Unit 3: Chemical Kinetics

1. Rate of Reactions

The rate of a chemical reaction quantifies how fast reactants are converted into products. It can be expressed as an average rate over a time interval or as an instantaneous rate at a specific moment.

Average rate: change in concentration of a reactant or product divided by the time interval.
```
Average rate = -Δ[A]/Δt = Δ[B]/Δt
```
where [A] and [B] are molar concentrations (mol L⁻¹) and Δt is the elapsed time (s). The negative sign for reactants ensures a positive rate.
Instantaneous rate: the rate at a particular instant, obtained as the derivative of concentration with respect to time.
```
Instantaneous rate = -d[A]/dt = d[B]/dt
```
This is the slope of the tangent to the concentration‑time curve at that point.

Example 1 (Average rate): For the reaction A → B, the concentration of A falls from 0.80 M to 0.50 M in 120 s. Calculate the average rate.

Solution:
Δ[A] = 0.50 M – 0.80 M = –0.30 M
Δt = 120 s
Average rate = –(–0.30 M)/120 s = 0.0025 M s⁻¹

2. Rate Law and Its Expressions

The rate law relates the reaction rate to the concentrations of reactants raised to experimentally determined powers.

Rate = k [A]^m [B]^n

where:

k = rate constant (depends on temperature)
m and n = reaction orders with respect to A and B (determined experimentally, not necessarily equal to stoichiometric coefficients)

Units of the rate constant k vary with overall reaction order.

Overall Order	Rate Law Form	Units of k
Zero	Rate = k	mol L⁻¹ s⁻¹
First	Rate = k[A]	s⁻¹
Second	Rate = k[A]² or k[A][B]	L mol⁻¹ s⁻¹
Third	Rate = k[A]³	L² mol⁻² s⁻¹

Example 2 (Determining order from initial rates): The following initial‑rate data were obtained for the reaction 2NO + O₂ → 2NO₂ at constant temperature.

Experiment	[NO] (M)	[O₂] (M)	Initial Rate (M s⁻¹)
1	0.10	0.10	2.0 × 10⁻³
2	0.20	0.10	8.0 × 10⁻³
3	0.10	0.20	4.0 × 10⁻³

Comparing experiments 1 and 2 (constant [O₂]), doubling [NO] quadruples the rate → order in NO = 2. Comparing 1 and 3 (constant [NO]), doubling [O₂] doubles the rate → order in O₂ = 1. Hence the rate law is Rate = k[NO]²[O₂] and the overall order is 3.

3. Order and Molecularity

Order (experimental) = sum of the exponents in the rate law. Molecularity (theoretical) = number of reactant particles that collide in an elementary step.

Zero order: Rate = k (independent of concentration). Example: decomposition of NH₃ on a platinum surface.
First order: Rate = k[A]. Example: radioactive decay, N₂O₅ → 2NO₂ + ½O₂.
Second order: Rate = k[A]² or k[A][B]. Example: 2HI → H₂ + I₂.

Molecularity can be unimolecular (one molecule), bimolecular (two molecules), or termolecular (three molecules). Termolecular steps are rare due to low probability of three‑body collisions.

Example 3 (Identifying order from half‑life): For a reaction, the half‑life is observed to double when the initial concentration is halved. What is the order?

Solution:
For a reaction of order n, t₁/₂ ∝ [A]₀^{1‑n}.
If t₁/₂ doubles when [A]₀ is halved:
(½)^{1‑n} = 2 → 2^{n‑1} = 2 → n‑1 = 1 → n = 2.
Thus the reaction is second order.

4. Integrated Rate Equations

Integrating the differential rate law gives concentration‑time relationships.

Zero‑order reaction

[A] = [A]₀ – kt

Half‑life:

t₁/₂ = [A]₀ / (2k)

First‑order reaction

ln[A] = ln[A]₀ – kt

Alternative base‑10 form:

k = (2.303 / t) log([A]₀ / [A])

Half‑life:

t₁/₂ = 0.693 / k

Second‑order reaction (single reactant)

1/[A] = 1/[A]₀ + kt