frac{d[\mathrm{C}]}{dt} = \frac{1}{d}\frac{d[\mathrm{D}]}{dt}$

The negative sign indicates that reactant concentrations decrease over time.

Average Rate vs Instantaneous Rate

• Average rate: total change divided by total time (gradient of secant on concentration-time graph).
• Instantaneous rate: rate at a specific moment (gradient of tangent on concentration-time graph).

Units

Rate is typically expressed in mol dm$^{-3}$ s$^{-1}$ (change in concentration per unit time).

---

Measuring Rates of Reaction

Methods

Method	Measured Quantity	Suitable Reactions
Collection of gas	Volume of gas over time	Reactions producing gas ($\mathrm{CO}_2$, $\mathrm{H}_2$)
Mass loss	Mass of flask over time	Reactions producing gas
Titration	Concentration at time intervals	Reactions where aliquots can be quenched
Colorimetry	Absorbance over time	Reactions involving colour change
Conductivity	Conductance over time	Reactions producing/ consuming ions
Clock reaction	Time for observable change	Initial rate determination

Gas Collection Methods

1. Gas syringe: Direct measurement of gas volume. Accurate but limited range.
2. Inverted burette over water: Useful for insoluble gases. Volume read from burette.
3. Mass loss method: The flask is placed on a balance; the mass decreases as gas escapes.

warning

The mass loss method only works if the gas escapes the flask. If the flask is sealed, no mass change occurs. Also, ensure the reaction does not produce heat that would affect the balance reading.

Concentration-Time Graphs

For reactants: concentration decreases over time (negative gradient).

For products: concentration increases over time (positive gradient).

The gradient at any point gives the instantaneous rate. The gradient decreases over time as reactants are consumed and rate slows down.

Initial Rate

The initial rate is the rate at $t = 0$, determined from the gradient of the tangent at the origin of a concentration-time graph. It is the fastest rate in the reaction.

---

Factors Affecting the Rate of Reaction

Concentration (or Pressure for Gases)

Increasing concentration increases the frequency of collisions between particles, increasing the rate of reaction.

For gaseous reactants, increasing pressure has the same effect as increasing concentration.

Temperature

Increasing temperature increases the kinetic energy of particles. More particles have energy $\geq E_a$ (activation energy), so more successful collisions occur per unit time. This is the dominant effect. The increased collision frequency is a secondary, minor effect.

Approximate rule: Rate roughly doubles for every $10^\circ\mathrm{C}$ rise in temperature.

Surface Area of Solid Reactants

Increasing surface area (e.g., using a powder instead of a lump) increases the number of particles exposed for collision, increasing the rate.

Catalyst

A catalyst provides an alternative reaction pathway with a lower activation energy. It increases the rate without being consumed. It does not affect the position of equilibrium or the equilibrium yield.

Factor	Effect on Rate	Effect on $E_a$	Effect on Equilibrium
Concentration	Increases	No change	Position may shift
Temperature	Increases	No change	Position may shift
Surface area	Increases	No change	No effect
Catalyst	Increases	Decreases	No effect on position

---

Collision Theory

Successful Collisions

For a reaction to occur, particles must:

1. Collide with each other (with the correct orientation).
2. Have kinetic energy $\geq E_a$ (activation energy).

Only a small fraction of collisions lead to reaction because most particles do not have sufficient energy at any given temperature.

Activation Energy ($E_a$)

The minimum energy that colliding particles must have for a successful reaction.

$\mathrm{Reactants} \xrightarrow{E_a} \mathrm{Products}$

On an energy profile diagram, $E_a$ is the energy difference between the reactants and the peak (transition state).

Effect of a Catalyst on Activation Energy

A catalyst provides an alternative pathway with lower $E_a$:

$\mathrm{Reactants} \xrightarrow{E_a \mathrm{ (uncatalysed)}} \mathrm{Products}$ $\mathrm{Reactants} \xrightarrow{E_a \mathrm{ (catalysed)}} \mathrm{Products}$

The catalysed pathway has a lower energy barrier, so more particles have sufficient energy to react.

