Given that for a reaction of \( n^{\text {th }} \) order, the integrated rate equation is: \[ K=...
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0Given that for a reaction of \( n^{\text {th }} \) order, the integrated rate equation is:
\[
K=\frac{1}{t(n-1)}\left[\frac{1}{C^{n-1}}-\frac{1}{C_{0}^{n-1}}\right]
\]
where, \( C \) and \( C_{0} \) are the concentration of reactant at time \( t \) and initially respectively. The \( t_{3 / 4} \) and \( t_{1 / 2} \) are related as : \( \left(t_{3 / 4}\right. \) is time required for \( C \) to become \( \left.C_{1} / 4\right) \)
(A) \( t_{3 / 4}=t_{1 / 2}\left[2^{n-1}+1\right] \)
(B) \( t_{3 / 4}=t_{1 / 2}\left[2^{n-1}-1\right] \)
(C) \( t_{3 / 4}=t_{1 / 2}\left[2^{n+1}+1\right] \)
(D) \( t_{3 / 4}=t_{1 / 2}\left[2^{n+1}-1\right] \)
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