Chapter 5: Chemical Kinetics

Imagine you are timing how fast ice melts in a glass of water. You could measure how much ice is left every five minutes and calculate an average melting rate. Or you could zoom in on one specific moment and ask: “How fast is the ice melting right now?” Both approaches give you useful information, but they answer slightly different questions.

Chemical reaction rates work the same way. We can measure an average rate over a time interval, or we can determine the instantaneous rate at a specific point in time.

What Is Reaction Rate?

The reaction rate measures how quickly the concentration of a reactant decreases or a product increases over time. Rate is always positive, and it is expressed in units of molarity per second (M/s or mol/L·s).

For reactants, concentration decreases over time, so the change in concentration is negative. We add a negative sign to make the rate positive. For products, concentration increases, so no sign correction is needed.

Average Rate vs. Instantaneous Rate

The average rate is calculated over a time interval. If the concentration of reactant A drops from 0.50 M to 0.30 M over 10 seconds:

Average rate = (0.50 - 0.30) / 10 = 0.020 M/s

The instantaneous rate is the rate at one specific moment in time. Mathematically, it is the slope of the tangent line to the concentration-vs-time curve at that point. On the MCAT, you will most often work with initial rates - the instantaneous rate at the very beginning of the reaction (t = 0).

Stoichiometry and Rate

Different species in a reaction may disappear or appear at different rates because of their stoichiometric coefficients. Consider the reaction:

2 A + B → C

Two moles of A are consumed for every one mole of B. So A disappears twice as fast as B. To define a single “rate of reaction” that is the same regardless of which species you track, we divide each rate of change by the stoichiometric coefficient.

For example, in the reaction N₂ + 3 H₂ → 2 NH₃:

• H₂ disappears 3 times faster than N₂
• NH₃ appears 2 times faster than N₂
• But the rate of reaction is the same no matter which species you track, once you divide by the coefficient

Units of Rate

Rate is always expressed as concentration per unit time. On the MCAT, this is almost always M/s (molarity per second), sometimes written as mol/(L·s).

If reaction rate is the speedometer, the rate law is the owner’s manual that explains what controls the speed. The rate law is a mathematical equation that relates the reaction rate to the concentrations of the reactants. It is the single most important equation in this chapter.

The Rate Constant (k)

The rate constant k is a proportionality constant that is specific to a particular reaction at a particular temperature. A larger k means a faster reaction. Two key facts about k:

1. k changes with temperature. Increase the temperature and k increases (the reaction gets faster). This relationship is described by the Arrhenius equation, which we cover in Section 5.7.
2. k does not change with concentration. For a given reaction at a fixed temperature, k is truly constant regardless of how much reactant you add.

Reaction Orders

The exponents x and y in the rate law are called the reaction orders. They tell you how sensitive the rate is to changes in each reactant’s concentration.

• Order 0: Rate is independent of that reactant’s concentration. Doubling [A] has no effect on rate.
• Order 1: Rate is directly proportional to concentration. Double [A], double the rate.
• Order 2: Rate is proportional to the square of concentration. Double [A], quadruple the rate.

The overall reaction order is the sum of all individual orders. For rate = k[A]²[B]¹, the overall order is 2 + 1 = 3 (third-order overall).

Two Exceptions Where Orders Match Coefficients

There are exactly two situations where the orders in the rate law match the stoichiometric coefficients:

1. The reaction occurs in a single elementary step. If the balanced equation represents the actual molecular event, then the coefficients are the orders.
2. You are given the mechanism and told which step is rate-determining. The coefficients of the reactants in the slow step become the orders.

In every other case, the rate law must be determined experimentally.

Rate Law vs. Equilibrium Expression

This is one of the most common traps on the MCAT. The rate law and the equilibrium expression look similar but mean completely different things.

Units of k

The units of k depend on the overall reaction order. This is because the rate always has units of M/s, so k must have whatever units are needed to make the math work.

The MCAT will not ask you to derive a rate law from first principles. Instead, it will hand you a table of experimental data and ask you to figure out the rate law from the numbers. This is one of the most predictable question types on the exam, and the method is the same every time.

