Which plot turns the data into a straight line?
"Which plot turns the data into a straight line?" is a free, interactive Class 12 Chemistry simulation for NEET and JEE, built on the NCERT section "First-order reactions" in Chemical Kinetics. One set of concentration-time data plotted four ways ([R], ln[R], log([R]₀/[R]), 1/[R] against t); only the right axis gives a straight line, and its slope gives k (−k or k/2.303) NEET has asked about this idea 6 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 First-order reactions in Chemical Kinetics. Change the values and watch what happens.
What it shows
NCERT §3.3.2
One set of concentration-time data plotted four ways ([R], ln[R], log([R]₀/[R]), 1/[R] against t); only the right axis gives a straight line, and its slope gives k (−k or k/2.303)
Controls
4 to try
- Toggletrue order of the hidden data 0 / 1 / 2
- Toggleplot [R] / ln[R] / log([R]₀/[R]) / 1/[R] vs t
- Dragslope line on the plot
- Readoutsslope, k from slope, residual
Try this
For first-order data, pick the plot that is straight and drag the line to read k within 5% (remember the 2.303)
Common mistake
Dropping the 2.303 when switching between ln and log.
In NEET
10 years of papers
Asked 6 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 §3.3.2
- 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.
Questions
5 answers
- What does the "Which plot turns the data into a straight line?" simulation show?
- One set of concentration-time data plotted four ways ([R], ln[R], log([R]₀/[R]), 1/[R] against t); only the right axis gives a straight line, and its slope gives k (−k or k/2.303)
- What mistake does the "Which plot turns the data into a straight line?" simulation fix?
- A common mistake it clears up: dropping the 2.303 when switching between ln and log.
- Has NEET asked about this idea?
- Yes. NEET has asked about this idea 6 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 "First-order reactions" in Chemical Kinetics (Class 12 Chemistry).
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