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, This reaction follows first order kine...

, This reaction follows first order kinetics. The rate constant at particular temperature is `2.303xx10^(-2)"hour"^(-1)`. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? (log 2 = 0.3010)

A

0.125 M

B

0.215 M

C

`0.25xx2.303M`

D

0.05 M

Text Solution

Verified by Experts

The correct Answer is:
B

`k=(2.303/t)log(([A_(0)])/([A]))`
`rArr2.303xx10^(-2)"hour"^(-1)=(2.303/(1806 min))log(0.25/([A]))`
`rArr((2.303xx10^(-2)"hour"^(-1)xx1806min)/2.303)=log(0.25/([A]))rArr((1806xx10^(-2))/60)=log(0.25/([A]))`
`rArr0.301=log(0.25/([A]))rArrlog2=log(0.25/([A]))rArr2=(0.25/([A]))`
`[A]=(0.25/2)=0.125M`
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