Match the rate expressions in List-I for the | vedclass

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(A) $(I)$ $	ext{rate} = \frac{k[X]}{X_s + [X]}$. As $[X]$ increases,rate increases and approaches $k$ (saturation). This matches profile $(R)$.$(II)$

Vedclass · Accepted Answer

$I$ $\rightarrow R; II$ $\rightarrow T; III$ $\rightarrow S; IV$ $\rightarrow Q$

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(A) $(I)$ $	ext{rate} = \frac{k[X]}{X_s + [X]}$. As $[X]$ increases,rate increases and approaches $k$ (saturation). This matches profile $(R)$.$(II)$ If $[X] \ll X_s$,then $	ext{rate} \approx \frac{k[X]}{X_s} = k'[X]$. This is a first-order reaction. For first-order,$\ln[X]$ vs time is linear $(T)$ and $t_{1/2}$ is constant $(Q)$. Given the options,$(T)$ is the standard kinetic profile.$(III)$ If $[X] \gg X_s$,then $	ext{rate} \approx \frac{k[X]}{[X]} = k$. This is a zero-order reaction. For zero-order,$[X]$ vs time is linear $(S)$.$(IV)$ If $[X] \gg X_s$,then $	ext{rate} \approx \frac{k[X]^2}{[X]} = k[X]$. This is a first-order reaction. For first-order,$t_{1/2}$ is independent of initial concentration $(Q)$.

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.

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