Reaction A to B — How to Find Rate Law, Order of Reaction and Activation Energy

The rate law for A → B is: Rate = k[A]ⁿ

To find the order n, compare two experimental runs where [A] is changed while keeping other conditions the same:

Rate₁ / Rate₂ = ([A]₁ / [A]₂)ⁿ

Example data:

Comparing Exp 1 and 2: Rate doubles when [A] doubles → n = 1 (first order) Rate 2/Rate 1 = (0.20/0.10)ⁿ → 2 = 2ⁿ → n = 1

Rate law: Rate = k[A]¹ = k[A]

Once the order n is known, substitute any experimental data point into the rate law to find k:

For first-order A → B: Rate = k[A] k = Rate / [A]

Using Experiment 1: k = (2.0 × 10⁻³ mol/L·s) / (0.10 mol/L) = 2.0 × 10⁻² s⁻¹

Units of k depend on the order:

• Zero order: mol/L·s or M·s⁻¹
• First order: s⁻¹
• Second order: L/mol·s or M⁻¹s⁻¹
• nth order: M^(1-n)·s⁻¹

The rate constant k is temperature-dependent (increases with temperature).

The Arrhenius equation relates the rate constant k to temperature T:

k = A·e^(−Ea/RT)

Where:

• k = rate constant
• A = pre-exponential factor (frequency factor)
• Ea = activation energy (J/mol)
• R = 8.314 J/mol·K (gas constant)
• T = temperature in Kelvin

Logarithmic form: ln k = ln A − Ea/RT

To find Ea from two temperatures T₁ and T₂: ln(k₂/k₁) = (Ea/R) × (1/T₁ − 1/T₂)

or: log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁ × T₂)

Example: If k₁ = 2.0 × 10⁻² s⁻¹ at 300 K and k₂ = 8.0 × 10⁻² s⁻¹ at 320 K: ln(8.0/2.0) = (Ea/8.314) × (1/300 − 1/320) ln 4 = (Ea/8.314) × (320 − 300)/(300 × 320) 1.386 = (Ea/8.314) × (20/96000) Ea = 1.386 × 8.314 × 96000/20 = 55,440 J/mol ≈ 55.4 kJ/mol

The integrated rate law gives concentration as a function of time:

Zero order (n = 0): [A]t = [A]₀ − kt Half-life: t₁/₂ = [A]₀/(2k)

First order (n = 1): ln[A]t = ln[A]₀ − kt or [A]t = [A]₀·e^(−kt) Half-life: t₁/₂ = 0.693/k (constant, independent of [A]₀)

Second order (n = 2): 1/[A]t = 1/[A]₀ + kt Half-life: t₁/₂ = 1/(k·[A]₀)

Graphical determination:

• Plot [A] vs time → straight line → zero order
• Plot ln[A] vs time → straight line → first order (slope = −k)
• Plot 1/[A] vs time → straight line → second order (slope = k)

Effect of concentration: For A → B with rate = k[A]ⁿ:

• Doubling [A]: rate increases by 2ⁿ fold
• For n=1: rate doubles
• For n=2: rate quadruples
• For n=0: rate is unchanged

Effect of temperature:

• A 10°C rise in temperature approximately doubles the reaction rate (rule of thumb)
• Precisely described by the Arrhenius equation
• Higher temperature → more molecules have energy ≥ Ea → more successful collisions
• Activation energy Ea: minimum energy required for a reaction to occur
• Lower Ea = faster reaction at a given temperature

Catalysts: Lower Ea without being consumed, increasing the rate constant k at a given temperature.