Chemical Kinetics

Chemical KineticsNEET Chemistry · Class 12 · NCERT Chapter 3

Overview Notes NEET Questions Interactive Learning

3 interactive concept widgets for Chemical Kinetics. Drag any slider, change any number, and watch the formula and the answer update live. Built so you understand how each NEET problem actually works, not just the final number.

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Rate orders: concentration vs time table for zero/first/second order, half-life, query any time

Half-life: zero/first/second order, fraction remaining, n half-lives calculation

Arrhenius: k₂/k₁ from Ea and temperatures, or Ea from two k values — full working shown

Rate orders explorer

Compare zero, first, and second-order kinetics. Adjust initial concentration and rate constant, read concentration at every half-life interval, and query any time point.

Chemical kinetics

Rate orders explorer

Compare zero, first, and second-order kinetics. Adjust [A]₀ and k, read off concentrations at any half-life interval, and query any time point.

Explore how concentration changes over time for zero, first, and second-order reactions. Adjust the initial concentration and rate constant, then query any time point.

Zero-order

First-order

Second-order

Integrated rate law

[A] = [A]₀ × e^(−k·t)

Initial concentration [A]₀: 1.00 mol/L

Rate constant k: 0.05 s⁻¹

Half-life calculation

t½ = 0.693 / k = 0.693 / 0.05 = 13.86 s

Time to reach 25% of [A]₀: 27.73 s

Concentration at each half-life interval

Time (s)

[A] (mol/L)

% remaining

0.00

1.0000

100.0%

13.86

0.5001

50.0%

27.72

0.2501

25.0%

41.58

0.1251

12.5%

55.44

0.0625

6.3%

Query concentration at any time

t = 10 s

[A] at t = 10s

0.6065 mol/L

(60.7% of [A]₀)

Try this

Switch between orders and notice how the half-life formula changes. First-order t½ is always constant; zero-order and second-order t½ depend on [A]₀.
For first-order, try doubling k and observe that t½ halves. For zero-order, double [A]₀ and t½ doubles.
Try to find the time when [A] = 0.10 mol/L by dragging the query slider. Then verify with the formula.

Half-life calculator

Find t½ and the time for [A] to reach any fraction of [A]₀ for all three reaction orders. First-order mode lets you enter either k or t½ directly.

Chemical kinetics

Half-life calculator

Calculate t½ and time for [A] to fall to any fraction of [A]₀ for zero, first, and second-order reactions. First-order mode lets you enter either k or t½ directly.

Calculate half-life and time to reach any fraction of [A]₀ for zero, first, or second-order reactions. For first-order you can enter either k or t½ directly.

Zero-order

First-order

Second-order

t½ = 0.693 / k

Half-life is independent of [A]₀ for first-order reactions.

k (s⁻¹)

Half-life result

t½ = 0.693 / k = 0.693 / 0.0500 = 13.8600 s

t½ = 13.8600 s

Time for [A] to reach a fraction of [A]₀

½ (50%)

¼ (25%)

⅛ (12.5%)

1/16 (6.25%)

1/10 (10%)

1/100 (1%)

Custom fraction (0–1)

e.g. 0.33 for 33%

Calculation

t = ln([A]₀/[A]) / k = ln(1/0.25) / 0.0500 = 1.3863 / 0.0500 = 27.7259 s

Time required

27.7259 s

In half-lives

= 2.00 half-lives

Try this

First-order, k = 0.0693 s⁻¹: verify t½ = 10 s, then check how long for 12.5% remaining (should be 30 s = 3 half-lives).
Try second-order with k = 0.5 L mol⁻¹ s⁻¹ and [A]₀ = 2.0 mol/L. Change [A]₀ to 1.0 mol/L and notice the half-life doubles.
Switch to "Enter t½" for first-order and put t½ = 693 s. The calculator derives k = 0.001 s⁻¹ for you.

Arrhenius equation calculator

Two modes: find the rate constant ratio k₂/k₁ from Ea and two temperatures, or find Ea from rate constants at two temperatures. Full step-by-step working with presets.

Chemical kinetics

Arrhenius equation calculator

Two modes: find the rate constant ratio k₂/k₁ given activation energy and two temperatures, or find Ea from rate constants at two temperatures. Full step-by-step working shown.

Use the Arrhenius equation in its two-temperature form: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂). Switch between finding the rate constant ratio or the activation energy.

Ea (kJ/mol)

T₁ (K)

Lower temperature

T₂ (K)

Higher temperature

Step-by-step working

Ea = 55 kJ/mol = 55000 J/mol
1/T₁ - 1/T₂ = 1/300 - 1/320 = 20.8333 × 10⁻⁵ K⁻¹
ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂)
= (55000/8.314) × 2.083e-4
= 1.3782

k₂/k₁ = 3.968

The rate increases by a factor of 3.97 when temperature rises from 300 K to 320 K.

Load a preset reaction

Acid hydrolysis (55 kJ/mol)

Enzyme reaction (40 kJ/mol)

High Ea reaction (250 kJ/mol)

Low Ea reaction (20 kJ/mol)

Remember: Always use T in Kelvin. Convert from Celsius by adding 273 (or 273.15). R = 8.314 J mol⁻¹ K⁻¹. Ea is in J/mol when using R in these units. Divide by 1000 to get kJ/mol.

Try this

Try Ea = 100 kJ/mol, T₁ = 300 K, T₂ = 310 K. Notice the rate nearly doubles for just a 10 K rise.
Try Ea = 20 kJ/mol vs Ea = 200 kJ/mol with the same temperatures. See how a higher Ea makes temperature a much stronger lever on rate.
In "Find Ea" mode, enter k₁ = 0.02, k₂ = 0.07, T₁ = 500, T₂ = 700. Verify against the NEET worked problem in your notes.

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