The Method of Initial Rates

Here is the step-by-step process:

Step 1: Find two trials where only ONE reactant’s concentration changes while everything else stays the same.

Step 2: Compare how the rate changed when that concentration changed.

Step 3: Use the relationship Δrate = (Δconcentration)^order to solve for the order.

Step 4: Repeat for each reactant.

Step 5: Plug the orders back into the rate law and use any trial to solve for k.

Worked Example

Consider the reaction A + B → C + D at 300 K:

Finding the order with respect to B: Compare Trials 1 and 2. [A] is held constant at 1.00 M. [B] doubles from 1.00 to 2.00 M. The rate increases from 2.0 to 8.1 - approximately a factor of 4.

Since (2)^y = 4, y = 2. The reaction is second order with respect to B.

Finding the order with respect to A: Compare Trials 2 and 3. [B] is held constant at 2.00 M. [A] doubles from 1.00 to 2.00 M. The rate increases from 8.1 to 15.9 - approximately a factor of 2.

Since (2)^x = 2, x = 1. The reaction is first order with respect to A.

The rate law: rate = k[A]¹[B]² = k[A][B]²

Finding k: Plug in values from Trial 1:

2.0 = k(1.00)(1.00)²

k = 2.0 M⁻²·s⁻¹

Quick Reference: Common Patterns

What If No Two Trials Hold Everything Else Constant?

Sometimes the MCAT will give you data where no pair of trials isolates a single variable perfectly. In that case, you may need to:

1. Use the rate law algebraically. Write the rate law for two different trials and divide one equation by the other to cancel k.
2. If you already know one order, substitute it in and solve for the remaining unknown.

This is less common, but it does appear in harder questions.

A zero-order reaction is the simplest type of kinetics: the rate does not depend on the concentration of reactants at all. No matter how much reactant you add, the reaction proceeds at the same constant speed.

Rate Law

For a zero-order reaction:

Integrated Rate Law

The integrated rate law lets you calculate the concentration of a reactant at any time t:

Graphical Analysis

Plotting [A] vs. time for a zero-order reaction gives a straight line with:

• Slope = -k (negative because concentration decreases)
• y-intercept = [A]₀

This is the simplest graph you will see in kinetics. If [A] vs. t is linear and decreasing, the reaction is zero order.

Half-Life

The half-life of a zero-order reaction is the time it takes for the concentration to drop to half its initial value:

What Can Change the Rate?

Since the rate equals k, the only ways to change the rate of a zero-order reaction are:

1. Change the temperature - this changes k via the Arrhenius equation
2. Add a catalyst - this lowers the activation energy, increasing k

Changing reactant concentrations has no effect. This is the defining feature of zero-order kinetics.

Enzyme kinetics is the classic place to see both orders in one curve. Slide the
substrate concentration: at low [S] the rate climbs with concentration, which is
first-order behavior, and once every active site is occupied the rate plateaus at
V_max, which is zero-order.

First-order reactions are the most important reaction order for the MCAT. Radioactive decay, many drug metabolism pathways, and numerous biological processes follow first-order kinetics. If you learn only one reaction order thoroughly, make it this one.

Rate Law

Integrated Rate Law

Graphical Analysis

• [A] vs. time: Curved (exponential decay). NOT linear.
• ln[A] vs. time: Straight line with slope = -k and y-intercept = ln[A]₀.

If you plot the data and ln[A] vs. t gives a straight line, you have confirmed first-order kinetics. This is the diagnostic test for first-order reactions.

Half-Life: The Star of First-Order Kinetics

The half-life is the time required for the concentration to drop to half its initial value. For first-order reactions:

Calculating with Half-Lives

After n half-lives, the fraction of the original sample remaining is ($\frac{1}{2}$)^n:

Radioactive Decay

Radioactive decay is the classic example of a first-order process. The rate of decay depends only on the amount of radioactive isotope present, not on temperature, pressure, or chemical environment.

