Part of PC-08 — Chemical Kinetics

Zero-Order Reactions: Deep Dive

by Notetube Official204 words4 views

Characteristics

In a zero-order reaction: rate = k (constant), independent of concentration.

Why zero order occurs:

  • Catalyst surface saturation (Langmuir-Hinshelwood at high pressure)
  • Rate determined by a step that doesn't involve the reactant (e.g., light intensity in photochemical reactions)
  • Enzyme saturation (Michaelis-Menten at high substrate concentrations → zero order in substrate)

Classic Example

2NH3(g)Pt surface, high PN2(g)+3H2(g)2\text{NH}_3(g) \xrightarrow{\text{Pt surface, high P}} \text{N}_2(g) + 3\text{H}_2(g)

At high pressure: all Pt surface sites occupied → rate = k (zero order in NH3NH_{3}) At low pressure: fewer sites occupied → rate ≈ k[NH3NH_{3}] (first order)

Key Features of Zero-Order Kinetics

  1. [A] vs t graph: Straight line; slope = −k; intercept = [A]_{0}
  2. Rate: Constant = k throughout
  3. Half-life: t_{1}/{2} = [A]{0}/(2k) — proportional to [A]_{0}
  4. Each successive half-life: Shorter than the previous (as [A]_{0} decreases)
  5. Completion time: t = [A]_{0}/k (finite! Unlike first order)
  6. Number of half-lives to completion: Always exactly 2

Worked Example

k = 0.01 mol/(L·s), [A]_{0} = 0.5 M:

tcomplete=0.50.01=50 st_{\text{complete}} = \frac{0.5}{0.01} = 50 \text{ s}

t1/2=0.52×0.01=25 st_{1/2} = \frac{0.5}{2 \times 0.01} = 25 \text{ s}

After t = 30 s: [A] = 0.5 − 0.01 × 30 = 0.5 − 0.3 = 0.2 M

Like these notes? Save your own copy and start studying with NoteTube's AI tools.

Sign up free to clone these notes