For the first-order reaction,T_{av} (average | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a first$-order$ reaction,the half$-life$ is $T_{50} = \frac{0.693}{k}$.Average life is $T_{av} = \frac{1}{k} \approx 1.44 	imes T_{50}$.For $

Vedclass · Accepted Answer

$T_{50} < T_{av} < T_{75}$

Vedclass · Answer

(A) For a first$-order$ reaction,the half$-life$ is $T_{50} = \frac{0.693}{k}$.Average life is $T_{av} = \frac{1}{k} \approx 1.44 	imes T_{50}$.For $75\%$ completion,the time taken is $T_{75} = \frac{2.303}{k} \log(\frac{100}{100-75}) = \frac{2.303}{k} \log(4) = \frac{2.303 	imes 0.602}{k} \approx 2 	imes T_{50}$.Comparing the values: $T_{50} \approx 1.00 	imes T_{50}$,$T_{av} \approx 1.44 	imes T_{50}$,and $T_{75} \approx 2.00 	imes T_{50}$.Therefore,the increasing order is $T_{50} < T_{av} < T_{75}$.

For the first$-order$ reaction,$T_{av}$ (average life),$T_{50}$ and $T_{75}$ in the increasing order are:

Similar Questions

For which of the following graphs of a first-order reaction will the value of the slope be $\frac{K}{2.303}$?

If the rate law is $r = K[A]$,then the concentration of the reactant remaining after time $t = 1/k$ is: ($[A]_0$ is the concentration of the reactant at $t = 0$)

What is the value of the rate constant for a first-order reaction if the slope for the graph of rate versus concentration is $2.5 \times 10^{-3}$?

If the concentration of a first order reaction is increased by $x$ times,then the rate constant $(k)$ becomes

Vedclass Test Series

Exam Paper Generator

Online Exam Module