Pharmacokinetics Connection

Most drugs are eliminated from the body via first-order kinetics. The half-life of a drug tells physicians how often to dose: if the half-life is 6 hours, giving a dose every 6 hours maintains a relatively stable blood concentration. This is why “take every 4-6 hours” appears on medication labels.

A second-order reaction is one where the rate depends on the square of one reactant’s concentration, or on the product of two reactants’ concentrations. Physically, this often corresponds to a reaction that requires two molecules to collide - a bimolecular event.

Rate Law

Second-order reactions can take two forms:

When the rate law is k[A][B] with first-order dependence on each of two reactants, it suggests a collision between one molecule of A and one molecule of B as the rate-determining step.

Integrated Rate Law

For a reaction that is second order with respect to a single reactant:

Graphical Analysis

• [A] vs. time: Curved (not linear). The curve decreases steeply at first, then levels off.
• ln[A] vs. time: Also curved (NOT linear). This rules out first-order.
• 1/[A] vs. time: Straight line with slope = +k and y-intercept = 1/[A]₀.

If 1/[A] vs. t gives a straight line, the reaction is second order. Note that the slope is positive (unlike zero-order and first-order, where the slope is negative).

Half-Life

This makes intuitive sense with the dance partner analogy: as the party empties out, it takes longer and longer to find someone to dance with.

Master Comparison Table

We know that increasing the temperature speeds up nearly all reactions. But by how much? The Arrhenius equation gives us the precise mathematical relationship between the rate constant k, the temperature T, and the activation energy $E_{a}$.

The Equation

Understanding the Variables

The frequency factor (A) represents how often molecules collide with the correct orientation. It depends on the physical characteristics of the molecules (size, shape) and the concentration. A larger A means more frequent, correctly oriented collisions.

The exponential term (e^(-$E_{a}$/RT)) represents the fraction of molecules that have enough energy to overcome the activation energy barrier. This fraction is always between 0 and 1.

How Temperature Affects k

The key to understanding the Arrhenius equation on the MCAT is the exponent -$E_{a}$/RT:

• Increasing T makes the denominator RT larger, making $E_{a}$/RT smaller, making -$E_{a}$/RT less negative (closer to zero). Since e⁰ = 1, the exponential term gets larger, so k increases.
• Increasing $E_{a}$ makes $E_{a}$/RT larger, making -$E_{a}$/RT more negative. Since e^(-large number) is tiny, k decreases.
• As T approaches infinity, -$E_{a}$/RT approaches zero, and k approaches A (the maximum possible rate constant).

The Arrhenius Plot

Taking the natural log of both sides of the Arrhenius equation gives a linear form:

Plotting ln(k) vs. 1/T gives a straight line. The slope is -$E_{a}$/R, so you can determine the activation energy from the slope:

$E_{a}$ = -slope x R

A steeper (more negative) slope means a higher activation energy. The reaction is more sensitive to temperature changes.

The Two-Point Form

If you know the rate constant at two different temperatures, you can find $E_{a}$ without plotting:

Why do reactions happen at all? At the molecular level, the answer is simple: molecules must collide. But not just any collision will do. Collision theory lays out the rules for what makes a collision productive.

The Two Requirements for an Effective Collision

For a collision between reactant molecules to produce products, two conditions must be met simultaneously:

1. Sufficient energy. The colliding molecules must have a combined kinetic energy equal to or greater than the activation energy ($E_{a}$). If they collide too gently, they simply bounce off each other unchanged.

2. Correct orientation. The reactive parts of the molecules must be facing each other during the collision. A molecule of A might need to collide with the oxygen end of molecule B, not the nitrogen end.

Collisions that meet both requirements are called effective collisions (or productive collisions). Only effective collisions lead to product formation.

The Rate Equation from Collision Theory

Factors That Increase the Collision Rate

Collision theory directly explains why each of these factors increases reaction rate:

Concentration: More molecules in a given volume means more collisions per second (Z increases). This is why increasing concentration usually increases rate - except for zero-order reactions, where the bottleneck is elsewhere.