---

Maxwell-Boltzmann Distribution

Description

At a given temperature, the kinetic energies of particles in a gas or liquid follow a Maxwell-Boltzmann distribution:

• The curve starts at the origin (no particles have zero energy).
• It rises to a peak and then tails off exponentially.
• The peak represents the most probable energy.
• The area under the curve equals the total number of particles.

Effect of Temperature

When temperature increases:

• The curve shifts to the right (higher average energy).
• The peak lowers and broadens.
• The area under the curve remains constant (same number of particles).
• A larger fraction of particles has energy $\geq E_a$.

$\frac{\mathrm{Fraction with } E \geq E_a \mathrm{ at } T_2}{\mathrm{Fraction with } E \geq E_a \mathrm{ at } T_1} \gt 1 \quad \mathrm{when } T_2 \gt T_1$

Effect of a Catalyst

The $E_a$ line shifts to the left on the Maxwell-Boltzmann distribution. More particles now have energy $\geq E_a$ (the new, lower activation energy), so the rate increases.

info

A catalyst does NOT change the distribution itself. It only lowers the threshold. The curve shape remains the same at the same temperature.

Key DSE Exam Points

• The Maxwell-Boltzmann curve never touches the x-axis.
• The total area is always constant for the same number of particles.
• The most probable energy increases with temperature but the average energy increase is small compared to the dramatic increase in the fraction exceeding $E_a$.

---

Energy Profile Diagrams

Exothermic Reaction (Uncatalysed)

Energy
  |          /\
  |         /  \  Transition state
  |        /    \
  |  E_a  /      \
  |      /        \
  |-----/----------\---- Products (lower energy)
  |    /            |
  |   / Reactants   |  ΔH < 0 (exothermic)
  |  /              |
  | /               |
  |/________________|____________

Endothermic Reaction (Uncatalysed)

Energy
  |         /\
  |        /  \  Transition state
  |       /    \
  |      /      \
  |     /        \
  |    /          \---- Products (higher energy)
  |   /            |
  |  /  E_a        |  ΔH > 0 (endothermic)
  | /              |
  |/ Reactants     |
  |_________________|____________

Catalysed vs Uncatalysed

On the same diagram, the catalysed pathway has a lower peak (lower $E_a$). The reactant and product energy levels remain unchanged. $\Delta H$ is the same for both pathways.

---

Catalytic Converters

Purpose

Catalytic converters reduce the emission of toxic gases from car exhausts.

Reactions

1. Oxidation of CO:

$2\mathrm{CO} + \mathrm{O}_2 \xrightarrow{\mathrm{Pt, Pd}} 2\mathrm{CO}_2$

2. Oxidation of unburnt hydrocarbons (e.g., octane):

$2\mathrm{C}_8\mathrm{H}_{18} + 25\mathrm{O}_2 \xrightarrow{\mathrm{Pt, Pd}} 16\mathrm{CO}_2 + 18\mathrm{H}_2\mathrm{O}$

3. Reduction of nitrogen oxides:

$2\mathrm{NO} \xrightarrow{\mathrm{Rh}} \mathrm{N}_2 + \mathrm{O}_2$

$2\mathrm{NO}_2 \xrightarrow{\mathrm{Rh}} \mathrm{N}_2 + 2\mathrm{O}_2$

Catalysts Used

• Platinum (Pt) and palladium (Pd): for oxidation reactions.
• Rhodium (Rh): for reduction of NO$_x$.

Limitations

• Lead poisoning: Lead compounds in "leaded" petrol coat and deactivate the catalyst. This is why unleaded petrol is required for cars with catalytic converters.
• Temperature dependence: The converter only works efficiently at high temperatures (above about $300^\circ\mathrm{C}$). It is ineffective during cold starts.
• Sulfur poisoning: Sulfur compounds in fuel can also poison the catalyst.

---

Enzyme Catalysis

Enzymes as Biological Catalysts

Enzymes are protein molecules that catalyse specific biochemical reactions.

