Two substances A ( = 5 min) and B( = 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

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Khimraj · Answer

Initial conc of A = 4[B] initial conc of B = [B] conc of A after n1 half life = 4[B]/2 n1 conc of B after n2 half life = [B]/2 n2 for conc of A = conc of B 4[B]/2 n1 = [B]/2 n2 2 (n1 – n2) = 2 2 n1 – n2 = 2 – (1) also t = n1*5 = n2*15 so n1 = 3*n2 so 3*n2 – n2 = 2 so n2 = 1 so time taken = 1*15 = 15 min Hope it clears.....................

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Two substances A ( = 5 min) and B( = 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

Two substances A ( = 5 min) and B(= 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

Two substances A ( = 5 min) and B(= 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

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Two substances A ( = 5 min) and B( = 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

Two substances A ( = 5 min) and B(= 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

Two substances A ( = 5 min) and B(= 15 min) are taken in such a way that initially [A] = 4[B]. The time after which both the concentrations will be equal is : (assume that reaction is first order

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