Chemical Kinetics

Collision Theory of Reaction Rates

Chemical kinetics studies the speed at which chemical reactions occur and the factors that affect these rates. According to collision theory, for a reaction to proceed, reactant molecules must:

• Physically collide: Higher concentrations increase collision frequency, leading to faster reactions.

• Collide with sufficient energy: Collisions must overcome the activation energy barrier (Ea), the minimum energy required to break bonds and form new ones. This is associated with the transition state (activated complex).

• Collide with correct orientation: Molecules must be properly aligned for new bonds to form.

Activation Energy and Transition State

• Activation Energy (Ea): The energy barrier that must be overcome for reactants to become products.

• Transition State: The unstable, high-energy intermediate formed during a successful collision.

• Potential Energy Diagrams: Show energy changes during a reaction.

  • Exothermic: Products have lower energy than reactants; ΔH is negative.

  • Endothermic: Products have higher energy than reactants; ΔH is positive.

Reaction Mechanisms

A reaction mechanism is a sequence of elementary steps that describe how a reaction occurs at the molecular level.

• Elementary Step: A single molecular event. The rate law for an elementary step is determined by its molecularity (number of reactant molecules involved).

  • Unimolecular: A → product; Rate =

  • Bimolecular: A + B → product; Rate =

  • Bimolecular (same species): A + A → product; Rate =

  • Termolecular: A + B + C → product; Rate =

  • Termolecular (same species): A + A + A → product; Rate =

  • Termolecular (two same, one different): A + A + B → product; Rate =

• Intermediate: Produced in one step, consumed in another; does not appear in the overall equation.

• Rate-Limiting Step: The slowest step; determines the overall reaction rate.

Example Mechanism

• 2 NO(g) → N2(g) + O2(g)

• Step 1: 2 NO(g) → N2(g) + 2 O(g) (slow)

• Step 2: 2 O(g) → O2(g) (fast)

• Intermediate: O(g)

Mechanisms with Reversible Steps

When a mechanism includes a fast, reversible step, equilibrium principles are used to relate intermediate concentrations to reactants.

• For a fast equilibrium:

• For a slow step:

• Express the intermediate in terms of reactants using the equilibrium constant.

Reaction Rates and Temperature

• Increasing temperature increases the fraction of molecules with energy ≥ Ea, leading to more effective collisions and a faster reaction.

The Arrhenius Equation

The Arrhenius equation relates the rate constant (k) to temperature (T) and activation energy (Ea):

• Exponential form:

• Linear form:

• Dual temperature form:

• Where A = pre-exponential factor, R = 8.3145 J/(mol·K), T in Kelvin.

Reaction Rate and Catalysts

• Catalyst: Increases reaction rate by lowering Ea via an alternative mechanism.

• Homogeneous: Same phase as reactants (e.g., enzymes).

• Heterogeneous: Different phase (e.g., solid catalyst with gas reactants).

• Catalysts are not consumed, do not appear in the overall equation, and do not affect ΔH.

• They increase both forward and reverse rates equally.

Enzymes and Catalytic Converters

• Enzymes: Biological catalysts, highly specific, operate via a catalytic cycle.

• Catalytic converters: Use precious metals to convert toxic gases in vehicle exhaust.

Reaction Rates of Change

• Average rate: for

• For a general reaction:

• Overall rate:

• Instantaneous rate: Slope of tangent to [A] vs. time curve at a point.

Rate Laws

• General form:

• k = rate constant; x, y, z = reaction orders (determined experimentally).

• Overall order = x + y + z + ...

Method of Initial Rates Example

• Order with respect to A: 0 (rate does not change when [A] doubles)

• Order with respect to B: 1 (rate doubles when [B] doubles)

• Rate law:

• Calculate k:

Integrated Rate Laws

• Zero-order: ; (plot [A] vs. t is linear, slope = -k)

• First-order: ; (plot ln[A] vs. t is linear, slope = -k)

• Second-order: ; (plot 1/[A] vs. t is linear, slope = k)

Half-Lives (t1/2)

• Zero-order: (depends on initial concentration)

• First-order: (independent of initial concentration)

• Second-order: (inversely proportional to initial concentration)

Gases and Their Laws

Visualizing Gas Pressure

• Gases have no fixed volume or shape; they fill their container.

