JEE Advanced 2026 — Paper 2 — CHEM-II solution

JEE Advanced2026Paper 2CHEM-II

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
(A)
B
(B)
C
(C)
D
(D)

Solution

(A) $[P]$ vs $t$ . For a first-order reaction starting from $[R]_{0}$ and no product:

$[P]_{t} = [R]_{0} (1 - e^{- k t})$ .

This curve rises from zero and asymptotically approaches $[R]_{0}$ from below (concave down, saturating). The exponential-growth shape shown in option (A) is wrong. Incorrect.

(B) $\frac{d [ R ]}{d t}$ vs $[R]$ . Rate of disappearance: $- \frac{d [ R ]}{d t} = k [R]$ , so

$\frac{d [ R ]}{d t} = - k [R]$ .

The slope is negative, so the correct line lies in the fourth quadrant (negative $y$ , positive $x$ ), not the first. Incorrect.

(C) $\frac{d [ P ]}{d t}$ vs $t$ . Since $\frac{d [ P ]}{d t} = k [R] = k [R]_{0} e^{- k t}$ , this quantity starts at $k [R]_{0}$ and decays exponentially to zero. The shape matches. Correct.

(D) $k$ vs $t$ . The rate constant depends only on temperature, not on time, so the plot is a horizontal line. Correct.

$∴$ Correct options are (C) and (D).

← Back to full paper

Practice more CHEM-II

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →