ete59. In the following first order competing reactionsonA+ Reagent ->

1.	ete59. In the following first order competing reactionsonA+ Reagent -> Product, B + Reagent ->ProductThe ratio of K1 / K2 if only 50% of B will havebeen reacted. When 94% of A has been reactedis -(1) 4.06(3) 2.06ofion,(2) 0.246(4) 0.06
Answer» From the question , at a given time t for A , the [A] left is 100-94 = 6% so, K1 = ln(100/6)/t => t = ln(100/6)/k1....(1) and for B , the [B] is 50% so, K2 = ln(100/50)/t => t = ln2/k2.....(2) equating 1 and 2 we get , k1/k2 = ln(100/6)/ln2 = 2.81/0.6931 = 4.05889 = 4.06 answer option 1

ete59. In the following first order competing reactionsonA+ Reagent -> Product, B + Reagent ->ProductThe ratio of K1 / K2 if only 50% of B will havebeen reacted. When 94% of A has been reactedis -(1) 4.06(3) 2.06ofion,(2) 0.246(4) 0.06

Answer»

From the question , at a given time t

for A , the [A] left is 100-94 = 6% so, K1 = ln(100/6)/t => t = ln(100/6)/k1....(1)

and for B , the [B] is 50% so, K2 = ln(100/50)/t => t = ln2/k2.....(2)

equating 1 and 2

we get , k1/k2 = ln(100/6)/ln2 = 2.81/0.6931 = 4.05889 = 4.06 answer

option 1