Kinetic Chemistry (Chemical Kinetics)

The study of reaction rates, the factors affecting them, and the mechanisms by which reactions occur.

"Thermodynamics tell us what can occur during a process, while kinetics tell us what actually occurs"

Thermochemistry (enthalpy) tells us if a process is spontaneous; kinetic chemistry tells us how fast it actually happens. A reaction can be thermodynamically favourable yet kinetically frozen — e.g. diamond converting to graphite ($\Delta G^\circ_f = -2,900 \text{ kJ mol}^{-1}$) is spontaneous but occurs on geological timescales.

Reaction Rate

The change in the concentration of a reactant or product with time.

• Units: $\text{M min}^{-1}$ or $\text{M s}^{-1}$
• Inversely proportional to time; the shorter the time taken, the higher the rate

Types of Rate (Graphical)

Type	Definition	Graphical Method
Average Rate	Rate over a time period	Gradient of a straight line between two points on the concentration–time curve
Instantaneous Rate	Rate at a specific instant	Gradient of the tangent to the curve at that point
Initial Rate	Rate at the very beginning ($t = 0$)	Gradient of the tangent drawn at $t = 0$

Differential Rate Equation

For a general reaction $aA + bB \longrightarrow cC + dD$:

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

The negative sign indicates that reactants are disappearing.

Example — Ammonia Synthesis

N#N

[HH]

N

$$N_2(g) + 3H_2(g) \longrightarrow 2NH_3(g)$$

$$\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}$$

Example — HI Decomposition

I

[HH]

II

$$2HI \longrightarrow H_2 + I_2$$

If $\frac{d[I_2]}{dt} = 1.8 \times 10^{-6} \text{ M s}^{-1}$, then rate of disappearance of HI: $$-\frac{d[HI]}{dt} = 2 \times 1.8 \times 10^{-6} = 3.6 \times 10^{-6} \text{ M s}^{-1}$$

Rate Law (Rate Equation)

An expression relating the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers:

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

Where:

• $k$ = rate constant (temperature dependent)
• $m, n$ = reaction orders with respect to each reactant
• Overall order = $m + n$

Critical: $m$ and $n$ are determined experimentally and are not the stoichiometric coefficients $a$ and $b$. Order $\neq$ stoichiometric coefficient.

Rate Constant ($k$)

• Constant of proportionality between reaction rate and reactant concentration
• The unit of $k$ depends on the overall order of the reaction
• Greater $k$ → faster reaction rate

Order of Reaction

• The sum of the powers to which all reactant concentrations in the rate law are raised
• Defined in terms of reactant concentration
• Must be determined experimentally

Lecture Examples

Example A — Fluorine and Chlorine Dioxide

FF

O=Cl=O

O=Cl(=O)F

$$F_2(g) + 2ClO_2(g) \longrightarrow 2FClO_2(g)$$

Rate law: $\text{Rate} = k[F_2]^1[ClO_2]^1$

• Order wrt $F_2$ = 1, wrt $ClO_2$ = 1
• Overall order = 2 (second order)

Example B — Nitric Oxide and Oxygen

[O]N=O

O=O

O=[N+]([O-])[O-]

$$2NO(g) + O_2(g) \longrightarrow 2NO_2(g)$$

Rate law: $\text{rate} = k[NO]^2[O_2]^1$

• Order wrt $NO$ = 2, wrt $O_2$ = 1
• Overall order = 3 (third order)

Example C — Hydrogen Peroxide and Iodide

OO

[I-]

[O-][I+2][O-]

O

$$H_2O_2(aq) + 3I^-(aq) + 2H^+(aq) \longrightarrow I_3^-(aq) + 2H_2O(l)$$

Rate law: $\text{Rate} = k[H_2O_2]^1[I^-]^1$

• Order wrt $H_2O_2$ = 1, wrt $I^-$ = 1, wrt $H^+$ = 0 (does not appear in rate law)
• Overall order = 2 (second order)

Determining Reaction Order

Method of Initial Rates

Compare initial rates at different initial concentrations.