Properties

1. Specificity: Each enzyme catalyses only one reaction (or a small group of reactions).
2. Efficiency: Enzymes can increase reaction rates by factors of $10^6$ to $10^{12}$.
3. Optimum temperature: Most enzymes work best around $37^\circ\mathrm{C}$ (body temperature). Above about $45^\circ\mathrm{C}$, the enzyme denatures (loses its 3D structure) and activity drops sharply.
4. Optimum pH: Each enzyme has an optimal pH. Pepsin (stomach) works at pH ~2; trypsin (intestine) works at pH ~8.
5. Concentration dependence: Rate increases with enzyme concentration (at fixed substrate concentration) and with substrate concentration (at fixed enzyme concentration), until a maximum is reached.

Lock and Key Model

The substrate fits into the active site of the enzyme like a key in a lock. The enzyme-substrate complex forms, the reaction occurs, and products are released. The enzyme is unchanged and available for further catalysis.

Effect of Temperature on Enzyme Activity

• Below optimum: rate increases with temperature (standard kinetic effect).
• At optimum (~$37^\circ\mathrm{C}$): maximum rate.
• Above optimum: enzyme denatures, rate drops rapidly.
• At very high temperatures: complete loss of activity.

Effect of pH on Enzyme Activity

• Changes in pH alter the charges on amino acid residues at the active site.
• This changes the shape of the active site, reducing substrate binding.
• Extreme pH values cause permanent denaturation.

---

Rate Equations

Definition

For a reaction $a\mathrm{A} + b\mathrm{B} \to \mathrm{products}$, the rate equation is:

$\mathrm{Rate} = k[\mathrm{A}]^m[\mathrm{B}]^n$

Where:

• $k$ = rate constant (depends on temperature and catalyst)
• $m$ = order of reaction with respect to A
• $n$ = order of reaction with respect to B
• Overall order = $m + n$

warning

warning the stoichiometric coefficients $a$ and $b$.

Order of Reaction

Zero order: Rate is independent of concentration.

$\mathrm{Rate} = k$

Concentration decreases linearly with time: $[\mathrm{A}] = [\mathrm{A}]_0 - kt$

First order: Rate is directly proportional to concentration.

$\mathrm{Rate} = k[\mathrm{A}]$

Second order: Rate is proportional to the square of concentration.

$\mathrm{Rate} = k[\mathrm{A}]^2$

Determining Order from Initial Rates

Run experiments with different initial concentrations and measure the initial rate.

If doubling [A] doubles the rate: first order with respect to A. If doubling [A] quadruples the rate: second order with respect to A. If doubling [A] has no effect on the rate: zero order with respect to A.

Worked example 1: For the reaction $\mathrm{A} + \mathrm{B} \to \mathrm{C}$:

Experiment	$[\mathrm{A}]$ (mol/dm$^3$)	$[\mathrm{B}]$ (mol/dm$^3$)	Initial Rate (mol dm$^{-3}$ s$^{-1}$)
1	0.10	0.10	$2.0 \times 10^{-4}$
2	0.20	0.10	$4.0 \times 10^{-4}$
3	0.10	0.20	$8.0 \times 10^{-4}$

Determine the rate equation and the rate constant.

Answer

Comparing experiments 1 and 2: [A] doubles, [B] constant, rate doubles.

Order with respect to A = 1 (first order).

Comparing experiments 1 and 3: [A] constant, [B] doubles, rate quadruples.

Order with respect to B = 2 (second order).

Rate equation: $\mathrm{Rate} = k[\mathrm{A}][\mathrm{B}]^2$

Using experiment 1:

$k = \dfrac{\mathrm{Rate}}{[\mathrm{A}][\mathrm{B}]^2} = \dfrac{2.0 \times 10^{-4}}{0.10 \times (0.10)^2} = \dfrac{2.0 \times 10^{-4}}{0.0010} = 0.20 \mathrm{ dm}^6 \mathrm{ mol}^{-2} \mathrm{ s}^{-1}$

Check with experiment 3: $\mathrm{Rate} = 0.20 \times 0.10 \times (0.20)^2 = 0.20 \times 0.10 \times 0.040 = 8.0 \times 10^{-4}$. Correct.