Temperature: Temperature has a double effect. First, molecules move faster, so they collide more frequently (Z increases slightly). Second, and much more importantly, a higher fraction of molecules have enough energy to exceed $E_{a}$ (f increases dramatically). The temperature effect is dominated by the energy factor.

Surface area (for heterogeneous reactions): Grinding a solid into a fine powder exposes more surface for collisions. This is why powdered sugar dissolves faster than sugar cubes.

Medium and viscosity: The physical environment of the reaction affects how freely molecules can move and collide. In a low-viscosity medium like water, molecules diffuse quickly and collide often. In a high-viscosity medium like glycerol or a thick gel, molecular movement is restricted, collisions are less frequent, and the reaction slows down. This is one reason biological reactions rely on enzymes - in the crowded, viscous environment of the cytoplasm, uncatalyzed collisions would be too infrequent to sustain life.

The Maxwell-Boltzmann Distribution

At any given temperature, not all molecules move at the same speed. The Maxwell-Boltzmann distribution describes the range of kinetic energies in a sample of gas molecules.

At a given temperature, some molecules move slowly, most move at moderate speed, and a few move very fast. The area under the curve to the right of $E_{a}$ represents the fraction of molecules with enough energy to react.

When temperature increases:

• The peak of the distribution shifts to the right (higher average energy)
• The curve broadens and flattens
• The area to the right of $E_{a}$ increases significantly
• More molecules can clear the activation energy barrier

Collision Theory vs. Transition State Theory

Collision theory and transition state theory are two models that explain the same phenomenon from different perspectives:

Both theories require the activation energy to be overcome. They are complementary, not contradictory.

If collision theory is the “bird’s-eye view” of reactions (molecules smashing into each other), transition state theory zooms in to show exactly what happens during the collision itself. It tracks the energy of the system as reactants transform into products, step by step.

The Transition State (Activated Complex)

The transition state is the highest-energy point along the reaction pathway. At this point, old bonds are partially broken and new bonds are partially formed. The transition state is denoted by the symbol ‡ (double dagger).

Key properties of transition states:

• They exist at the energy maximum on the reaction coordinate diagram
• They are theoretical structures that cannot be isolated or observed directly
• They have an extremely brief lifetime (on the order of femtoseconds)
• Once formed, they can proceed forward to products OR fall back to reactants

Reading Reaction Coordinate Diagrams

A reaction coordinate diagram contains several important pieces of information:

Activation energy ($E_{a}$ forward): The energy difference between the reactants and the transition state peak. This is the barrier the forward reaction must overcome.

Activation energy ($E_{a}$ reverse): The energy difference between the products and the transition state peak. This is the barrier the reverse reaction must overcome.

Free energy of reaction (ΔG): The energy difference between products and reactants. This determines whether the reaction is exergonic or endergonic.

Exergonic vs. Endergonic Reactions

The Relationship Between $E_{a}$ and ΔG

A critical concept: activation energy and ΔG are independent. A reaction can be exergonic (thermodynamically favorable) but still have a high activation energy (kinetically slow). Diamond converting to graphite is thermodynamically favorable but kinetically so slow that your engagement ring is safe.

On the diagram:

• ΔG = $E_{\text{products}}$ - $E_{\text{reactants}}$ (where on the y-axis the reaction ends vs. starts)
• $E_{a}$ = $E_{\text{transition}}$ state - $E_{\text{reactants}}$ (how high you must climb)

Changing one does not automatically change the other. You can lower $E_{a}$ (with a catalyst) without changing ΔG.

Intermediates vs. Transition States

For multi-step reactions, the reaction coordinate diagram has multiple peaks and valleys:

An intermediate sits in a valley between two peaks. It is a real molecule with a finite lifetime, formed after one elementary step and consumed in the next.

A catalyst is a substance that speeds up a reaction without being consumed in the process. It participates in the reaction mechanism, but it is regenerated by the end - what goes in comes back out unchanged. Catalysts are arguably the most important concept connecting general chemistry kinetics to biochemistry, because enzymes are biological catalysts.