Integrated Rate Laws

Order	Differential Form	Integrated Form	Half-Life	Linear Plot (Conc vs Time)	Slope
0	$-\frac{d[A]}{dt} = k$	$[A]_t = [A]_0 - kt$	$t_{1/2} = [A]_0/2k$	$[A]$ vs $t$	$-k$
1	$-\frac{d[A]}{dt} = k[A]$	$ln[A]_t = ln[A]_0 - kt$	$t_{1/2} = 0.693/k$	$ln[A]$ vs $t$	$-k$
2	$-\frac{d[A]}{dt} = k[A]^2$	$\frac{1}{[A]_t} = \frac{1}{[A]_0} + kt$	$t_{1/2} = 1/k[A]_0$	$\frac{1}{[A]}$ vs $t$	$+k$

Key points:

• First-order reactions have constant half-life
• Second-order linear plot is the only one with positive slope ($+k$)
• Rate constant $k$ is obtained from the magnitude of the gradient

Rate vs Concentration Plots (p. 36–40)

An alternative way to identify order from the differential rate equation:

Order	Rate Law	Plot of Rate vs [A]	Appearance
0	Rate = k	Rate vs [A]	Horizontal line (independent of [A])
1	Rate = k[A]	Rate vs [A]	Straight line through origin (slope = k)
2	Rate = k[A]²	Rate vs [A]	Parabola (curved)

[!note] Lecture emphasis For zero order, the plot of reaction rate vs concentration is a straight line parallel to the x-axis. For first order, the plot of reaction rate vs concentration produces a linear line with slope = k.

Half-Life

Definition: Length of time required for the concentration of a reactant to decrease to half of its initial value.

Order	Half-Life Formula	Graphical Behavior
0	$t_{1/2} = \frac{[A]_0}{2k}$	Half-life decreases as reaction proceeds
1	$t_{1/2} = \frac{ln\ 2}{k} = \frac{0.693}{k}$	Half-life is constant (independent of concentration)
2	$t_{1/2} = \frac{1}{k[A]_0}$	Half-life increases as reaction proceeds

Graphical Identification

From a plot of $[A]$ vs time:

• Zero-order: Linear decay; successive half-lives get shorter
• First-order: Exponential decay; successive half-lives are equal
• Second-order: Curved decay; successive half-lives get longer

[!tip] Lecture memory aids

• 0th order: masa ↓ (half-life decreases)
• 1st order: Masa sama (half-life stays the same)
• 2nd order: Masa ↑ (half-life increases)

Worked Examples

Example 1: Zero-Order Decomposition of H₂O₂

OO

Given: $k = 3.66 \times 10^{-3}\ \text{Ms}^{-1}$, $[H_2O_2]_0 = 0.88\ \text{M}$

(a) Time when $[H_2O_2] = 0.600\ \text{M}$: $$t = \frac{0.88 - 0.600}{3.66 \times 10^{-3}} = 77.05\ \text{s}$$

(b) $[H_2O_2]$ after $225\ \text{s}$: $$[H_2O_2]_t = 0.88 - (3.66 \times 10^{-3})(225) = 0.8215\ \text{M}$$

[!note] Lecture working Lecture shows: $0.882 - [H_2O_2]_t = 0.0585$, giving $[H_2O_2]_t = 0.8235\ \text{M}$. Use lecture figures in exams.

Example 2: First-Order Reaction 2A → B

Given: $k = 2.5 \times 10^{-3}\ \text{s}^{-1}$. Time for 75% consumption of A:

$$t = \frac{ln(100/25)}{2.5 \times 10^{-3}} = \frac{ln(4)}{2.5 \times 10^{-3}} = 554.52\ \text{s}$$

Example 3: Second-Order Reaction 2A → B

Given: $k = 2.8 \times 10^{-2}\ \text{M}^{-1}\text{min}^{-1}$ at $80\ ^\circ\text{C}$

Time for $[A]$ to decrease from $0.88\ \text{M}$ to $0.14\ \text{M}$:

$$\frac{1}{0.14} - \frac{1}{0.88} = (2.8 \times 10^{-2})t$$ $$t = 214.52\ \text{min}$$

Example 4: Second-Order Iodine Recombination

[I]

[I][I]

Reaction: $I(g) + I(g) \longrightarrow I_2(g)$

Given: $k = 7.0 \times 10^{9}\ \text{M}^{-1}\text{s}^{-1}$ at $23\ ^\circ\text{C}$, $[I]_0 = 0.086\ \text{M}$