Worked example 2: For the reaction $\mathrm{X} + 2\mathrm{Y} \to \mathrm{Z}$:

Experiment	$[\mathrm{X}]$ (mol/dm$^3$)	$[\mathrm{Y}]$ (mol/dm$^3$)	Initial Rate (mol dm$^{-3}$ s$^{-1}$)
1	0.20	0.10	$6.0 \times 10^{-3}$
2	0.40	0.10	$1.2 \times 10^{-2}$
3	0.20	0.20	$6.0 \times 10^{-3}$

Answer

Exp 1 vs 2: [X] doubles, rate doubles. First order in X.

Exp 1 vs 3: [Y] doubles, rate unchanged. Zero order in Y.

Rate equation: $\mathrm{Rate} = k[\mathrm{X}]$

$k = \dfrac{6.0 \times 10^{-3}}{0.20} = 0.030 \mathrm{ s}^{-1}$

Units of the Rate Constant

The units of $k$ depend on the overall order of reaction.

Overall Order	Rate Equation	Units of $k$
0	Rate = $k$	mol dm$^{-3}$ s$^{-1}$
1	Rate = $k[\mathrm{A}]$	s$^{-1}$
2	Rate = $k[\mathrm{A}]^2$	dm$^3$ mol$^{-1}$ s$^{-1}$
3	Rate = $k[\mathrm{A}]^3$	dm$^6$ mol$^{-2}$ s$^{-1}$
1 + 2 = 3	Rate = $k[\mathrm{A}][\mathrm{B}]^2$	dm$^6$ mol$^{-2}$ s$^{-1}$

---

The Arrhenius Equation

Equation

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

Where:

• $k$ = rate constant
• $A$ = pre-exponential factor (frequency factor)
• $E_a$ = activation energy (J/mol)
• $R$ = gas constant = 8.314 J/(mol K)
• $T$ = temperature (K)

Logarithmic Form

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

or equivalently:

$\log_{10} k = \log_{10} A - \frac{E_a}{2.303RT}$

Plotting

A plot of $\ln k$ (y-axis) vs $1/T$ (x-axis) gives a straight line:

• Gradient = $-E_a / R$
• y-intercept = $\ln A$

From the gradient: $E_a = -\mathrm{gradient} \times R$

Two-Temperature Form

If the rate constant is known at two temperatures:

$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Worked example 3: The rate constant for a reaction is $3.46 \times 10^{-5} \mathrm{ s}^{-1}$ at $298 \mathrm{ K}$ and $1.50 \times 10^{-3} \mathrm{ s}^{-1}$ at $350 \mathrm{ K}$. Calculate $E_a$.

Answer

$\ln\left(\frac{1.50 \times 10^{-3}}{3.46 \times 10^{-5}}\right) = \frac{E_a}{8.314}\left(\frac{1}{298} - \frac{1}{350}\right)$

$\ln(43.35) = \frac{E_a}{8.314}(0.003356 - 0.002857)$

$3.770 = \frac{E_a}{8.314}(0.000499)$

$E_a = \frac{3.770 \times 8.314}{0.000499} = \frac{31.34}{0.000499} = 62800 \mathrm{ J/mol} = 62.8 \mathrm{ kJ/mol}$

Worked example 4: A reaction has $E_a = 75.0 \mathrm{ kJ/mol}$. The rate constant at $300 \mathrm{ K}$ is $2.50 \times 10^{-3} \mathrm{ s}^{-1}$. Calculate the rate constant at $320 \mathrm{ K}$.

Answer

$\ln\left(\frac{k_2}{2.50 \times 10^{-3}}\right) = \frac{75000}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)$

$\ln\left(\frac{k_2}{2.50 \times 10^{-3}}\right) = 9022 \times (0.003333 - 0.003125)$

$\ln\left(\frac{k_2}{2.50 \times 10^{-3}}\right) = 9022 \times 0.000208 = 1.877$

$\frac{k_2}{2.50 \times 10^{-3}} = e^{1.877} = 6.534$

$k_2 = 6.534 \times 2.50 \times 10^{-3} = 1.63 \times 10^{-2} \mathrm{ s}^{-1}$

---

Iodine Clock Reaction

Overview

The iodine clock reaction is a classic experiment for measuring initial rates. The reaction involves the oxidation of iodide ions by an oxidising agent (e.g., peroxydisulfate):

