Reaction Order Formula

Determine how concentration affects reaction rate through experimental methods

Rate Law & Order

Rate = k[A]m[B]n

m = order with respect to A

n = order with respect to B

m + n = overall reaction order

Orders are determined EXPERIMENTALLY, not from stoichiometry!

Method 1: Method of Initial Rates

Procedure

  1. Measure initial rate for different initial concentrations
  2. Compare trials where only ONE reactant concentration changes
  3. Calculate ratio: Rate₂/Rate₁ = ([A]₂/[A]₁)m
  4. Solve for m (order with respect to A)
  5. Repeat for other reactants

Order Determination

If [A] doubles and...Then order m =Math
Rate stays same0 (zero order)20 = 1
Rate doubles1 (first order)21 = 2
Rate quadruples2 (second order)22 = 4
Rate increases 8×3 (third order)23 = 8

Method 2: Graphical Analysis

Zero Order

Linear plot:

[A] vs t

Slope = -k

First Order

Linear plot:

ln[A] vs t

Slope = -k

Second Order

Linear plot:

1/[A] vs t

Slope = k

Worked Examples

Example 1: Method of Initial Rates

Reaction: 2NO(g) + O₂(g) → 2NO₂(g)

Trial[NO]₀ (M)[O₂]₀ (M)Initial Rate (M/s)
10.0100.0102.5×10⁻⁵
20.0200.0101.0×10⁻⁴
30.0100.0205.0×10⁻⁵

Solution:

Find order with respect to NO (compare trials 1 & 2):

Rate₂/Rate₁ = (1.0×10⁻⁴)/(2.5×10⁻⁵) = 4

[NO]₂/[NO]₁ = 0.020/0.010 = 2

4 = 2m → m = 2

Order with respect to NO = 2

Find order with respect to O₂ (compare trials 1 & 3):

Rate₃/Rate₁ = (5.0×10⁻⁵)/(2.5×10⁻⁵) = 2

[O₂]₃/[O₂]₁ = 0.020/0.010 = 2

2 = 2n → n = 1

Order with respect to O₂ = 1

Overall:

Rate = k[NO]²[O₂]

Overall order = 2 + 1 = 3 (third order)

Example 2: Calculate Rate Constant

Problem: Using data from Example 1, calculate k.

Solution:

Rate = k[NO]²[O₂]

Using Trial 1:

2.5×10⁻⁵ = k(0.010)²(0.010)

2.5×10⁻⁵ = k(1.0×10⁻⁶)

k = 2.5×10⁻⁵ / 1.0×10⁻⁶

k = 25 M⁻²s⁻¹

Units: M⁻²s⁻¹ for third order (overall order 3)

Example 3: Fractional Orders

Problem: When [A] triples, rate increases by factor of 5.2. Find order.

Solution:

Rate₂/Rate₁ = ([A]₂/[A]₁)m

5.2 = 3m

log(5.2) = m × log(3)

0.716 = m × 0.477

m = 0.716 / 0.477

m = 1.5 (order = 3/2)

Fractional orders are possible! Often indicates complex mechanisms.

Example 4: Half-Life Method

Problem: For decomposition of A, t1/2 is constant = 15.0 min. What is the order?

Solution:

Compare half-life behavior:

  • Zero order: t1/2 ∝ [A]₀ (decreases with [A])
  • First order: t1/2 = constant ✓
  • Second order: t1/2 ∝ 1/[A]₀ (increases as [A] decreases)

First order reaction

k = 0.693/t1/2 = 0.693/15.0 = 0.0462 min⁻¹

Common Mistakes

⚠️

Using Stoichiometric Coefficients

Order ≠ coefficient! For 2NO + O₂ → products, rate ≠ k[NO]²[O₂]. Must determine experimentally!

⚠️

Comparing Wrong Trials

Only compare trials where ONE concentration changes and others stay constant!

⚠️

Ignoring Units of k

Units depend on overall order: 0th (M/s), 1st (s⁻¹), 2nd (M⁻¹s⁻¹), 3rd (M⁻²s⁻¹)

💡

Negative Orders Possible

If rate decreases when [A] increases, order can be negative! Example: inhibition.