After $t = 2.0\ \text{min} = 120\ \text{s}$:

$$[I]_t = 1.19 \times 10^{-12}\ \text{M}$$

Example 5: Half-Life of N₂O₅

O=[N+]([O-])O[N+](=O)[O-]

Given: first-order decomposition, $k = 5.7 \times 10^{-4}\ \text{s}^{-1}$

$$t_{1/2} = \frac{ln\ 2}{5.7 \times 10^{-4}} = 1216.05\ \text{s}$$

Example 6: Radioactive Element K (First-Order)

Given: $[K]0 = 3.5 \times 10^{-6}\ \text{M}$, $t{1/2} = 0.22\ \text{years}$

$$k = \frac{ln\ 2}{0.22} = 3.15\ \text{years}^{-1}$$

(a) $[K]$ after 6 months ($0.5$ year): $$[K]_t = 7.2 \times 10^{-7}\ \text{M}$$

(b) Time to reduce to $1.75 \times 10^{-6}\ \text{M}$ (half of initial): $$t = 0.22\ \text{years}$$

Example 7: Graphical Order Determination

For $S \rightarrow T + V$:

Time (min)	0	10	20	30	40	50	60
[S] (mol dm⁻³)	4.5	3.10	2.38	1.92	1.60	1.40	1.22

Since $\frac{1}{[S]}$ vs $t$ is linear → second order

• $k = 0.01\ \text{M}^{-1}\text{min}^{-1}$
• $t_{1/2} = 22.2\ \text{min}$
• At $1500\ \text{s} = 25\ \text{min}$: $[S] = 2.13\ \text{M}$

Example 8: Method of Initial Rates — S₂O₈²⁻ + I⁻ (p. 21)

O=S(=O)([O-])OOS(=O)(=O)[O-]

[I-]

[O-]S(=O)(=O)[O-]

I[I-]I

Reaction: S₂O₈²⁻(aq) + 3I⁻(aq) → 2SO₄²⁻(aq) + I₃⁻(aq)

Exp	[S₂O₈²⁻]	[I⁻]	Initial Rate (M s⁻¹)
1	0.08	0.034	2.2 × 10⁻⁴
2	0.08	0.017	1.1 × 10⁻⁴
3	0.16	0.017	2.2 × 10⁻⁴

• Comparing Exp 1 and 2: [I⁻] halved → rate halves → order w.r.t [I⁻] = 1
• Comparing Exp 2 and 3: [S₂O₈²⁻] doubled → rate doubles → order w.r.t [S₂O₈²⁻] = 1

Rate law: Rate = k[S₂O₈²⁻][I⁻] (overall order = 2)

Example 9: Determining Order from Rate Change (p. 31)

For A → Product, rate increases 8-fold when [A] doubles:

$$Rate = k[A]^x \Rightarrow 8 = 2^x \Rightarrow x = 3$$

Third order reaction.

Example 10: Multi-Reactant System (p. 33)

For aA + bB → cC:

Exp	[A]/M	[B]/M	Initial Rate (M min⁻¹)
1	1.0	1.0	2.20 × 10⁻⁴
2	2.0	2.0	1.70 × 10⁻³
3	1.0	3.0	6.60 × 10⁻⁴
4	3.0	2.0	3.96 × 10⁻³

• Order w.r.t [B] = 1, order w.r.t [A] = 2
• Rate law: Rate = k[A]²[B] (overall order = 3)

Factors Affecting Reaction Rate

The rate of a reaction can be speed up or slowed down by four factors:

1. Concentration

Raising the concentration makes the reaction happen at a faster rate.

Mechanism:

1. Higher concentration → more particles per unit volume
2. More particles → frequency of collision increases
3. More collisions → frequency of effective collision increases
4. Therefore, rate of reaction increases

Rate Law: $Rate = k[A]^m$

[!important] Exception Rate of reaction for zero-order reaction is not affected by concentration.

2. Pressure (for gaseous reactions)

[!tip] Skema exam — This is the exam marking scheme format

The effect of changing pressure is like changing the concentration.