How Catalysts Work

Catalysts lower the activation energy (E_a) of a reaction. They do this by providing an alternative reaction pathway - a different mechanism with a lower energy barrier.

What catalysts DO:

• Lower E_a for both the forward AND reverse reactions (by the same amount)
• Increase the rate constant k (because k depends on E_a through the Arrhenius equation)
• Speed up the approach to equilibrium
• Interact with reactants (through adsorption or intermediate formation), then are regenerated

What catalysts DO NOT do:

• Change ΔG or ΔH of the reaction
• Change the equilibrium position (K_eq is unchanged)
• Change the concentrations of reactants or products at equilibrium
• Get consumed in the overall reaction

Types of Catalysts

Homogeneous Catalysis

In homogeneous catalysis, the catalyst is in the same phase as the reactants. For example, an acid catalyst dissolved in an aqueous solution with aqueous reactants.

Example: The decomposition of hydrogen peroxide (H₂O₂) is catalyzed by iodide ions (I⁻) in aqueous solution. Both the catalyst and reactant are dissolved in water.

Heterogeneous Catalysis

In heterogeneous catalysis, the catalyst is in a different phase from the reactants. The most common example is a solid catalyst with gaseous or liquid reactants.

Example: A catalytic converter in a car uses solid platinum and palladium metals to catalyze the conversion of toxic exhaust gases (CO, NO) into less harmful products (CO₂, N₂). The solid metal surface provides sites where gas molecules can adsorb, react, and then desorb as products.

Heterogeneous catalysts work through adsorption - reactant molecules bind to the surface of the catalyst, which weakens their bonds and brings them into close proximity, making it easier for them to react.

Enzymatic Catalysis

Enzymes are biological catalysts - proteins that catalyze specific reactions in living organisms. They are extraordinarily efficient, often increasing reaction rates by factors of $10^6$ to $10^{12}$ compared to the uncatalyzed reaction.

Key enzyme concepts for kinetics:

• Enzymes lower E_a by stabilizing the transition state
• At saturation (all active sites occupied), enzyme-catalyzed reactions become zero-order with respect to substrate
• Enzymes do not change ΔG - they only speed up reactions that are already thermodynamically favorable

Catalyst Effect on Reaction Coordinate Diagrams

On a reaction coordinate diagram, adding a catalyst creates a new curve that:

• Starts at the same energy level (same reactants)
• Ends at the same energy level (same products)
• Has a lower peak (reduced activation energy)
• May have a different shape (different mechanism, possibly more steps)

The ΔG remains identical. Only the barrier height changes.

Most chemical reactions do not happen in a single collision. They proceed through a series of simpler steps, each involving only one or two molecules at a time. The complete sequence of these steps is the reaction mechanism - it is the molecular-level story of how reactants become products.

Elementary Steps

An elementary step is a single molecular event - one collision, one bond-breaking, one rearrangement. Unlike the overall balanced equation, elementary steps describe what actually happens at the molecular level.

For an elementary step (and ONLY for elementary steps), the rate law can be written directly from the stoichiometric coefficients. This is the one exception to the rule that “you can’t get the rate law from the equation.”

Molecularity

Molecularity describes how many reactant molecules participate in an elementary step:

The Rate-Determining Step (RDS)

The rate-determining step is the slowest elementary step in the mechanism. It acts as a kinetic bottleneck.

Key facts about the RDS:

• The overall reaction rate equals the rate of the RDS
• The rate law for the overall reaction is determined by the RDS
• On a reaction coordinate diagram, the RDS corresponds to the highest energy barrier (tallest peak)
• Speeding up the RDS speeds up the entire reaction

Deriving Rate Laws from Mechanisms

When the MCAT gives you a mechanism and asks for the rate law:

Step 1: Identify the slow (rate-determining) step.

Step 2: Write the rate law using the reactants and coefficients of that step. Since the RDS is an elementary step, you CAN use the coefficients as orders.

Step 3: Check whether any intermediates appear in your rate law. If so, you need to eliminate them (see Section 5.12 on the steady-state approximation).