$\mathrm{S}_2\mathrm{O}_8^{2-} + 2\mathrm{I}^- \to 2\mathrm{SO}_4^{2-} + \mathrm{I}_2$

A small, fixed amount of sodium thiosulfate and starch are added. The thiosulfate reacts with iodine as it forms:

$\mathrm{I}_2 + 2\mathrm{S}_2\mathrm{O}_3^{2-} \to 2\mathrm{I}^- + \mathrm{S}_4\mathrm{O}_6^{2-}$

Once all the thiosulfate is consumed, free iodine accumulates and reacts with starch to produce a blue-black colour.

Determining the Rate

The time for the colour change ($t$) is measured. The rate is proportional to $1/t$:

$\mathrm{Rate} \propto \frac{1}{t}$

By varying the concentration of one reactant while keeping others constant, the order with respect to each reactant can be determined.

Example Data Analysis

Experiment	$[\mathrm{S}_2\mathrm{O}_8^{2-}]$ (mol/dm$^3$)	$[\mathrm{I}^-]$ (mol/dm$^3$)	Time (s)
1	0.040	0.040	48
2	0.080	0.040	24
3	0.040	0.080	24

Answer

Exp 1 vs 2: $[\mathrm{S}_2\mathrm{O}_8^{2-}]$ doubles, time halves, rate doubles. First order in $\mathrm{S}_2\mathrm{O}_8^{2-}$.

Exp 1 vs 3: $[\mathrm{I}^-]$ doubles, time halves, rate doubles. First order in $\mathrm{I}^-$.

Rate equation: $\mathrm{Rate} = k[\mathrm{S}_2\mathrm{O}_8^{2-}][\mathrm{I}^-]$

Overall order = 2.

---

Common Pitfalls

1. Assuming order equals stoichiometry: The orders in the rate equation must be determined experimentally. They are not necessarily the same as the coefficients in the balanced equation.

2. Confusing rate and rate constant: The rate depends on concentrations; the rate constant ($k$) does not (it depends only on temperature and the presence of a catalyst).

3. Wrong units for the rate constant: Always derive the units of $k$ from the rate equation. A first-order rate constant has units s$^{-1}$, not mol dm$^{-3}$ s$^{-1}$.

4. Forgetting to convert temperature to Kelvin: The Arrhenius equation requires absolute temperature in Kelvin, not degrees Celsius.

5. Misinterpreting Maxwell-Boltzmann diagrams: Increasing temperature shifts the peak right AND lowers it, but the total area stays the same. The curve never touches the x-axis.

6. Catalyst misconceptions: A catalyst does NOT start a reaction, does NOT increase the amount of product, and does NOT change the enthalpy change of the reaction.

7. Using the wrong $R$ value: In the Arrhenius equation, use $R = 8.314$ J/(mol K) when $E_a$ is in joules. Convert kJ to J before substituting.

---

Practice Problems

Problem 1

For the reaction $2\mathrm{NO}_{(g)} + 2\mathrm{H}_{2(g)} \to \mathrm{N}_{2(g)} + 2\mathrm{H}_2\mathrm{O}_{(g)}$, the following data were obtained:

Experiment	$[\mathrm{NO}]$ (mol/dm$^3$)	$[\mathrm{H}_2]$ (mol/dm$^3$)	Initial Rate (mol dm$^{-3}$ s$^{-1}$)
1	0.010	0.010	$1.2 \times 10^{-5}$
2	0.020	0.010	$4.8 \times 10^{-5}$
3	0.010	0.020	$2.4 \times 10^{-5}$

Determine the rate equation, the rate constant, and the overall order.

Answer

Exp 1 vs 2: [NO] doubles, rate quadruples ($1.2 \times 10^{-5} \to 4.8 \times 10^{-5}$). Second order in NO.

Exp 1 vs 3: $[\mathrm{H}_2]$ doubles, rate doubles. First order in $\mathrm{H}_2$.

