Chemical Kinetics

Chemical Kinetics

Part 1 of 5 — Rate Laws & Reaction Order

Reaction Rate

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

For $a\text{A} + b\text{B} \to c\text{C}$:

$\text{rate} = -\frac{1}{a}\frac{\Delta[\text{A}]}{\Delta t} = -\frac{1}{b}\frac{\Delta[\text{B}]}{\Delta t} = +\frac{1}{c}\frac{\Delta[\text{C}]}{\Delta t}$

Rate is always expressed as a positive number; the minus sign accounts for consumption of reactants.

Rate Law

The rate law relates reaction rate to reactant concentrations:

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

• $k$ = rate constant (depends on temperature, not concentration)
• $m$ = order with respect to A
• $n$ = order with respect to B
• Overall reaction order = $m + n$

Rate laws must be determined experimentally — you cannot read them from a balanced equation (unless it is an elementary step).

Determining Order from Experimental Data

Compare experiments where all concentrations except one are held constant:

$\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[\text{A}]_2}{[\text{A}]_1}\right)^m$

Example:

From Exp 1 vs 2: doubling [A] doubles rate → 1st order in A ($m = 1$)

From Exp 1 vs 3: tripling [B] triples rate → 1st order in B ($n = 1$)

Overall: 2nd order; rate $= k[\text{A}][\text{B}]$

$k = \frac{\text{rate}}{[\text{A}][\text{B}]} = \frac{2.0 \times 10^{-4}}{(0.10)(0.10)} = 2.0 \times 10^{-2}\text{ M}^{-1}\text{s}^{-1}$

Units of $k$ by Reaction Order

MCAT tip: The units of $k$ tell you the reaction order!

Zero-Order Reactions

Rate is constant, independent of concentration: $\text{rate} = k$

Example: enzyme-catalyzed reactions at substrate saturation; zero-order because the active sites are all occupied.

Rate Laws & Reaction Order 🎯

Key Takeaways — Part 1

• Rate laws must be determined experimentally from rate/concentration data
• To find order: compare runs where only one concentration changes; use ratio $\text{rate}_2/\text{rate}_1 = ([\text{X}]_2/[\text{X}]_1)^m$
• Units of $k$: 0th order = M/s; 1st order = $s^{-1}$; 2nd order = $M^{-1}s^{-1}$
• Zero-order reactions: rate is constant; seen in enzyme kinetics at Vmax
• Rate laws are determined experimentally — not from balanced equations (unless elementary step)

Chemical Kinetics

Part 2 of 5 — Integrated Rate Laws & Half-Life

Integrated Rate Laws

Integrated rate laws give concentration as a function of time — essential for predicting how much reactant remains after a given time.

Zero-Order Integrated Rate Law

$[\text{A}]_t = [\text{A}]_0 - kt$

Graph: $[\text{A}]$ vs $t$ is a straight line with slope $= -k$.

Half-life: $t_{1/2} = \frac{[\text{A}]_0}{2k}$ (depends on initial concentration)

First-Order Integrated Rate Law

$[\text{A}]_t = [\text{A}]_0 e^{-kt}$

Equivalently: $\ln[\text{A}]_t = \ln[\text{A}]_0 - kt$

Graph: $\ln[\text{A}]$ vs $t$ is a straight line with slope $= -k$.

Half-life:

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

Half-life is constant — independent of $[\text{A}]_0$. This is the hallmark of 1st-order kinetics.

Applications: Radioactive decay, drug elimination from the body.

Second-Order Integrated Rate Law

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

Graph: $1/[\text{A}]$ vs $t$ is a straight line with slope $= +k$.

Half-life: $t_{1/2} = \frac{1}{k[\text{A}]_0}$ (depends on initial concentration; increases over time)

Summary Table

Half-Life Applications (MCAT)

For first-order processes:

$\frac{[\text{A}]_t}{[\text{A}]_0} = \left(\frac{1}{2}\right)^n \quad \text{where } n = \frac{t}{t_{1/2}}$

Example: $k = 0.0693\text{ min}^{-1}$ (1st order). What fraction remains after 30 min?

$t_{1/2} = 0.693/0.0693 = 10\text{ min}$. After 30 min = 3 half-lives.

