Lesson 6 of 11 · 6 min
First-order reactions
NCERT §3.3.2
The club's syringe readings never fall along a straight line. Yet every 10 minutes the gas still to come halves: 80, 40, 20, 10 mL. Which kind of law gives that pattern?
The lesson in notes
In short
For first order, rate = −d[R]/dt = k[R]. Integration gives ln([R]₀/[R]) = kt, or [R] = [R]₀e^(−kt).
In common logs: k = (2.303/t) log([R]₀/[R]).
A plot of ln[R] against t is a straight line of slope −k; a plot of log([R]₀/[R]) against t is a straight line through the origin with slope k/2.303.
Between two times t₁ and t₂: k = 2.303/(t₂ − t₁) × log([R]₁/[R]₂).
Examples: hydrogenation of ethene (rate = k[C₂H₄]), natural radioactive decay such as that of radium, and the decomposition of N₂O₅ and N₂O.
Worked number: [N₂O₅] falls from 1.24 × 10⁻² to 0.20 × 10⁻² mol L⁻¹ in 60 min at 318 K. k = (2.303/60) log 6.2 = 0.0304 min⁻¹.
For a gas-phase reaction A(g) → B(g) + C(g) at constant volume, pressure follows the progress. If the initial pressure is pᵢ and the total pressure at time t is pₜ, then p_A = 2pᵢ − pₜ and k = (2.303/t) log(pᵢ/(2pᵢ − pₜ)).
First-order decay is exponential: the reactant never quite reaches zero, but falls by the same fraction in every equal time interval.
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Zero, first and second order rate laws
The Organic Chemistry Tutor · English · Lecture · Open on YouTube