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  • For a first order reaction, the slope of the graph between  <img alt="\mathrm{\ln _e\left([A]_0 / [A]\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cln%20_e%

For a first order reaction, the slope of the graph between  \mathrm{\ln _e\left([A]_0 / [A]\right)} (on y-axis) and time (min) (on x-axis) is found out to be \mathrm{0.5 min{^{-1}}} . The time at which 25 % of the initial reactant is remaining is:

Here,
\mathrm{[A]_0 } is the initial concentration
\mathrm{[A]} is the concentration at time t

Option: 1

1.38 min


Option: 2

2.77 min


Option: 3

0.693 min


Option: 4

0.5 min


Answers (1)

best_answer

For a first order:
\mathrm{[A]=[A]_0 e^{-k t}}
\mathrm{ \ln _e\left(\frac{[A]_0}{[A]}\right)=k t }
Comparing with

y=mx

Here slope is the rate constant.
So,

\mathrm{ k=0.5 \ min ^ {-1} }

At
\mathrm{ \ln _e\left(\frac{[A]_0}{[A]_0}\right)=k t \\ }

\mathrm{ \ln _e 4=0.5 \times t \\ }
\mathrm{ t=\frac{2 \times 0.693}{0.5} \mathrm{~min} \\ }

\mathrm{ t=2.77 \mathrm{~min} }.
 

Posted by

Ritika Kankaria

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