Can you find a rate law from initial rates alone?
"Can you find a rate law from initial rates alone?" is a free, interactive Class 12 Chemistry simulation for NEET and JEE, built on the NCERT section "Rate law and rate constant" in Chemical Kinetics. A virtual bench for 2A + B → products with hidden orders: set starting concentrations, run, read the initial rate, and deduce the rate law and the units of k. NEET has asked about this idea 2 times in 10 years of papers. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Rate law and rate constant in Chemical Kinetics. Change the values and watch what happens.
What it shows
NCERT §3.2; §3.2.1–3.2.3
A virtual bench for 2A + B → products with hidden orders: set starting concentrations, run, read the initial rate, and deduce the rate law and the units of k.
Controls
4 to try
- Slider[A]₀ 0.05–0.40 mol L⁻¹
- Slider[B]₀ 0.05–0.40 mol L⁻¹
- Buttonrun trial (adds a row to the results table)
- Segmentedyour answer for order in A and B (0, ½, 1, 2)
Try this
Run at most four trials and identify both orders; the sim checks your answer and the units of k.
Common mistake
Reading the order off the stoichiometric coefficients.
In NEET
10 years of papers
Asked 2 times in 10 years of NEET.
See the past questions: Chemical Kinetics, Chemical Kinetics (chapter)
In JEE Main
42 shifts analysed
48 questions from Chemical Kinetics across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §3.2; §3.2.1; §3.2.2
- Rate depends on the concentration of reactants (pressure for gases), temperature and catalyst.
- The rate law (rate equation) expresses rate as k times concentrations raised to powers: for aA + bB → products, rate = k[A]^x[B]^y.
- The powers x and y are found by experiment. They need not equal the coefficients a and b of the balanced equation.
- The standard method compares initial rates. For 2NO + O₂ → 2NO₂, doubling [NO] at fixed [O₂] makes the rate four times larger and doubling [O₂] at fixed [NO] makes it twice as large, so rate = k[NO]²[O₂].
- Two reactions from the book whose powers do not follow the equation: CHCl₃ + Cl₂ → CCl₄ + HCl has rate = k[CHCl₃][Cl₂]^½, and the acid hydrolysis of an ester has rate = k[ester][H₂O]⁰ in excess water.
- k, the rate constant, equals the rate when every concentration in the rate law is 1 mol L⁻¹. At a given temperature it is fixed for the reaction; it changes with temperature and with a catalyst, not with concentration.
- Rate falls as the reaction proceeds because concentrations fall; k stays the same throughout.
Questions
5 answers
- What does the "Can you find a rate law from initial rates alone?" simulation show?
- A virtual bench for 2A + B → products with hidden orders: set starting concentrations, run, read the initial rate, and deduce the rate law and the units of k.
- What mistake does the "Can you find a rate law from initial rates alone?" simulation fix?
- A common mistake it clears up: reading the order off the stoichiometric coefficients.
- Has NEET asked about this idea?
- Yes. NEET has asked about this idea 2 times in 10 years of papers.
- Does JEE Main ask questions from Chemical Kinetics?
- Yes. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Rate law and rate constant" in Chemical Kinetics (Class 12 Chemistry).
Embed this simulation
Free, with a link back
Teaching this idea? Put the simulation on your own page or LMS with this code.