Relationship: $\uparrow P = \downarrow V$ (inverse relationship)

Mechanism:

1. Increase in pressure decreases the volume of container
2. Gas particles are pushed closer to each other
3. More gas per unit volume → frequency of collision increases
4. Frequency of effective collision increases
5. Thus, rate of reaction increases

3. Particle Size / Surface Area

[!note] Key concept: "tak boleh bergerak" (cannot move)

Solid particles are immobile but rely on liquid or gas particles colliding with solid's surface to react.

Key Points:

• In a solid-liquid reaction, the surface area of the solid impacts how fast the reaction occurs
• Liquid and solid can only bump into each other at the liquid-solid interface (surface of the solid)
• Solid molecules trapped within the body of the solid cannot react

Mechanism:

1. Breaking solid into smaller pieces → larger total surface area
2. Greater total area exposed → more solid molecules exposed
3. Frequency of collision increases
4. Frequency of effective collision increases
5. Rate of reaction increases

4. Temperature

• Higher T → more molecules with E ≥ Ea → faster rate
• Illustrated using Maxwell-Boltzmann distribution curve

Mechanism:

1. When temperature increases, kinetic energy of molecules increases
2. More collisions occur at a given time
3. Higher kinetic energy → higher energy of collisions
4. More collisions achieve the activation energy, $E_a$
5. Frequency of effective collision increases
6. Rate of reaction increases

5. Catalyst

Definition: A substance that increases the rate of reaction by providing an alternative pathway with lower activation energy.

Key Properties:

• Catalyst will not affect the enthalpy of reactant or product
• It only decreases the activation energy, $E_a$
• Types: Homogeneous, Heterogeneous, Enzymes

Mechanism: $$\downarrow E_a \rightarrow \uparrow \text{particles with } E \geq E_a \rightarrow \uparrow \text{frequency of effective collision} \rightarrow \uparrow \text{rate}$$

Effect of Concentration Change by Order (p. 22–25)

For A → Products with Rate = k[A]ⁿ, the effect of doubling [A]:

Order (n)	Rate Law	Effect of Doubling [A]	Factor
0	Rate = k	No change	2⁰ = 1
1	Rate = k[A]	Rate doubles	2¹ = 2
2	Rate = k[A]²	Rate quadruples	2² = 4

General rule: If [A] is multiplied by factor f, rate changes by factor fⁿ.

[!tip] Memory aid from lecture 0th order: Rate is flat — doubling [A] does nothing. 1st order: Rate is proportional — doubling [A] doubles rate. 2nd order: Rate is squared — doubling [A] gives 4× rate.

Arrhenius Equation

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

Where:

• k = rate constant
• A = Arrhenius constant (collision frequency factor)
• Eₐ = activation energy (J/mol)
• R = gas constant (8.314 J mol⁻¹ K⁻¹)
• T = absolute temperature (K)

Logarithmic Forms

Natural logarithm form: $\ln k = \ln A - \frac{E_a}{RT}$

Linear (straight-line) form — plot ln k vs 1/T: $\ln k = \frac{-E_a}{R}\left(\frac{1}{T}\right) + \ln A$

• Slope, m = -Eₐ/R
• Y-intercept = ln A

Two-Point Form

For the same reaction at two temperatures: $\ln\frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

Factors affecting $k$:

• Higher temperature → greater rate constant
• Catalyst present → lowers $E_a$ → greater rate constant

Worked Example: Graphical Determination of $E_a$

For the decomposition of $\ce{N2O5}$:

O=[N+]([O-])O[N+](=O)[O-]

$T$ (K)	$k$ (s⁻¹)	$1/T$ (K⁻¹)	$\ln k$
298	$3.46 \times 10^{-5}$	$3.36 \times 10^{-3}$	$-10.27$
308	$1.35 \times 10^{-4}$	$3.25 \times 10^{-3}$	$-8.91$
318	$4.98 \times 10^{-4}$	$3.14 \times 10^{-3}$	$-7.61$
328	$1.44 \times 10^{-3}$	$3.05 \times 10^{-3}$	$-6.54$
338	$4.87 \times 10^{-3}$	$2.96 \times 10^{-3}$	$-5.32$

Slope of $\ln k$ vs $1/T$ ≈ $-12,400$ K. Therefore: $$E_a = -\text{slope} \times R = 12,400 \times 8.314 = 103,100\text{ J·mol}^{-1} = 103.1\text{ kJ·mol}^{-1}$$

Worked Example: Finding $k$ at a New Temperature

For $\ce{2HI(g) -> H2(g) + I2(g)}$, $E_a = 183$ kJ·mol⁻¹, $k_1 = 3.02 \times 10^{-5}$ M⁻¹·s⁻¹ at $T_1 = 629$ K. Find $k_2$ at $T_2 = 700$ K.

