Chemical KineticsTopic #27 of 40

Rate Laws

Determining rate constants and reaction orders from experimental data.

Overview

Rate laws (rate equations) express the relationship between reaction rate and reactant concentrations. They must be determined experimentally and help predict how reactions proceed.

General Form

For reaction: aA+bBProductsa\text{A} + b\text{B} \rightarrow \text{Products}

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

Where:

  • kk = rate constant
  • [A],[B][\text{A}], [\text{B}] = reactant concentrations
  • m,nm, n = reaction orders (determined experimentally)

Reaction Order

Definition

The power to which concentration is raised in the rate law.

Types

  • Zero order (n=0n = 0): Rate independent of concentration
  • First order (n=1n = 1): Rate directly proportional to concentration
  • Second order (n=2n = 2): Rate proportional to concentration squared

Overall Order

Sum of all individual orders:

Overall Order=m+n+\text{Overall Order} = m + n + \cdots

Determining Order from Data

Method of Initial Rates

Compare experiments where only one concentration changes:

Exp[A][B]Initial Rate
10.100.102.0×1032.0 \times 10^{-3}
20.200.108.0×1038.0 \times 10^{-3}
30.100.204.0×1034.0 \times 10^{-3}

Finding order with respect to A (compare Exp 1 and 2):

[A] doubles (×2\times 2), Rate quadruples (×4\times 4)

2m=4    m=2 (second order in A)2^m = 4 \implies m = 2 \text{ (second order in A)}

Finding order with respect to B (compare Exp 1 and 3):

[B] doubles (×2\times 2), Rate doubles (×2\times 2)

2n=2    n=1 (first order in B)2^n = 2 \implies n = 1 \text{ (first order in B)}

Rate Law:

Rate=k[A]2[B]\text{Rate} = k[\text{A}]^2[\text{B}]

Overall order = 3

Finding k:

k=Rate[A]2[B]=2.0×103(0.10)2(0.10)=2.0 M2s1k = \frac{\text{Rate}}{[\text{A}]^2[\text{B}]} = \frac{2.0 \times 10^{-3}}{(0.10)^2(0.10)} = 2.0 \text{ M}^{-2}\text{s}^{-1}

Integrated Rate Laws

Zero Order

[A]=[A]0kt[\text{A}] = [\text{A}]_0 - kt
  • Linear plot: [A] vs t
  • Half-life: t1/2=[A]02kt_{1/2} = \frac{[\text{A}]_0}{2k}
  • Units of k: M/s

First Order

ln[A]=ln[A]0kt\ln[\text{A}] = \ln[\text{A}]_0 - kt [A]=[A]0ekt[\text{A}] = [\text{A}]_0 e^{-kt}
  • Linear plot: ln[A]\ln[\text{A}] vs t
  • Half-life: t1/2=0.693kt_{1/2} = \frac{0.693}{k}
  • Units of k: s⁻¹

Second Order

1[A]=1[A]0+kt\frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + kt
  • Linear plot: 1/[A]1/[\text{A}] vs t
  • Half-life: t1/2=1k[A]0t_{1/2} = \frac{1}{k[\text{A}]_0}
  • Units of k: M⁻¹s⁻¹

Summary Table

OrderRate LawIntegrated LawLinear PlotHalf-lifek Units
0Rate=k\text{Rate} = k[A]=[A]0kt[\text{A}] = [\text{A}]_0 - kt[A] vs t[A]02k\frac{[\text{A}]_0}{2k}M/s
1Rate=k[A]\text{Rate} = k[\text{A}]ln[A]=ln[A]0kt\ln[\text{A}] = \ln[\text{A}]_0 - ktln[A]\ln[\text{A}] vs t0.693k\frac{0.693}{k}s⁻¹
2Rate=k[A]2\text{Rate} = k[\text{A}]^21[A]=1[A]0+kt\frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + kt1[A]\frac{1}{[\text{A}]} vs t1k[A]0\frac{1}{k[\text{A}]_0}M⁻¹s⁻¹

Determining Order from Graphs

Plot data and see which gives a straight line:

  • [A] vs t is linear → Zero order
  • ln[A]\ln[\text{A}] vs t is linear → First order
  • 1/[A]1/[\text{A}] vs t is linear → Second order

Examples

Example 1: First Order Decay

A first-order reaction has k=0.035k = 0.035 s⁻¹. How long for [A] to drop from 0.50 M to 0.10 M?

ln[A]=ln[A]0kt\ln[\text{A}] = \ln[\text{A}]_0 - kt ln(0.10)=ln(0.50)(0.035)t\ln(0.10) = \ln(0.50) - (0.035)t 2.303=0.6930.035t-2.303 = -0.693 - 0.035t t=46 st = 46 \text{ s}

Example 2: Finding Half-Life

For a first-order reaction with k=5.0×103k = 5.0 \times 10^{-3} s⁻¹:

t1/2=0.693k=0.6935.0×103=139 st_{1/2} = \frac{0.693}{k} = \frac{0.693}{5.0 \times 10^{-3}} = 139 \text{ s}

Example 3: Second Order

For 2AProducts2\text{A} \rightarrow \text{Products}, k=0.50k = 0.50 M⁻¹s⁻¹, [A]0=0.20[\text{A}]_0 = 0.20 M

Find [A] after 5.0 s.

1[A]=1[A]0+kt\frac{1}{[\text{A}]} = \frac{1}{[\text{A}]_0} + kt 1[A]=10.20+(0.50)(5.0)=5.0+2.5=7.5\frac{1}{[\text{A}]} = \frac{1}{0.20} + (0.50)(5.0) = 5.0 + 2.5 = 7.5 [A]=0.13 M[\text{A}] = 0.13 \text{ M}

Pseudo-First-Order Reactions

When one reactant is in large excess, its concentration effectively doesn't change.

For A+BProducts\text{A} + \text{B} \rightarrow \text{Products} with [B][A][\text{B}] \gg [\text{A}]:

Rate=k[A][B]k[A]\text{Rate} = k[\text{A}][\text{B}] \approx k'[\text{A}]

where k=k[B]k' = k[\text{B}] (pseudo-first-order rate constant)

Elementary vs Overall Reactions

Elementary Reaction

Single-step reaction. Order = molecularity (number of molecules colliding).

Overall Reaction

May consist of multiple elementary steps. Order must be determined experimentally.

Molecularity

MolecularityNameExample
1UnimolecularAProducts\text{A} \rightarrow \text{Products}
2BimolecularA+BProducts\text{A} + \text{B} \rightarrow \text{Products}
3TermolecularA+B+CProducts\text{A} + \text{B} + \text{C} \rightarrow \text{Products} (rare)