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2.3 Differential rate laws

3 min readjuly 22, 2024

Chemical kinetics is all about how fast reactions happen. Differential rate laws are the key to understanding reaction speeds. They show how concentrations of reactants affect the rate, using a and reaction orders.

These laws help us predict how quickly chemicals will react or form. By integrating them, we can figure out concentrations at any time and calculate half-lives for first-order reactions. It's like having a crystal ball for chemical reactions!

Differential Rate Laws

Differential rate law derivation

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  • Expresses reaction rate in terms of reactant concentrations and rate constant
    • General reaction: aA+bBcC+dDaA + bB \rightarrow cC + dD
    • : Rate=k[A]m[B]n\text{Rate} = k[A]^m[B]^n
      • kk: rate constant, specific to reaction at given temperature
      • mm, nn: reaction orders for reactants A and B
  • for each reactant determined experimentally, not from balanced equation
    • Overall reaction order: sum of individual reaction orders m+nm + n
  • Examples:
    • with respect to A: Rate=k[A]\text{Rate} = k[A]
    • with respect to B: Rate=k[B]2\text{Rate} = k[B]^2

Rate expression from concentration changes

  • Rate expressed as change in reactant or product concentration over time
    • For reactant: Rate=1ad[A]dt=1bd[B]dt\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt}
    • For product: Rate=1cd[C]dt=1dd[D]dt\text{Rate} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}
      • Negative sign for reactants: concentrations decrease over time
      • Coefficients aa, bb, cc, dd: normalize rate expression based on balanced equation
  • Examples:
    • Reactant A disappearing at 0.5 M/s: Rate=d[A]dt=0.5 M/s\text{Rate} = -\frac{d[A]}{dt} = 0.5 \text{ M/s}
    • Product C forming at 0.2 M/min: Rate=d[C]dt=0.2 M/min\text{Rate} = \frac{d[C]}{dt} = 0.2 \text{ M/min}

Rate law components and relationships

  • Differential rate law combines reaction orders and rate constant
  • Reaction order for each reactant: exponent of its concentration term
    • First-order with respect to A: m=1m = 1, Rate=k[A]\text{Rate} = k[A]
    • Second-order with respect to A: m=2m = 2, Rate=k[A]2\text{Rate} = k[A]^2
  • Rate constant kk: specific to reaction and temperature
    • Represents intrinsic reactivity of reactants, determined experimentally
    • Units of kk depend on overall reaction order, ensure rate has units of concentration/time
  • Examples:
    • First-order reaction: Rate=k[A]\text{Rate} = k[A], kk has units of s1\text{s}^{-1}
    • Second-order reaction: Rate=k[A][B]\text{Rate} = k[A][B], kk has units of M1s1\text{M}^{-1}\text{s}^{-1}

Integrated rate law for first-order reactions

  • Obtained by integrating differential rate law
  • Allows determination of reactant or product concentrations at any time
  • For first-order reaction, differential rate law: Rate=k[A]\text{Rate} = k[A]
    1. Rearrange and integrate with respect to time: [A]0[A]td[A][A]=k0tdt\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -k \int_0^t dt
    2. Integrated rate law: ln[A]t[A]0=kt\ln\frac{[A]_t}{[A]_0} = -kt
    • [A]0[A]_0: initial concentration of reactant A
    • [A]t[A]_t: concentration at time tt
  • Uses of integrated rate law:
    • Determine at any time tt
    • Calculate half-life of first-order reaction, independent of initial concentration: t1/2=ln2kt_{1/2} = \frac{\ln 2}{k}
  • Examples:
    • Reactant A with initial concentration 1.0 M, rate constant 0.1 s1\text{s}^{-1}, concentration after 10 s: [A]10=[A]0ekt=1.0 M×e(0.1 s1)(10 s)=0.37 M[A]_{10} = [A]_0 e^{-kt} = 1.0 \text{ M} \times e^{-(0.1 \text{ s}^{-1})(10 \text{ s})} = 0.37 \text{ M}
    • Half-life of first-order reaction with rate constant 0.05 min1\text{min}^{-1}: t1/2=ln20.05 min1=13.9 mint_{1/2} = \frac{\ln 2}{0.05 \text{ min}^{-1}} = 13.9 \text{ min}
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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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