ADV-2026-P2-CHE-Q05 · First-Order Kinetics Virtual Lab

01Original question — verbatim

For a first-order reaction R ⟶ P at a given temperature, k is the rate constant. For this reaction, at the given temperature, the concentrations of R and P at a time t are [R] and [P], respectively. The correct graphical representation(s) for this reaction is(are)

(A)

[P] against t — rising, concave up
(B)

d[R]/dt against [R] — straight line, positive slope through the origin
(C)

d[P]/dt against t — decaying from a maximum
(D)

k against t — horizontal line

Multiple options may be correct (+4 / partial / −1). The four graphs are redrawn from the printed paper. Don't guess — go and measure them.

02The reactor

A vessel of R decaying to P. You set the starting concentration and the rate constant — either directly or through the temperature, via the Arrhenius equation — then press Run. Every value on this page is computed from the same [R]₀, k and t: the particles, the meters, the equation and the graphs can never disagree.

Vessel

R (reactant) P (product) particle count ∝ concentration · jiggle speed ∝ temperature

Controls

What sets k?

Only one of them can be the independent variable — that is the constraint model.

Initial concentration [R]₀ 1.000 M

mol L⁻¹

0.10 to 5.00 M · try 1/2, 2.5 or 5e-1

Rate constant k 0.300 s⁻¹

s⁻¹

0.05 to 2.00 s⁻¹ · first order, so the unit is s⁻¹

Run

Time t (drag to scrub) 0.00 s

Live readings

[R]

—

mol L⁻¹

[P]

—

mol L⁻¹

d[R]/dt

—

M s⁻¹

d[P]/dt

—

M s⁻¹

rate constant k

—

s⁻¹

half-life t_½

—

conversion

—

samples logged

this run

[R] [P] previous run (ghost) ◦ your logged samples

03Plot your own data on each option's axes

This is the question itself. Pick a set of axes and your logged samples are replotted on them, with the exact shape the printed option shows drawn behind for comparison. If your measurements land on that shape the option is right; if they don't, it is wrong — and you will be able to see why.

what the data actually does the shape option shows ◦ your logged samples

Axes

04Half-life trials — the first-order signature

Each completed run is logged below with the half-life measured from your own samples, not calculated for you. Change [R]₀ between runs and watch what happens to t_½; then change k and watch again. The pattern tells you what "first order" means.

Trials

No runs yet. Press ▶ Run in the reactor and let it finish.

Observations

Your notebook fills itself as your runs actually demonstrate something.

05Verdict on each option

Commit to a prediction for each graph, then check it against what you measured. The verdict column is computed from your logged samples — not from a stored answer key.

Option (A) — [P] against t. Is the printed graph correct?

06Step-by-step solution

Earn it first

Complete 0/2 runs and inspect 0/4 sets of axes, and the derivation unlocks with your own measured numbers in it.

Write the rate law and integrate it

First order in R means the rate is proportional to [R], with a single constant of proportionality k:

rate = −d[R]/dt = k[R] ⟹ [R] = [R]₀ e^−kt

R only becomes P, so whatever R loses, P gains:

[P] = [R]₀ − [R] = [R]₀(1 − e^−kt)
(A) [P] against t — the curvature is wrong

d[P]/dt = k[R]₀e^−kt · d²[P]/dt² = −k²[R]₀e^−kt < 0 for all t

The slope is largest at t = 0 and decays to zero, so [P] rises fast at first and then flattens onto the plateau [R]₀ — concave down. The printed graph starts flat and gets steeper, which is concave up. (A) is incorrect.
(B) d[R]/dt against [R] — the sign is wrong

d[R]/dt = −k[R]

That is a straight line through the origin, so the printed shape is right — but its slope is −k, a negative number. R is being consumed, so d[R]/dt is negative at every instant and the line must fall below the axis. The printed line rises. (B) is incorrect. Had the axis been labelled −d[R]/dt, or rate, it would have been right — which is exactly the trap.
(C) d[P]/dt against t — correct

d[P]/dt = +k[R] = k[R]₀e^−kt

Positive throughout, starting at its maximum k[R]₀ and decaying exponentially to zero as R runs out — an exponential decay curve, exactly as printed. (C) is correct.
(D) k against t — correct

The question fixes the temperature. The rate constant depends on temperature through k = A e^−Ea/RT and on nothing else — time does not appear. It is a constant of the reaction, so plotting it against t gives a horizontal line. (D) is correct.

the rate falls as the reaction proceeds; the rate constant does not
Answer

(C) and (D)

Marking: +4 only for choosing both; +2 for either one alone; −1 for including (A) or (B).

✓ Independent verification

Curvature computed numerically: for [R]₀ = 1 M, k = 0.30 s⁻¹ the second derivative of [P] is negative at every sampled point (maximum value −2×10⁻⁴), confirming concave down and killing (A).
Sign swept over the whole run: the largest value d[R]/dt reaches is −7×10⁻⁴ M s⁻¹ — it is never positive, so the line in (B) cannot rise.
Linearity check for (B): a least-squares fit of d[R]/dt against [R] returns slope −0.3000 = −k exactly, so the printed straight-line-through-the-origin shape is right and only the sign is wrong.
Exponential decay check for (C): the ratio of d[P]/dt at t = 0 to its value at t = 1/k is 2.7196 ≈ e, the signature of a true exponential; monotonically decreasing and positive throughout.
Half-life independence: computed for [R]₀ = 0.5, 1, 2 and 5 M, the half-life is 2.3105 s every time — matching ln2/k and confirming genuine first-order behaviour across the whole input range, not just at the default values.
Dimensional check: k[R] has units s⁻¹ × mol L⁻¹ = mol L⁻¹ s⁻¹, which is a rate. A first-order k is s⁻¹ and carries no concentration unit, which is why it cannot drift with time or with [R].
Distractor audit: (A) confuses the shape of [P] with the shape of an exponential growth; (B) is the classic missing-minus-sign — the magnitude plot −d[R]/dt vs [R] would have been correct. Both are single, believable slips, which is why the sign and the curvature both have to be checked explicitly.
Live model: this page derives every verdict from the samples you logged, using the same integrated rate law — it holds no stored answer key.

07Concepts and classification

Relations used

−d[R]/dt = k[R] (first order)

[R] = [R]₀ e^−kt · ln[R] = ln[R]₀ − kt

[P] = [R]₀(1 − e^−kt) · [R] + [P] = [R]₀

d[P]/dt = −d[R]/dt = k[R]

t_½ = ln2 / k — independent of [R]₀

k = A e^−Ea/RT — depends on T, never on t

Classification

Simulation ID	`ADV-2026-P2-CHE-Q05`
Exam / Year / Paper	JEE Advanced · 2026 · Paper 2
Subject	Chemistry (Physical)
Chapter	Chemical Kinetics
Topic	First-order reactions
Subtopics	Integrated first-order rate law · Rate vs rate constant · Sign conventions for d[R]/dt and d[P]/dt · Half-life independence from concentration · Arrhenius temperature dependence · Reading kinetics graphs
Question type	Multiple correct MCQ (+4 / partial / −1)
Difficulty	Moderate
Answer	— run the experiment

01Original question — verbatim

02The reactor

Vessel

Controls

Live readings

03Plot your own data on each option's axes

Axes

04Half-life trials — the first-order signature

Trials

Observations

05Verdict on each option

06Step-by-step solution

Earn it first

Write the rate law and integrate it

(A) [P] against t — the curvature is wrong

(B) d[R]/dt against [R] — the sign is wrong

(C) d[P]/dt against t — correct

(D) k against t — correct

Answer

✓ Independent verification

07Concepts and classification

Relations used

Classification