Rate equation: $\mathrm{Rate} = k[\mathrm{NO}]^2[\mathrm{H}_2]$

$k = \dfrac{1.2 \times 10^{-5}}{(0.010)^2 \times 0.010} = \dfrac{1.2 \times 10^{-5}}{1.0 \times 10^{-6}} = 12.0 \mathrm{ dm}^6 \mathrm{ mol}^{-2} \mathrm{ s}^{-1}$

Overall order = $2 + 1 = 3$

Problem 2

The rate constant for a first-order reaction is $5.00 \times 10^{-4} \mathrm{ s}^{-1}$ at $300 \mathrm{ K}$. The activation energy is $65.0 \mathrm{ kJ/mol}$. Calculate the rate constant at $320 \mathrm{ K}$.

Answer

$\ln\left(\frac{k_2}{5.00 \times 10^{-4}}\right) = \frac{65000}{8.314}\left(\frac{1}{300} - \frac{1}{320}\right)$

$= 7820 \times (0.003333 - 0.003125) = 7820 \times 0.000208 = 1.627$

$\frac{k_2}{5.00 \times 10^{-4}} = e^{1.627} = 5.089$

$k_2 = 5.089 \times 5.00 \times 10^{-4} = 2.54 \times 10^{-3} \mathrm{ s}^{-1}$

Problem 3

In an iodine clock experiment, the following data were collected:

Experiment	$[\mathrm{S}_2\mathrm{O}_8^{2-}]$ (mol/dm$^3$)	$[\mathrm{I}^-]$ (mol/dm$^3$)	Time (s)
1	0.020	0.020	60
2	0.040	0.020	30
3	0.020	0.040	15

Determine the order with respect to each reactant and the overall order.

Answer

Exp 1 vs 2: $[\mathrm{S}_2\mathrm{O}_8^{2-}]$ doubles, time halves. Rate doubles. First order in $\mathrm{S}_2\mathrm{O}_8^{2-}$.

Exp 1 vs 3: $[\mathrm{I}^-]$ doubles, time becomes $60/4 = 15$ s. Rate quadruples. Second order in $\mathrm{I}^-$.

Rate equation: $\mathrm{Rate} = k[\mathrm{S}_2\mathrm{O}_8^{2-}][\mathrm{I}^-]^2$

Overall order = $1 + 2 = 3$

Problem 4

Sketch and label the Maxwell-Boltzmann distribution curves at $T_1 = 300 \mathrm{ K}$ and $T_2 = 400 \mathrm{ K}$ on the same axes. Mark the activation energy $E_a$ and shade the area representing particles with energy $\geq E_a$ at each temperature. Explain why the rate increases with temperature in terms of the distribution.

Answer

At $T_1$ (300 K): the curve peaks at lower energy, with a smaller fraction of particles exceeding $E_a$.

At $T_2$ (400 K): the curve is broader and shifted right, peaking at higher energy. A significantly larger fraction of particles now exceeds $E_a$.

The rate increase with temperature is primarily because more particles have kinetic energy $\geq E_a$, leading to a larger proportion of successful collisions. The increase in collision frequency is a secondary, smaller effect.

Both curves start at the origin, never touch the x-axis, and enclose the same total area.

Problem 5

Explain, with reference to collision theory and the Maxwell-Boltzmann distribution, why adding a catalyst increases the rate of a reaction but does not increase the yield of products at equilibrium.

Answer

A catalyst provides an alternative reaction pathway with a lower activation energy ($E_a$). On the Maxwell-Boltzmann distribution, the $E_a$ threshold moves to the left. A larger fraction of particles now has energy $\geq E_a$, so the proportion of successful collisions increases, and the rate increases.

The catalyst lowers $E_a$ equally for both the forward and reverse reactions. Therefore, both the forward and reverse rates increase by the same factor. The ratio of forward to reverse rates (the equilibrium constant $K$) remains unchanged. Since $K$ is unchanged, the position of equilibrium and the equilibrium yield of products remain the same.

Problem 6

A reaction has the rate equation $\mathrm{Rate} = k[\mathrm{P}]^2[\mathrm{Q}]$. At a certain temperature, when $[\mathrm{P}] = 0.30 \mathrm{ mol/dm}^3$ and $[\mathrm{Q}] = 0.20 \mathrm{ mol/dm}^3$, the rate is $1.08 \times 10^{-2} \mathrm{ mol dm}^{-3} \mathrm{ s}^{-1}$. Calculate the rate constant and its units.