$\text{fraction remaining} = (1/2)^3 = 1/8 = 12.5\%$

Identifying Reaction Order from Graphs

Integrated Rate Laws & Half-Life 🎯

Key Takeaways — Part 2

• 0th order: $[\text{A}]$ vs $t$ linear; $t_{1/2} = [\text{A}]_0/2k$ (concentration-dependent)
• 1st order: $\ln[\text{A}]$ vs $t$ linear; $t_{1/2} = 0.693/k$ (constant)
• 2nd order: $1/[\text{A}]$ vs $t$ linear; $t_{1/2} = 1/k[\text{A}]_0$ (inversely proportional to concentration)
• 1st-order fraction remaining: $(1/2)^n$ where $n$ = number of half-lives elapsed
• Radioactive decay, drug elimination = 1st-order processes
• Linear graph identity: the order of the reaction is identified by which plot gives a straight line

Chemical Kinetics

Part 3 of 5 — Activation Energy, Arrhenius Equation & Collision Theory

Activation Energy ($E_a$)

The activation energy is the minimum energy that colliding molecules must have for a reaction to occur. The transition state (activated complex) sits at the energy maximum.

Energy diagram key points:

• Reactants → Transition state → Products
• $E_a(\text{forward})$ = energy gap from reactants to peak
• $E_a(\text{reverse})$ = energy gap from products to peak
• $\Delta H = E_a(\text{forward}) - E_a(\text{reverse})$
• Exothermic: products at lower energy than reactants ($\Delta H < 0$)
• Endothermic: products at higher energy than reactants ($\Delta H > 0$)

A catalyst lowers $E_a$ (provides an alternative pathway) without changing $\Delta H$ or the equilibrium constant.

Arrhenius Equation

The Arrhenius equation describes how the rate constant $k$ depends on temperature:

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

where $A$ = frequency factor (pre-exponential factor accounting for collision frequency and orientation), $R = 8.314$ J/mol·K, $T$ = temperature in Kelvin.

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

Two-temperature form (MCAT-useful): $\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$

Qualitative Rule of Thumb: For many reactions near room temperature, a 10°C rise in temperature approximately doubles the rate constant (though the precise factor depends on $E_a$).

Collision Theory

For a reaction to occur, molecules must:

1. Collide with enough energy ($\geq E_a$)
2. Collide with the correct orientation

The fraction of collisions with energy $\geq E_a$:

$f = e^{-E_a/RT}$

This fraction increases exponentially as $T$ increases or $E_a$ decreases, which is why:

• Higher temperature → faster rate
• Catalyst (lower $E_a$) → faster rate

Effect of Temperature on Rate

MCAT application: The Arrhenius equation explains why biological enzymes are so effective — they dramatically lower $E_a$ for biochemical reactions.

Transition State Theory

The transition state (activated complex) is at the peak of the reaction coordinate diagram. It is not a stable product and cannot be isolated.

Intermediate vs. transition state: An intermediate lies in an energy well (local minimum) between two transition states. A transition state is at an energy peak (maximum).

Activation Energy, Arrhenius & Transition State 🎯

Key Takeaways — Part 3

• $E_a$ = minimum energy for a successful reaction; sits at transition state peak
• Arrhenius: $k = Ae^{-E_a/RT}$; larger $E_a$ or lower $T$ → smaller $k$
• Catalyst lowers $E_a$, increases rate, but does NOT change $\Delta H$ or $K_{eq}$
• Two-temperature Arrhenius: $\ln(k_2/k_1) = (E_a/R)(1/T_1 - 1/T_2)$
• Rough rule: 10°C rise ≈ doubles reaction rate (for moderate $E_a$)
• Intermediate: energy valley (local minimum) between two transition states
• Transition state: energy peak (maximum), cannot be isolated

Chemical Kinetics

Part 4 of 5 — Reaction Mechanisms, Rate-Determining Step & Catalysis

Elementary Steps & Molecularity

A reaction mechanism is a proposed sequence of elementary steps that sum to give the overall balanced equation.

Key distinction: For elementary steps only, the rate law can be written directly from the stoichiometry.

Rate-Determining Step (RDS)

The slowest elementary step controls the overall rate. The rate law for the overall reaction reflects the stoichiometry of the RDS.

Mechanism example:

Step 1 (slow, RDS): $\text{NO}_2 + \text{NO}_2 \to \text{NO}_3 + \text{NO}$

Step 2 (fast): $\text{NO}_3 + \text{CO} \to \text{NO}_2 + \text{CO}_2$

From the RDS (Step 1, bimolecular): $\text{rate} = k[\text{NO}_2]^2$

This matches the experimentally observed rate law — the mechanism is consistent.