• Key variables: P (pressure), V (volume), T (temperature).

• Pressure: (force per unit area)

• Common units: 1 atm = 760 mm Hg = 101,325 Pa = 101.3 kPa = 14.7 psi

Boyle's Law

• At constant T and n, P and V are inversely proportional:

• As volume decreases, pressure increases.

Charles's Law

• At constant P and n, V and T are directly proportional:

• Temperature must be in Kelvin.

• Explains hot air balloon operation and the concept of absolute zero (0 K = -273.15°C).

Combined Gas Law

• Combines Boyle's and Charles's Laws:

• Use when P, V, and T all change but n is constant.

Avogadro's Law

• At constant P and T, V and n are directly proportional:

• Equal volumes of gases at the same T and P contain equal numbers of molecules.

Ideal Gas Law

• Relates P, V, n, and T:

• R = 0.0821 L·atm/(mol·K) (use consistent units)

Standard Temperature and Pressure (STP) and Standard Molar Volume

• STP: 273.15 K (0°C), 1.00 bar (≈1 atm)

• Standard molar volume: 1 mol gas = 22.4 L at STP

Density of Gases and Molar Masses

• Density:

• Molar mass:

• Density depends on both molar mass and conditions (P, T).

Dalton's Law of Partial Pressures

• Total pressure = sum of partial pressures:

• Partial pressure: , where

Stoichiometric Calculations for Gas Reactions

• Use balanced equations and the ideal gas law to relate volumes, moles, and masses.

• General steps: identify known/unknown, convert to moles, use mole ratios, convert to desired units.

Example Problem

How many grams of NaN3(s) are needed to produce 75.0 L of N2 at 25°C and 0.9842 atm?

1. Convert T to K: 25°C + 273.15 = 298.15 K

2. Use : mol N2

3. Mole ratio: 6 NaN3 : 9 N2 → mol NaN3

4. Molar mass NaN3: 65.02 g/mol

5. Mass: g NaN3

Thermochemistry Principles

The First Law of Thermodynamics

• Energy is conserved: it can be transferred as heat (q) or work (w), but not created or destroyed.

• System: the chemicals reacting; surroundings: everything else.

Heat and Temperature

• Heat (q): Energy transfer due to temperature difference.

• Temperature: Average kinetic energy of particles.

Exothermic and Endothermic Processes

• Exothermic: Releases heat; ΔH < 0; surroundings warm up.

• Endothermic: Absorbs heat; ΔH > 0; surroundings cool down.

Heat Capacity and Specific Heat Capacity

• Heat capacity (C): (J/°C or J/K)

• Specific heat capacity (c): (J/g·°C)

• Molar heat capacity: Heat needed to raise 1 mol by 1°C or 1 K.

Calorimetry

• Measures heat changes using calorimeters (coffee-cup for constant P, bomb for constant V).

• For coffee-cup calorimeter:

• If solution is water: , J/g·°C

Enthalpy (H) and Change in Enthalpy (ΔH)

• Enthalpy: (U = internal energy, P = pressure, V = volume)

• Change in enthalpy: (heat at constant pressure)

• Sign convention: ΔH < 0 (exothermic), ΔH > 0 (endothermic)

Calculating Heat from ΔH

• Use stoichiometry:

• Example: For , kJ

• Heat released for 1 mol Fe2O3: kJ

• Heat released for 1 g Fe: mol; kJ

Hess's Law

• If a reaction is the sum of several steps, is the sum of for each step.

• If a reaction is reversed, change the sign of ; if multiplied, multiply $\Delta H$ by the same factor.

Example: Calculate for

• Reverse and multiply reactions as needed, sum values to get kJ

Standard Enthalpy of Formation ()

• Enthalpy change when 1 mol of compound forms from elements in standard states at 25°C, 1 atm.

• for elements in standard state = 0

Calculating from

• Example: Combustion of glucose:

• Given: kJ/mol, kJ/mol, kJ/mol,

• Calculation: kJ