JEE Advanced 2026 Paper 2, Chemistry Section 2 Q1: Reactions of Various Orders

(A) $[\mathbf{P}]$ vs $t$. For a first-order reaction starting from $[\mathbf{R}{]}_0$ and no product:

$[\mathbf{P}{]}_t=[\mathbf{R}{]}_0(1-e^{-kt})$.

This curve rises from zero and asymptotically approaches $[\mathbf{R}{]}_0$ from below (concave down, saturating). The exponential-growth shape shown in option (A) is wrong. Incorrect.

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

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

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

(C) $\frac{d[\mathbf{P}]}{dt}$ vs $t$. Since $\frac{d[\mathbf{P}]}{dt}=k[\mathbf{R}]=k[\mathbf{R}{]}_0{e}^{-kt}$, this quantity starts at $k[\mathbf{R}{]}_0$ 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.

$\therefore$ Correct options are (C) and (D).