JEE Advanced 2022 Paper 1 · Q15 · Rate Laws & Order
Match the rate expressions in LIST-I for the decomposition of X with the corresponding profiles provided in LIST-II. X$_s$ and k are constants having appropriate units.
LIST-I: (I) rate = k[X]/(X$_s$ + [X]) under all possible initial concentrations of X (II) rate = k[X]/(X$_s$ + [X]) where initial concentrations of X are much less than X$_s$ (III) rate = k[X]/(X$_s$ + [X]) where initial concentrations of X are much higher than X$_s$ (IV) rate = k[X]$^2$/(X$_s$ + [X]) where initial concentration of X is much higher than X$_s$
LIST-II (graphical profiles): (P) half-life (t$_{1/2}$) increases linearly with initial concentration of X (zero-order plot of t$_{1/2}$ vs [X]$_0$). (Q) half-life t$_{1/2}$ is constant (independent of initial concentration), characteristic of first-order kinetics. (R) rate vs initial concentration of X — rate increases linearly with [X]$_0$ (first-order). (S) rate vs initial concentration of X — rate is constant (independent of [X]$_0$), characteristic of zero-order. (T) plot of [X] (or ln[X]) vs time — linear decay characteristic of zero-order.
Reveal answer + step-by-step solution
Correct answer:A
Solution
Analyse each rate law:
(I) rate = k[X]/(X$_s$ + [X]), all [X]. The half-life behaviour interpolates between zero-order (large [X]) and first-order (small [X]). For the half-life vs [X]$_0$ plot, when [X]$_0$ is small the kinetics are first order (constant t$_{1/2}$), and when [X]$_0$ is large the kinetics are zero order (t$_{1/2}$ varies linearly with [X]$_0$). The combined plot shows a constant region at low [X]$_0$ that transitions to a linear-rising region at high [X]$_0$. This corresponds to profile (P).
(II) When [X] $\ll$ X$_s$, the denominator $\approx$ X$_s$ and rate $\approx$ (k/X$_s$)[X], which is first-order in [X]. First-order kinetics have a constant half-life independent of [X]$_0$. This is profile (Q).
(III) When [X] $\gg$ X$_s$, the denominator $\approx$ [X] and rate $\approx$ k (zero-order — independent of [X]). For zero-order kinetics, the rate-vs-[X]$_0$ plot is a flat horizontal line. This is profile (S).
(IV) rate = k[X]$^2$/(X$_s$ + [X]) with [X] $\gg$ X$_s$ simplifies to rate $\approx$ k[X], which is first-order. For first-order kinetics, [X] decays exponentially, so a plot of [X] vs time is exponential; FIITJEE matches this to profile … [truncated]
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