Kinetics: Energy, Catalysis & Mechanisms

Kinetics: Energy, Catalysis & Mechanisms

Part 1 of 3 — Activation Energy, Arrhenius Equation & Collision Theory

Where this fits. The Kinetics: Rate Laws & Half-Lives lesson covered rate laws (rate $= k[\text{A}]^m[\text{B}]^n$, with orders found by experiment), the rate constant $k$ and half-lives. This lesson explains what sets $k$: activation energy, the Arrhenius equation and collision theory, then how multistep mechanisms, the rate-determining step and catalysts, enzymes included, shape the rate law.

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.

Worked Example — Barriers, ΔH and a Catalyst

Problem. A reaction has $E_a(\text{fwd}) = 80$ kJ/mol and $\Delta H = -30$ kJ/mol. A catalyst lowers the forward barrier to 60 kJ/mol.

• (a) Is the reaction exothermic or endothermic, and where do the products sit?
• (b) What is $E_a(\text{rev})$ without the catalyst?
• (c) What are the forward and reverse barriers with the catalyst?
• (d) Does the catalyst change $\Delta H$ or $K$?

(a) Sign of ΔH. $\Delta H < 0$, so the reaction is exothermic: the products sit 30 kJ/mol below the reactants.

(b) Reverse barrier. From $\Delta H = E_a(\text{fwd}) - E_a(\text{rev})$: $E_a(\text{rev}) = 80 - (-30) = 110$ kJ/mol. Climbing back from the lower product plateau takes more energy.

(c) Catalyzed barriers. The catalyst lowers the single peak by 20 kJ/mol, so BOTH barriers fall by 20: $E_a(\text{fwd}) = 60$ kJ/mol and $E_a(\text{rev}) = 110 - 20 = 90$ kJ/mol.

(d) Thermodynamics. No. With the barriers from (c), $\Delta H = 60 - 90 = -30$ kJ/mol, unchanged: the plateaus never move. Both directions speed up equally, so equilibrium arrives sooner with the same $K$.

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.

Collision Theory

A reaction needs molecules to collide with enough energy ($\geq E_a$) AND in the correct orientation. The fraction with enough energy is about $f = e^{-E_a/RT}$, which grows exponentially as $T$ rises or a catalyst lowers $E_a$.

Anything that raises the collision frequency also speeds the reaction, without changing that fraction:

• Concentration: more reactant (or a higher gas partial pressure) means more collisions per second.
• Surface area: a powdered solid reacts faster than a lump.
• Phase: reactants mixed in one phase collide far more often than separate phases meeting at an interface.

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.

In a rate law, rate $= k[\text{A}]^m[\text{B}]^n$, the concentrations change as the reaction runs at a fixed $k$; only temperature or a catalyst (through $E_a$) changes $k$ itself.

Logarithmic form: a plot of $\ln k$ against $1/T$ is a straight line with slope $-E_a/R$ and intercept $\ln A$; the steeper the line, the larger $E_a$.

$\ln k = \ln A - \frac{E_a}{R}\cdot\frac{1}{T}$

Two-temperature form (MCAT-useful): compares the rate constants at two temperatures directly.

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

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 $E_a$).

Worked Example — Arrhenius and a Fever

Problem. A reaction with $E_a = 50$ kJ/mol runs at body temperature, 37 °C; a fever raises it to 40 °C. ($R = 8.314$ J/mol·K)

• (a) What are the two temperatures in kelvin?
• (b) What is $\ln(k_2/k_1)$?
• (c) By what factor does $k$ rise?

(a) Kelvin. $T_1 = 37 + 273 = 310$ K and $T_2 = 40 + 273 = 313$ K. Arrhenius always needs absolute temperature.

(b) Two-temperature form. With the temperatures from (a):

$\ln\frac{k_2}{k_1} = \frac{50{,}000}{8.314}\left(\frac{1}{310} - \frac{1}{313}\right) = 6014 \times 3.09 \times 10^{-5} \approx 0.186$

(c) Factor. $k_2/k_1 = e^{0.186} \approx 1.20$ (for small $x$, $e^x \approx 1 + x$): about 20% faster, although the absolute temperature rose by only 1%. That exponential sensitivity is why small temperature changes matter so much.

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.

Kinetic vs. Thermodynamic Control

When one reactant can form two products, the kinetic product has the LOWER $E_a$ (it forms faster), while the thermodynamic product is the more STABLE one, whatever its barrier. Barrier height is kinetics; product energy is thermodynamics.

• Kinetic control: at low temperature, over short times, or when the steps are irreversible, neither product can revert, so the faster-forming product wins.
• Thermodynamic control: when the steps are reversible (higher temperature, long times), the mixture equilibrates and the more stable product wins.