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

$$k_2 = 1.05 \times 10^{-3}\text{ M}^{-1}\text{s}^{-1}$$

A temperature increase of 71 K (~11%) increased $k$ by a factor of ~35 — demonstrating the exponential sensitivity of rate to temperature.

Collision Theory

$$Rate \propto \frac{\text{Number of effective collisions}}{time}$$

Collision Theory explains the rate of chemical reaction based on two key ideas:

1. Molecules must collide to react
2. The collision must be an effective collision

Effective Collision

Definition: A collision that leads to the formation of product.

Requirements:

Requirement	Explanation
Minimum Kinetic Energy	Colliding molecules must possess a certain minimum kinetic energy (activation energy, $E_a$) to initiate the chemical reaction
Proper Orientation	Molecules must collide in the right orientation

Collision Outcomes:

Type	Condition	Result
Effective collision	Proper orientation + sufficient energy	Product formed
Ineffective collision	Wrong orientation or insufficient energy	No reaction ("Takde tindak balas")

Rate Constant Equation: $$k = p·Z·e^{-E_a/RT}$$

Where:

• p = steric factor (orientation)
• Z = collision frequency

Maxwell-Boltzmann Distribution

The Maxwell-Boltzmann distribution curve explains why temperature affects reaction rate:

• Plot of the fraction of molecules having a certain kinetic energy vs kinetic energy
• Different temperatures produce different curve shapes (T₂ > T₁ produces a broader, flatter curve shifted to higher energies)
• The area under the graph represents the total number of molecules in the reaction
• Only collisions with energy greater than Eₐ can react
• At higher temperature (T₂), the area above Eₐ is larger → more particles possess sufficient energy to react → rate increases

I.I>>[H][H].[I][I]

(This illustrates that raising temperature increases the fraction of HI molecules with enough energy to decompose.)

Transition State Theory

Transition State: The configuration of atoms of the colliding species at the time of the collision.

Activation Complex ($AB^{#}$): Species formed at the transition state — at the peak of the energy barrier.

Key Relationships

$$\Delta H = E_{a\ \text{forward}} - E_{a\ \text{reverse}}$$

Activated Complex Characteristics

Characteristic	Description
Stability	Very unstable and has a short half-life (sangat cepat / very fast)
Potential Energy	Greater than reactants or products
Equilibrium	The activated complex and reactants are in chemical equilibrium
Decomposition	Decomposes to form products or reactants (boleh berlaku tindak balas berbalik / reversible)

Example: Nucleophilic Substitution

Hydroxide + Bromomethane → Transition State (imaginary) → Methanol + Bromide

[OH-]

CBr

CO

[Br-]

Energy Profile Diagrams

[!warning] Selalu keluar exam — Always appears in exams

Energy profile diagrams show the energy changes during a chemical reaction, including activation energies and enthalpy change.

Exothermic Reaction

• Reactant energy > Product energy
• $\Delta H$ is negative (heat released)
• Example: $\text{CO (g)} + \text{NO}_2 \text{ (g)} \longrightarrow \text{CO}_2 \text{ (g)} + \text{NO (g)}$

$$E_{a\ \text{reverse}} = E_a + \Delta H$$

Endothermic Reaction

• Reactant energy < Product energy
• $\Delta H$ is positive (heat absorbed)
• Example: $\text{CO}_2 \text{ (g)} + \text{NO (g)} \longrightarrow \text{CO (g)} + \text{NO}_2 \text{ (g)}$

$$E_{a\ \text{reverse}} = E_a - \Delta H$$

[C-]#[O+]

N(=O)[O]

O=C=O

[N]=O

Worked Example: Free-Radical Substitution

Problem: For the reaction $\text{P}\bullet + \text{Q}_2 \longrightarrow \text{PQ} + \text{Q}\bullet$

Given:

