ChemistryChemical kineticsClass 12

Reaction kinetics & Arrhenius activation: interactive simulation

Chemical kinetics governs how fast a reaction proceeds and through what molecular pathway. Reaction order reflects how rate scales with reactant concentrations, and each order has a distinct integrated mathematical signature.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards · 3 modes in one lab

What you can explore

  1. 1

    Integrated Rate Laws & Half-life

    Zero, first and second order concentration curves and half-life analysis.
  2. 2

    Arrhenius & Catalyst Energy Barrier

    Potential energy diagram with activation energy barrier and catalyst acceleration.
  3. 3

    Find the order from graphs

    Which of [A], ln[A] and 1/[A] is linear tells you the order. Radioactive decay stays in the nuclear physics lab.

Try this

Explore concentration-time curves, see why first-order half-life is completely independent of initial reactant concentration, and test how catalysts lower Arrhenius activation energy barriers to speed up reactions.

  1. 1

    In Rate Laws, double the initial concentration [A]₀ for First Order vs Zero Order.

    What you should see

    For First Order, t₁/₂ remains identical! For Zero Order, t₁/₂ doubles.

  2. 2

    In Arrhenius sim, turn on Catalyst.

    What you should see

    Activation energy drops by 25 kJ/mol, causing reaction rate to accelerate by over 20,000× at room temperature!

  3. 3

    In Find the order, switch the true order from zero to first to second.

    What you should see

    Only one of the three plots is a straight line: [A] for zero, ln[A] for first, and 1/[A] for second.

Formulas behind it

First-order integrated rate law

ln⁡[A]=ln⁡[A]0−kt,t1/2=0.693k\ln [A] = \ln [A]_0 - k t, \quad t_{1/2} = \frac{0.693}{k}

Zero-order rate law

[A]=[A]0−kt,t1/2=[A]02k[A] = [A]_0 - k t, \quad t_{1/2} = \frac{[A]_0}{2k}

Arrhenius equation

k=Ae−Ea/(RT),ln⁡(k2k1)=EaR(1T1−1T2)k = A e^{-E_a / (RT)}, \quad \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)

Second-order rate law

1[A]=1[A]0+kt,t1/2=1k[A]0\frac{1}{[A]} = \frac{1}{[A]_0} + k t, \quad t_{1/2} = \frac{1}{k [A]_0}

Common mistakes

  • Thinking molecularity and order are always the same. Order is purely experimental and can be fractional or zero; molecularity is a whole number for elementary steps.
  • Saying catalyst increases yield. A catalyst speeds up reaction rate to reach equilibrium faster, but cannot alter equilibrium yield or ΔG°.
  • Assuming temperature increases reaction rate by increasing collision frequency alone — the dominant factor is the exponential surge in Boltzmann fraction e^(-Ea/RT).
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FAQ

Why is the unit of k different for different reaction orders?

Units of k are (mol/L)^(1-n) s⁻¹, where n is the overall reaction order: mol L⁻¹ s⁻¹ for zero order, s⁻¹ for first order, L mol⁻¹ s⁻¹ for second order.

What is pseudo first-order reaction?

A higher-order reaction where one reactant (e.g. water in ester hydrolysis) is present in large excess, making its concentration effectively constant.

What is the temperature coefficient of a reaction?

The ratio of rate constants at temperatures differing by 10 °C (usually k(T+10)/k(T) ≈ 2 to 3).

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