Example: A Two-Step Mechanism

Consider:

Step 1: NO₂ + NO₂ → NO₃ + NO (slow)

Step 2: NO₃ + CO → NO₂ + CO₂ (fast)

Overall: NO₂ + CO → NO + CO₂

The slow step is Step 1, so the rate law is:

rate = k[NO₂]²

Notice that CO does not appear in the rate law even though it is a reactant in the overall equation. This is because CO only participates in the fast step, which is not rate-limiting. The MCAT loves this type of question.

Identifying Intermediates

An intermediate is a species that:

• Is produced in one step and consumed in a later step
• Does not appear in the overall balanced equation
• Appears as a product in one elementary step and a reactant in another

In the example above, NO₃ is an intermediate. It is produced in Step 1 and consumed in Step 2. Cancel it out when adding the steps together, and it disappears from the overall equation.

Valid Mechanisms

For a proposed mechanism to be valid, it must satisfy two criteria:

1. The elementary steps must add up to the overall balanced equation. All intermediates must cancel out.
2. The rate law derived from the mechanism must match the experimentally determined rate law. If the predicted and observed rate laws disagree, the mechanism is wrong.

Sometimes the rate-determining step of a mechanism involves an intermediate - a species that does not appear in the overall reaction. Since intermediates are not something you can easily measure or control in an experiment, their concentrations cannot appear in the final rate law. You need a way to replace the intermediate with reactant concentrations. That is what the steady-state and pre-equilibrium approximations do.

The Problem: Intermediates in the Rate Law

Consider this mechanism:

Step 1: A + B ⇌ C (fast, reversible)

Step 2: C + D → E (slow)

The slow step determines the rate law: rate = k₂[C][D]

But C is an intermediate. We cannot report a rate law with [C] in it because C is not a reactant we can measure directly. We need to express [C] in terms of [A], [B], and other measurable quantities.

The Pre-Equilibrium Approximation

The pre-equilibrium approximation applies when a fast, reversible step occurs before the slow step. Since Step 1 is fast and reversible, it reaches equilibrium much faster than Step 2 consumes C. We can treat Step 1 as being at equilibrium:

K_eq = k₁/k₋₁ = [C] / ([A][B])

Solving for [C]: [C] = (k₁/k₋₁)[A][B]

Substituting into the rate law:

rate = k₂[C][D] = k₂(k₁/k₋₁)[A][B][D]

Combining constants: rate = $k_{\text{obs}}$[A][B][D]

where $k_{\text{obs}}$ = k₁k₂/k₋₁

When to Use Pre-Equilibrium

Use the pre-equilibrium approximation when:

• A fast, reversible step occurs before the slow step
• The fast step has time to reach equilibrium because the slow step is much slower

This is the most common scenario on the MCAT.

The Steady-State Approximation

The steady-state approximation is a more general approach. It assumes that after a brief initial period, the concentration of an intermediate remains approximately constant - its rate of production equals its rate of consumption.

For the intermediate C:

Rate of production of C = Rate of consumption of C

k₁[A][B] = k₋₁[C] + k₂[C][D]

Solve for [C] and substitute into the rate law for the slow step.

The steady-state approximation is more mathematically involved than pre-equilibrium. On the MCAT, pre-equilibrium is far more common, but you should recognize the steady-state concept if it appears in a passage.

Comparison of Approaches

Worked Example

Consider the mechanism for the reaction 2 NO + O₂ → 2 NO₂:

Step 1: 2 NO ⇌ N₂O₂ (fast, reversible)

Step 2: N₂O₂ + O₂ → 2 NO₂ (slow)

Rate from slow step: rate = k₂[N₂O₂][O₂]

N₂O₂ is an intermediate. Using pre-equilibrium on Step 1:

K = [N₂O₂] / [NO]²

[N₂O₂] = K[NO]²

Substituting: rate = k₂ · K · [NO]² · [O₂] = $k_{\text{obs}}$[NO]²[O₂]

The experimentally observed rate law is rate = k[NO]²[O₂], which matches. This confirms the proposed mechanism is consistent with the data.