• $E_{a\ \text{forward}} = 127\ \text{kJ mol}^{-1}$
• $E_{a\ \text{reverse}} = 15\ \text{kJ mol}^{-1}$

Analysis:

• Forward $E_a$ > Reverse $E_a$ → Endothermic
• $\Delta H = E_{a\ \text{forward}} - E_{a\ \text{reverse}} = 127 - 15 = \mathbf{112\ \text{kJ mol}^{-1}}$

Energy Profile Diagram Labels:

• $E_{a\ (forward)}$ — activation energy for forward reaction
• $E_{a\ (backward)}$ — activation energy for reverse reaction
• $\Delta H$ — enthalpy change
• $AB^{#}$ — activated complex at the peak

Reaction Mechanisms

The step-by-step sequence of elementary reactions by which an overall reaction occurs.

Elementary Reaction vs Overall Reaction

• An elementary reaction is a single-step process where reactants directly form products in one molecular event — it cannot be broken down further
• An overall reaction represents the net change but may proceed through multiple elementary steps (the reaction mechanism)
• Reaction intermediates are formed and consumed within the mechanism, not appearing in the overall equation

Molecularity

Molecularity is the number of reactant species in an elementary step (theoretical, integer values only):

Molecularity	Rate Law (from mechanism)	Description	Example
Unimolecular	Rate $= k[A]$	Single molecule rearranges or decomposes; first order	$\ce{CH3NC -> CH3CN}$, $\ce{N2O5 -> NO2 + NO3}$
Bimolecular	Rate $= k[A][B]$ or $k[A]^2$	Two molecules collide; second order; most common	$\ce{NO + O3 -> NO2 + O2}$
Termolecular	Rate $= k[A][B][C]$ or $k[A]^2[B]$ or $k[A]^3$	Three molecules collide simultaneously; third order; extremely rare	$\ce{2NO + O2 -> 2NO2}$

Critical distinction: Molecularity ≠ Reaction Order

• Molecularity is theoretical (from proposed mechanism), always integer (1, 2, 3)
• Reaction order is experimental (from rate data), can be fractional, zero, or integer
• Molecularity applies only to elementary steps; order applies to overall reactions

Rate-Determining Step (RDS)

The slowest elementary step in the mechanism. It acts as a bottleneck — the overall rate cannot exceed the rate of this step. The rate law of the overall reaction is derived from the rate-determining step.

Determining Rate Law from Mechanism

Procedure:

1. Identify the slowest step (RDS)
2. Write its rate law based on molecularity
3. If the RDS involves intermediates, express their concentrations in terms of reactants using preceding fast equilibrium steps

Mechanism Validation Example: $\ce{2NO(g) + 2H2(g) -> N2(g) + 2H2O(g)}$

[O]N=O

[HH]

N#N

Experimentally determined rate law: Rate $= k[\ce{NO}]^2[\ce{H2}]$

Mechanism A (valid):

Step	Reaction	Rate Law
Step 1 (slow)	$\ce{2NO + H2 -> N2O + H2O}$	Rate $= k_1[\ce{NO}]^2[\ce{H2}]$
Step 2 (fast)	$\ce{N2O + H2 -> N2 + H2O}$	—

Step 1 is RDS; its rate law matches experiment directly. ✓

Mechanism B (also valid):

Step	Reaction
Step 1 (fast, eq)	$\ce{2NO <=> N2O2}$
Step 2 (slow)	$\ce{N2O2 + H2 -> N2O + H2O}$
Step 3 (fast)	$\ce{N2O + H2 -> N2 + H2O}$

From Step 1 equilibrium: $K = [\ce{N2O2}]/[\ce{NO}]^2$ → $[\ce{N2O2}] = K[\ce{NO}]^2$

RDS rate law: Rate $= k_2[\ce{N2O2}][\ce{H2}] = k_2K[\ce{NO}]^2[\ce{H2}] = k_{\text{obs}}[\ce{NO}]^2[\ce{H2}]$ ✓

Multiple mechanisms can be consistent with the same experimental rate law. Additional evidence (detection of intermediates, isotopic labelling) is needed to distinguish between them.

