In a certain chemical reaction the rate of conversion of asubstance is proprotional to the amount of the substance stillunreacted at that time. After 10 min one-third of the originalamount of the sustance has been reacted and 20g has been reactedafter 15min. What was the original amount of the substance?

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krolaniaN · Accepted Answer

let the original amount be P0 then dPdt=kP where k is proportionality constant. separating the variables. dPP=kdt integrating both sides ∫1PdP=∫kdt ln⁡P=kt+cI P=ekt+cI P=Cekt where C=ecI when t=0 min then amount =P0 then P0=C P=P0ekt when t=10 then amount =P03 then P03=P0e10k e10k=13⇒k=110ln⁡(13)≈−0.10986 P=P0e−0.10986t P0=20e0.10986×15≈104g

In a certain chemical reaction the rate of conversion of asubstance is proprotional to the amount of the substance stillunreacted at that time. After

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