If the concentration of P is tripled and the concentration of Q is doubled, by what factor does the rate increase?

Answer

$k = \frac{\mathrm{Rate}}{[\mathrm{P}]^2[\mathrm{Q}]} = \frac{1.08 \times 10^{-2}}{(0.30)^2 \times 0.20} = \frac{1.08 \times 10^{-2}}{0.09 \times 0.20} = \frac{1.08 \times 10^{-2}}{0.018} = 0.60 \mathrm{ dm}^6 \mathrm{ mol}^{-2} \mathrm{ s}^{-1}$

New rate factor: $[\mathrm{P}]$ triples (factor of $3^2 = 9$), $[\mathrm{Q}]$ doubles (factor of 2).

Overall factor = $9 \times 2 = 18$.

The rate increases by a factor of 18.

Problem 7

Explain why a small increase in temperature (e.g., from 298 K to 308 K) causes a much larger increase in reaction rate than would be predicted by the increase in collision frequency alone.

Answer

The collision frequency increases only slightly with temperature (proportional to $\sqrt{T}$), which would predict roughly a $\sqrt{308/298} \approx 1.017$, or about 1.7% increase.

However, the actual rate increase is much larger (approximately doubling for a 10 K increase) because the exponential dependence on $E_a / RT$ in the Arrhenius equation means that even a small temperature increase dramatically increases the fraction of molecules exceeding $E_a$.

On the Maxwell-Boltzmann distribution, the area under the curve beyond $E_a$ increases disproportionately as temperature rises. This is the dominant factor in the rate increase, not the collision frequency.

---

Practical Techniques for Measuring Rates

Continuous Monitoring Methods

These methods measure the concentration (or a quantity proportional to it) continuously throughout the reaction.

Gas syringe method: As the reaction produces gas, the volume is recorded at regular time intervals. A graph of volume vs time gives a curve whose gradient at any point equals the rate.

Mass loss method: The reaction flask is placed on a balance. As gas escapes, the mass decreases. The rate of mass loss at any point equals the reaction rate.

Colorimetry: For reactions involving a coloured species, a colorimeter measures absorbance over time. Absorbance is proportional to concentration (Beer-Lambert law), so the rate can be determined from the absorbance-time graph.

Conductivity method: For reactions that change the number or type of ions in solution (e.g., hydrolysis of an ester producing a carboxylic acid), the conductivity is measured over time.

Discontinuous Monitoring Methods

These methods involve taking samples (aliquots) from the reaction at regular intervals and analysing them.

Titration: A sample is withdrawn, the reaction is quenched (e.g., by cooling or adding a reactant that stops the reaction), and the concentration is determined by titration.

Quenching techniques: Common methods include rapid cooling, dilution with cold water, or adding a chemical that reacts with a catalyst or reactant.

Choosing the Right Method

Scenario	Best Method
Gas-producing reaction	Gas syringe or mass loss
Colour change	Colorimetry
Change in ion concentration	Conductivity
Acid-base reaction	Titration of aliquots
Slow reaction	Titration of aliquots
Fast reaction	Continuous monitoring (colorimetry)

---

Half-Life of Reactions

Definition

The half-life ($t_{1/2}$) is the time taken for the concentration of a reactant to decrease to half its initial value.

Zero-Order Half-Life

$t_{1/2} = \frac{[\mathrm{A}]_0}{2k}$

The half-life depends on the initial concentration. It increases as concentration decreases.

First-Order Half-Life

$t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$

The half-life is independent of the initial concentration. This is a defining characteristic of first-order reactions.

Radioactive decay is a first-order process.

Second-Order Half-Life

$t_{1/2} = \frac{1}{k[\mathrm{A}]_0}$

The half-life depends on the initial concentration. It increases as concentration decreases.

Using Half-Life to Determine Order

Order	Effect of doubling initial concentration on $t_{1/2}$
Zero	Doubles
First	No change
Second	Halves

Worked example 8: In a first-order decomposition, the concentration falls from 0.800 mol/dm$^3$ to 0.200 mol/dm$^3$ in 120 s. Calculate the half-life and the rate constant.