Reaction Intermediates

• Formed in one elementary step, consumed in a subsequent step
• Do not appear in the overall balanced equation
• Often radicals, carbocations, or unstable complexes
• Their concentration is typically low and they are hard to detect directly

Catalysts

• Increase reaction rate without being consumed (regenerated in the mechanism)
• Provide alternative reaction pathway with lower activation energy
• Do not affect $\Delta H$, equilibrium position, or thermodynamic spontaneity ($\Delta G$ unchanged)
• Reactants and products energy levels remain unchanged — only the barrier height changes

Energy Profile Comparison:

• $E_a$(with catalyst) < $E_a$(without catalyst)
• Transition state energy is lower for the catalyzed pathway
• More particles possess kinetic energy $\geq E_a$ → frequency of effective collisions increases → rate increases

Homogeneous Catalysis

Catalyst is in the same phase as the reactants:

• Example: Iodide-catalyzed decomposition of $\ce{H2O2}$
  • $\ce{H2O2 + I- -> H2O + IO-}$ (slow)
  • $\ce{H2O2 + IO- -> H2O + O2 + I-}$ (fast)
  • $\ce{I-}$ consumed in Step 1, regenerated in Step 2
• Example: Acid-catalyzed esterification

OO

[I-]

Heterogeneous Catalysis

Catalyst is in a different phase from the reactants (typically solid catalyst, gaseous/liquid reactants):

• Reaction occurs at the catalyst surface
• Mechanism: Adsorption → Surface reaction → Desorption
• Examples:
  • Haber process: $\ce{N2 + 3H2 ->[Fe(s)] 2NH3}$
  • Catalytic converters: Pt/Pd/Rh convert $\ce{CO}$, $\ce{NO_x}$, hydrocarbons
  • Hydrogenation: $\ce{C2H4 + H2 ->[Ni(s)] C2H6}$

Comparison: Homogeneous vs Heterogeneous

Feature	Homogeneous	Heterogeneous
Phase (catalyst vs reactants)	Same phase	Different phase
Active site	Individual molecules/ions	Solid surface sites
Separation after reaction	Difficult (distillation, extraction)	Easy (filtration)
Selectivity	Typically high	Moderate
Temperature tolerance	Limited (thermal decomposition)	High

Enzyme Kinetics

Enzymes are biological catalysts — typically proteins — that dramatically accelerate biochemical reactions with extraordinary specificity.

Key Terminology

Term	Definition
Substrate (S)	The reactant molecule that binds to the enzyme
Active site	The specific region of the enzyme where substrate binds and reaction occurs
Enzyme-substrate complex (ES)	The intermediate formed when substrate binds to the enzyme active site
Product (P)	The molecule released after the reaction
Turnover number ($k_{\text{cat}}$)	Maximum number of substrate molecules converted per enzyme molecule per second

General Enzyme-Catalyzed Reaction

$$\ce{E + S <=>[k_1][k_{-1}] ES ->[k_2] E + P}$$

Michaelis-Menten Kinetics

Leonor Michaelis and Maud Menten (1913) developed the fundamental model of enzyme kinetics.

Key assumptions:

1. Enzyme and substrate form an ES complex
2. ES formation is in rapid equilibrium (or steady state)
3. Breakdown of ES to product is the rate-determining step
4. $[\ce{S}] \gg [\ce{E}]$ (substrate in large excess)

The Michaelis-Menten Equation

$$v = \frac{V_{\text{max}}[S]}{K_m + [S]}$$

Where:

• $v$ = initial reaction velocity
• $V_{\text{max}}$ = maximum velocity (all enzyme active sites saturated)
• $[S]$ = substrate concentration
• $K_m$ = Michaelis constant = $[S]$ at which $v = V_{\text{max}}/2$

Interpreting $K_m$

$$K_m = \frac{k_{-1} + k_2}{k_1}$$

$K_m$ measures enzyme-substrate affinity:

• Low $K_m$ → high affinity (tight binding; reaches half-maximum velocity at low $[S]$)
• High $K_m$ → low affinity (weak binding; needs high $[S]$ to reach half-maximum velocity)

When $k_2 \ll k_{-1}$ (product formation is slow compared to ES dissociation), $K_m \approx k_{-1}/k_1 = K_d$ (the dissociation constant of the ES complex).

