t_{100%} is the time required for the 100% co | vedclass

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(B) $1$. For a zero-order reaction: The rate constant is $k = [A]_0 / t_{100\%}$. Also, $t_{1/2} = [A]_0 / (2k)$, which implies $[A]_0 = 2kt_{1/2}$. S

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$t_{100\%} = 2t_{1/2}$ and $t_{100\%} = \infty$

Vedclass · Answer

(B) $1$. For a zero-order reaction: The rate constant is $k = [A]_0 / t_{100\%}$. Also, $t_{1/2} = [A]_0 / (2k)$, which implies $[A]_0 = 2kt_{1/2}$. Substituting this into the first equation: $k = (2kt_{1/2}) / t_{100\%}$, which simplifies to $t_{100\%} = 2t_{1/2}$.$2$. For a first-order reaction: The concentration at time $t$ is given by $[A]_t = [A]_0 e^{-kt}$. For $100\%$ completion, $[A]_t = 0$. This implies $e^{-kt} = 0$, which occurs only when $t 	o \infty$. Thus, theoretically, a first-order reaction never reaches $100\%$ completion in finite time.

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