§ C-27 Reaction Orders (0th, 1st & 2nd Order), Rate Laws & Half-Life t₁/₂ (Grade 10–11 · Rate Laws & Half-Life)
Integrated rate laws of zero, first, and second order reactions and scaling of half-life t₁/₂. The kinetic order of a reaction reveals the true molecular mechanism hidden behind the overall stoichiometric equation.
Theory Summary & Scientific Mechanism
The reaction order n in the differential rate law v = -(d[A])/(dt) = k[A]^n dictates both the integrated concentration decay [A](t) and the scaling of the half-life t₁/2 with initial concentration [A]₀.
For zero-order kinetics (n=0) (e.g., surface-saturated catalysis), rate is constant (v = k), [A](t) = [A]₀ - kt, and t₁/2 = ([A]₀)/(2k) ∝ [A]₀. For first-order kinetics (n=1), decay is exponential [A](t) = [A]₀ e⁻kt and the half-life t₁/2 = (ln 2)/(k) is strictly independent of [A]₀.
For second-order kinetics (n=2) (bimolecular collisions 2A → P), (1)/([A](t)) = (1)/([A]₀) + kt (linear in 1/[A] vs. t) and the half-life scales inversely with initial concentration: t₁/2 = (1)/(k[A]₀).
Everyday Analogy
0th-order is like a single tollbooth line (t₁/2 ∝ [A]₀); 1st-order is like radioactive decay where half Always vanishes in a fixed time t₁/2 = (ln 2)/(k); 2nd-order requires partner collisions, so as molecules thin out, finding a partner takes much longer (t₁/2 ∝ 1/[A]₀)!
Interactive Question & Derivation
How does the half-life t₁/2 depend on the initial reactant concentration [A]₀ for reactions of zero (n=0), first (n=1), and second (n=2) kinetic order?
- A: For 1st-order reactions t₁/2 = (ln 2)/(k) is completely independent of [A]₀; for 0th-order t₁/2 = ([A]₀)/(2k) ∝ [A]₀; and for 2nd-order t₁/2 = (1)/(k[A]₀) ∝ (1)/([A]₀)
- B: For every reaction order, the half-life t₁/2 is always constant and independent of initial concentration [A]₀
- C: In a zero-order reaction, reactant concentration decays exponentially and never reaches zero
Study Hint: For 1st order t₁/2 = (ln 2)/(k) is independent of [A]₀, for 0th order t₁/2 = ([A]₀)/(2k), and for 2nd order t₁/2 = (1)/(k[A]₀).
Proof & Explanation: Integrating the differential rate law -(d[A])/(dt) = k[A]^n yields three distinct concentration profiles: 0th order (n=0, saturated heterogeneous catalysis): [A](t) = [A]₀ - kt, t₁/2 = ([A]₀)/(2k); 1st order (n=1, unimolecular decay): [A](t) = [A]₀ e⁻kt, t₁/2 = (ln 2)/(k) ≈ (0.693)/(k) (independent of [A]₀); 2nd order (n=2, bimolecular collisions): (1)/([A](t)) = (1)/([A]₀) + kt, t₁/2 = (1)/(k[A]₀).
Key Equations & 3D Simulation Parameters
-(d[A])/(dt) = k[A]^n ⇒ [A] = [A]₀ - kt, & t₁/2 = ([A]₀)/(2k) (n=0) ; [A] = [A]₀ e⁻kt, & t₁/2 = (ln 2)/(k) (n=1) ; (1)/([A]) = (1)/([A]₀) + kt, & t₁/2 = (1)/(k[A]₀) (n=2)
- Reaction order n (0 = zero, 1 = first, 2 = second): 0 … 2 (default: 1)
- Initial reactant concentration [A]₀: 0.5 … 3 mol/L (default: 2 mol/L)
- Rate constant k: 0.1 … 1 s⁻¹ (default: 0.35 s⁻¹)
- Reaction time t: 0 … 6 s (default: 2 s)