Kinetic Regimes

Condition	Velocity	Kinetic Order
$[S] \ll K_m$	$v \approx \frac{V_{\text{max}}}{K_m}[S]$	First-order in $[S]$
$[S] = K_m$	$v = V_{\text{max}}/2$	—
$[S] \gg K_m$	$v \approx V_{\text{max}}$	Zero-order in $[S]$ (saturated)

$V_{\text{max}}$ and Turnover Number

$$V_{\text{max}} = k_2[E]_{\text{total}}$$

$$k_{\text{cat}} = \frac{V_{\text{max}}}{[E]_{\text{total}}}$$

$k_{\text{cat}}$ (turnover number) ranges from $10^{-1}$ to $10^7$ s⁻¹. Carbonic anhydrase: ~$10^6$ s⁻¹ — one of the fastest known enzymes.

Lineweaver-Burk Plot (Double Reciprocal)

Taking reciprocals of the Michaelis-Menten equation:

$$\frac{1}{v} = \frac{K_m}{V_{\text{max}}}\frac{1}{[S]} + \frac{1}{V_{\text{max}}}$$

A plot of $1/v$ vs $1/[S]$ yields a straight line:

• Slope = $K_m/V_{\text{max}}$
• Y-intercept = $1/V_{\text{max}}$
• X-intercept = $-1/K_m$

Used experimentally to determine $K_m$ and $V_{\text{max}}$ from initial rate measurements.

Enzyme Inhibition

Type	Binding Site	Effect on $V_{\text{max}}$	Effect on $K_m$
Competitive	Active site (competes with S)	Unchanged	Increases
Non-competitive	Allosteric site (not active site)	Decreases	Unchanged
Uncompetitive	ES complex only	Decreases	Decreases

Biological Significance

• Specificity: Lock-and-key and induced-fit models
• Efficiency: Accelerate reactions by factors of $10^6$ to $10^{17}$
• Mild conditions: Operate at physiological temperature (37°C) and pH (~7.4)
• Regulation: Feedback inhibition, allosteric regulation, covalent modification

Worked Examples

Example: Temperature Effect on Rate Explain how temperature influences rate:

1. Kinetic energy is directly proportional to temperature
2. When temperature increases, kinetic energy increases
3. More molecules have energy ≥ activation energy
4. Effective collisions increase → rate increases

Example: Iodination of Acetone Reaction:

CC(=O)C.II>>CC(=O)CI.I

(CH₃COCH₃ + I₂ → ICH₂COCH₃ + HI)

Given rate constant = 1.8 × 10⁻³ M s⁻¹ (zero order). Ways to increase rate:

• Increase temperature (increases collision frequency and effective collisions)
• Add catalyst (lowers Eₐ)

Example: Arrhenius Calculation — Find k₂ k₁ = 3.46 × 10⁻² s⁻¹ at 298 K. Find k₂ at 350 K if Eₐ = 50.2 J/mol.

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

Solution: k₂ = 3.47 × 10⁻² s⁻¹

Example: Arrhenius Calculation — Find T₂ Eₐ = 50.2 kJ mol⁻¹, k₁ = 3.46 × 10⁻² s⁻¹ at 298 K. Find T₂ for k₂ = 6.30 × 10⁻⁴ s⁻¹.

Solution: T₂ = 248.81 K

Example: Arrhenius Graph — Determine Eₐ For 2HI(g) → H₂(g) + I₂(g), plot ln k vs 1/T. Slope = -Eₐ/R → calculate Eₐ in kJ/mol.

I.I>>[H][H].[I][I]

Related Topics

• Chemical Equilibrium — Kinetics relates to approach to equilibrium
• Thermochemistry — Activation energy and enthalpy

Sources

• FAD1018 W16 — Kinetic Chemistry — Part 1 (pages 1–80): reaction rates, rate laws, integrated rate laws, half-life, collision theory, transition state theory, energy profile diagrams, factors affecting reaction rate
• FAD1018 W16 — Chemical Kinetics Part 2 — Part 2: Arrhenius equation (detailed with worked examples), reaction mechanisms and molecularity, rate-determining step, mechanism validation, homogeneous vs heterogeneous catalysis, enzyme kinetics (Michaelis-Menten, $V_{\text{max}}$, $K_m$, turnover number, Lineweaver-Burk)
• FAD1018 - Basic Chemistry II