Activation Energy, Arrhenius & Transition State 🎯

Key Takeaways — Part 1

• ΔH from the barriers. $\Delta H = E_a(\text{fwd}) - E_a(\text{rev})$; exothermic products sit below the reactants.
• Arrhenius. $k = Ae^{-E_a/RT}$: only well-aimed collisions past $E_a$ react, so higher $T$ or lower $E_a$ raises $k$.
• Catalysts change kinetics only. A catalyst lowers $E_a$ both ways and speeds the reaction, leaving $\Delta H$ and $K$ unchanged.
• Kinetic vs. thermodynamic product. Cold, short or irreversible favors the lower-$E_a$ product; reversible or hot favors the more stable one.

Common traps

• Transition state as intermediate. A transition state is an energy peak that cannot be isolated; an intermediate sits in a valley.
• Celsius in Arrhenius. $T$ in $e^{-E_a/RT}$ must be in kelvin; plugging in °C gives wrong ratios.
• Catalyst shifting equilibrium. It speeds both directions equally, so equilibrium comes sooner with the same $K$.
• Linear temperature thinking. 300 K to 310 K raises $T$ by 3% yet nearly doubles $k$ (for $E_a \approx 50$ kJ/mol), because $k$ depends exponentially on $1/T$.

Kinetics: Energy, Catalysis & Mechanisms

Part 2 of 3 — 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: A rate law has the form rate $= k[\text{A}]^m[\text{B}]^n$. For an elementary step only, each exponent equals that reactant's coefficient in the step: $\text{A} + \text{B} \to$ products gives rate $= k[\text{A}][\text{B}]$, and $2\text{A} \to$ products gives rate $= k[\text{A}]^2$. For an overall reaction, the exponents must be measured.

Termolecular steps are rare, because three particles seldom collide at the same instant with the right energy and orientation.

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, once any intermediate in it is rewritten in terms of reactants (worked example below).

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

Where Kinetics Meets Equilibrium

• At equilibrium the forward and reverse rates are EQUAL: both reactions keep running, but the concentrations stop changing.
• For an elementary step $\text{A} + \text{B} \rightleftharpoons \text{C}$: $k_f[\text{A}][\text{B}] = k_r[\text{C}]$, so $K = \dfrac{[\text{C}]}{[\text{A}][\text{B}]} = \dfrac{k_f}{k_r}$.
• A catalyst raises $k_f$ and $k_r$ by the same factor, so $K$ is unchanged; equilibrium is simply reached sooner.
• Temperature changes $k_f$ and $k_r$ by DIFFERENT factors, because the two directions have different barriers, so $K$ itself moves with temperature. ($Q$ vs $K$ and Le Châtelier: Titrations, Equilibrium & Solubility, Part 2.)

Worked Example — Testing a Mechanism

Problem 1. Proposed mechanism: Step 1 (slow) $\text{NO}_2 + \text{NO}_2 \to \text{NO}_3 + \text{NO}$; Step 2 (fast) $\text{NO}_3 + \text{CO} \to \text{NO}_2 + \text{CO}_2$. Experiment gives rate $= k[\text{NO}_2]^2$.

• (a) What overall equation do the steps add up to?
• (b) Which species is an intermediate, and is there a catalyst?
• (c) What rate law does the mechanism predict?
• (d) Is the mechanism consistent with experiment?

(a) Overall equation. Add the steps and cancel what appears on both sides (NO₃ and one NO₂): $\text{NO}_2 + \text{CO} \to \text{NO} + \text{CO}_2$.

(b) Intermediate. NO₃ is made in Step 1 and used up in Step 2, so it is an intermediate. There is no catalyst: NO₂ is regenerated in Step 2 but consumed overall, since two enter Step 1.

(c) Predicted rate law. Step 1 is the RDS and elementary, bimolecular in NO₂, so rate $= k[\text{NO}_2]^2$. CO reacts only after the bottleneck, so it does not appear.

(d) Consistency. Yes: the steps sum to the overall equation from (a), and the predicted rate law from (c), $k[\text{NO}_2]^2$, matches experiment. Consistent means supported, not proven.

Problem 2. Proposed mechanism: Step 1 (fast, reversible; $k_1$ forward, $k_{-1}$ reverse) $\text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2$; Step 2 (slow, $k_2$) $\text{NOBr}_2 + \text{NO} \to 2\text{NOBr}$.

• (a) What is the overall reaction, and which species is the intermediate?
• (b) What rate law does the slow step give?
• (c) How can the intermediate's concentration be written in terms of reactants?
• (d) What is the rate law, and what is the observed $k$?
• (e) What is the overall order?