Checking a Proposed Mechanism

1. Steps must sum to give the overall balanced equation (intermediates cancel)
2. The rate law from the RDS must match the experimentally determined rate law
3. A catalyst that appears in one step must be regenerated in a later step

Intermediates vs. Catalysts in Mechanisms

• Intermediate: Produced in one step, consumed in a later step; does not appear in the overall equation
• Catalyst: Consumed in one step, regenerated in a later step; does not appear in the overall equation but speeds up the reaction

Types of Catalysis

Homogeneous Catalysis

Catalyst and reactants are in the same phase (e.g., acid catalysis in solution).

Example: Acid-catalyzed ester hydrolysis — $\text{H}^+$ speeds up the reaction and is regenerated.

Heterogeneous Catalysis

Catalyst and reactants are in different phases.

Example: Platinum catalyst in catalytic converters (solid Pt surface, gaseous reactants). Mechanism: reactant adsorption onto metal surface → reaction → product desorption.

Enzymatic Catalysis (Biological)

Enzymes are protein catalysts:

• Active site binds substrate (lock-and-key or induced-fit model)
• Michaelis-Menten kinetics: $v = V_{\text{max}}[\text{S}]/(K_m + [\text{S}])$
• $K_m$ = substrate concentration at which rate = $V_{\text{max}}/2$
• Lower $K_m$ → higher affinity for substrate
• At very high $[\text{S}]$: rate = $V_{\text{max}}$ (zero-order in substrate); all active sites saturated

Effect of Inhibitors on Enzyme Kinetics

Mechanisms, RDS & Catalysis 🎯

Key Takeaways — Part 4

• Rate law of overall reaction = rate law of the slowest elementary step (RDS)
• Intermediates: produced then consumed (cancel in overall equation)
• Catalysts: consumed then regenerated (cancel in overall equation)
• Homogeneous catalysis: same phase; Heterogeneous: different phases (e.g., solid metal + gas)
• Enzyme kinetics: Michaelis-Menten; $V_{\text{max}}$ at saturation; $K_m$ = affinity measure
• Competitive inhibitor: $K_m$ ↑, $V_{\text{max}}$ unchanged; Noncompetitive: $V_{\text{max}}$ ↓, $K_m$ unchanged
• A mechanism is valid only if its predicted rate law matches the observed experimental rate law

Chemical Kinetics

Part 5 of 5 — Mixed MCAT Review

High-Yield Checklist

• ✅ Rate law: experimentally determined; rate = $k[\text{A}]^m[\text{B}]^n$
• ✅ Units of $k$: 0th order = M/s; 1st order = $s^{-1}$; 2nd order = $M^{-1}s^{-1}$
• ✅ Integrated rate laws: 0th ($[A]$ vs $t$), 1st ($\ln[A]$ vs $t$), 2nd ($1/[A]$ vs $t$)
• ✅ First-order half-life: $t_{1/2} = 0.693/k$ (constant)
• ✅ Arrhenius: $k = Ae^{-E_a/RT}$; higher $T$ or lower $E_a$ → larger $k$
• ✅ Catalyst: lowers $E_a$; increases $k$; does not change $\Delta H$ or $K_{eq}$
• ✅ RDS: slowest step; determines overall rate law
• ✅ Intermediates cancel from overall equation; catalysts regenerate
• ✅ Enzyme kinetics: $V_{\text{max}}$, $K_m$, competitive vs noncompetitive inhibition

Mixed Kinetics — MCAT-Style Questions 🎯

Kinetics — Complete Topic Summary

Part 1: Rate laws, extracting order from experimental data, units of $k$, zero-order reactions.

Part 2: Integrated rate laws (0th, 1st, 2nd), graphical identification, half-life formulas, first-order $(1/2)^n$ calculations.

Part 3: Activation energy, Arrhenius equation, collision theory, temperature effect on $k$, transition state vs. intermediate.

Part 4: Elementary steps, rate-determining step, substituting intermediates into rate law, catalysis types (homo/hetero/enzymatic), Michaelis-Menten kinetics, inhibitor types.

Part 5: Integrated MCAT practice.

Most Common MCAT Pitfalls

1. Using stoichiometry to write rate laws instead of experimental data
2. Confusing intermediates with catalysts in mechanisms
3. Thinking catalysts change $\Delta H$ or the equilibrium constant (they don't)
4. Forgetting Km is the concentration at half-Vmax (not an affinity in M directly)
5. Misidentifying reaction order from graphs