Answer

The concentration halves twice: $0.800 \to 0.400 \to 0.200$.

Two half-lives = 120 s, so $t_{1/2} = 60 \mathrm{ s}$.

$k = \dfrac{0.693}{t_{1/2}} = \dfrac{0.693}{60} = 0.01155 \mathrm{ s}^{-1}$

---

Determining Order from Concentration-Time Graphs

Zero-Order

A straight line with negative gradient when concentration is plotted against time.

$[\mathrm{A}] = [\mathrm{A}]_0 - kt$

First-Order

A curve where a plot of $\ln[\mathrm{A}]$ vs time gives a straight line with gradient $-k$.

$\ln[\mathrm{A}] = \ln[\mathrm{A}]_0 - kt$

Alternatively, a plot of $\log_{10}[\mathrm{A}]$ vs time gives a straight line with gradient $-k/2.303$.

Second-Order

A curve where a plot of $1/[\mathrm{A}]$ vs time gives a straight line with gradient $+k$.

$\frac{1}{[\mathrm{A}]} = \frac{1}{[\mathrm{A}]_0} + kt$

Summary

Order	Linear Plot	Gradient
Zero	$[\mathrm{A}]$ vs $t$	$-k$
First	$\ln[\mathrm{A}]$ vs $t$	$-k$
Second	$1/[\mathrm{A}]$ vs $t$	$+k$

Worked example 9: The following data were collected for the decomposition of a substance X:

Time (s)	$[\mathrm{X}]$ (mol/dm$^3$)	$\ln[\mathrm{X}]$	$1/[\mathrm{X}]$
0	0.100	$-2.303$	10.0
30	0.0707	$-2.649$	14.1
60	0.0500	$-2.996$	20.0
90	0.0354	$-3.340$	28.3
120	0.0250	$-3.689$	40.0

Determine the order of reaction.

Answer

Check if $\ln[\mathrm{X}]$ vs $t$ is linear:

Gradient from first and last points: $\dfrac{-3.689 - (-2.303)}{120 - 0} = \dfrac{-1.386}{120} = -0.01155 \mathrm{ s}^{-1}$

Check intermediate: $\dfrac{-2.649 - (-2.303)}{30} = -0.01153$, $\dfrac{-2.996 - (-2.649)}{30} = -0.01157$.

The gradients are consistent (all approximately $-0.0116 \mathrm{ s}^{-1}$), confirming first order.

$k = 0.0116 \mathrm{ s}^{-1}$

---

Factors Affecting Rate in Real-World Contexts

Refrigeration and Food Preservation

Lowering temperature slows the rate of biochemical reactions (enzyme-catalysed decomposition) that cause food spoilage. Each $10^\circ$C reduction approximately halves the rate.

Pressure Cookers

Higher pressure raises the boiling point of water, allowing food to cook at temperatures above $100^\circ$C. The higher temperature dramatically increases the rate of cooking reactions.

Catalytic Converters Revisited

At room temperature, the conversion of CO and NO$_x$ is negligibly slow. The catalyst (Pt, Pd, Rh) lowers the activation energy so that the reactions proceed at useful rates at exhaust temperatures (300--600$^\circ$C).

Dissolving a Solid

Stirring increases the rate of dissolution by:

• Bringing fresh solvent into contact with the solid surface.
• Removing the saturated solution layer around the solid, maintaining a high concentration gradient.

Powdering the solid increases surface area, further increasing the rate.

---

Summary of Key Equations

Equation	Use
$\mathrm{Rate} = k[\mathrm{A}]^m[\mathrm{B}]^n$	Rate equation
$\ln k = \ln A - E_a/(RT)$	Arrhenius equation (linear form)
$\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$	Two-temperature Arrhenius
$t_{1/2} = 0.693/k$	First-order half-life
$\ln[\mathrm{A}] = \ln[\mathrm{A}]_0 - kt$	First-order integrated rate law
$\mathrm{Rate} \propto 1/t$	Clock reaction initial rate