(a) Overall and intermediate. Adding the steps cancels NOBr₂: $2\text{NO} + \text{Br}_2 \to 2\text{NOBr}$. NOBr₂ is made in Step 1 and used in Step 2, so it is the intermediate.

(b) Slow-step rate law. Step 2 is the RDS: rate $= k_2[\text{NOBr}_2][\text{NO}]$. It still contains the intermediate, which cannot appear in an experimental rate law.

(c) Fast pre-equilibrium. Step 1 stays at equilibrium, so its forward and reverse rates are equal: $k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2]$. So $[\text{NOBr}_2] = K_1[\text{NO}][\text{Br}_2]$, with $K_1 = k_1/k_{-1}$.

(d) Rate law. Substituting (c) into (b): rate $= k_2K_1[\text{NO}]^2[\text{Br}_2] = k[\text{NO}]^2[\text{Br}_2]$. The observed rate constant is the product $k = k_2K_1 = k_2k_1/k_{-1}$.

(e) Overall order. Second order in NO plus first in Br₂ (sum of exponents): third order overall. It happens to match the overall coefficients, but the reaction is not one termolecular step.

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 = $V_{\text{max}}/2$; a lower $K_m$ means higher affinity for the substrate
• At very high $[\text{S}]$: rate = $V_{\text{max}}$, zero order in substrate (the rate is independent of $[\text{S}]$) because all active sites are saturated

Effect of Inhibitors on Enzyme Kinetics

Mechanisms, RDS & Catalysis 🎯

Key Takeaways — Part 2

• The RDS sets the rate law. Write it from the slowest elementary step; species that react after it never appear.
• Intermediate vs. catalyst. An intermediate is made then used up; a catalyst is used then remade; neither is in the overall equation.
• Testing a mechanism. Its steps must sum to the overall equation and predict the observed rate law, or it is rejected.
• Inhibitor signatures. Competitive raises $K_m$ with $V_{\text{max}}$ unchanged; noncompetitive lowers $V_{\text{max}}$ with $K_m$ unchanged.

Common traps

• An intermediate in the rate law. Replace it using the fast pre-equilibrium, e.g. $[\text{B}] = K_1[\text{A}]$, so only reactants remain.
• Swapping intermediate and catalyst. Check the order: made-then-used is an intermediate; used-then-remade is a catalyst.
• Misjudging the phase. Catalyst in the reactants' phase is homogeneous; a solid Pt surface with gases is heterogeneous.
• A saturated enzyme as first order. At very high [S] the enzyme runs at $V_{\text{max}}$, zero order in substrate.

Kinetics: Energy, Catalysis & Mechanisms

Part 3 of 3 — Mixed MCAT Review

High-Yield Checklist

• 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 🎯

Key Takeaways — Part 3

• Barriers set the speed. $k = Ae^{-E_a/RT}$; a catalyst lowers $E_a$ without moving $\Delta H$ or $K$.
• The slow step rules. The RDS sets the rate law; intermediates cancel and catalysts are regenerated.
• Read the Arrhenius plot. $\ln k$ vs $1/T$ is a line of slope $-E_a/R$: steeper means a larger $E_a$; parallel lines differ only in $A$.
• Michaelis-Menten. $v = V_{\text{max}}[\text{S}]/(K_m + [\text{S}])$: at $[\text{S}] = K_m$ the rate is $V_{\text{max}}/2$; far above $K_m$ it is zero order in substrate.

Common traps

• Catalysts changing $K$. A catalyst speeds both directions equally; $\Delta H$ and $K$ stay the same.
• Misreading $K_m$. $K_m$ is the [S] that gives $V_{\text{max}}/2$; a lower $K_m$ means higher affinity.
• Kinetic product as the stable one. The kinetic product has the lower barrier; the thermodynamic product has the lower energy. They need not be the same species.
• Third order means a three-body step. A rate law like $k[\text{NO}]^2[\text{Br}_2]$ usually comes from a fast pre-equilibrium plus a slow bimolecular step, not one termolecular collision.

Energy, Catalysis & Mechanisms — Complete Topic Summary

Part 1: Activation energy, Arrhenius equation, collision theory, temperature effect on $k$, transition state vs. intermediate.

Part 2: Elementary steps, rate-determining step, substituting intermediates into rate law, catalysis types (homo/hetero/enzymatic), Michaelis-Menten kinetics, inhibitor types.

Part 3: Integrated MCAT practice.

Next up: the Acids, Bases & Buffers lesson — pH and pOH, strong and weak acids, $K_a$ and $K_b$, the Henderson-Hasselbalch equation and physiological buffers.