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Grade: 12
        

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

23 days ago

Answers : (1)

Khimraj
2468 Points
							
Initial conc of A = 4[B]
initial conc of B = [B]
conc of A after n1 half life = 4[B]/2n1
conc of B after n2 half life = [B]/2n2
for conc of A = conc of B
4[B]/2n1 = [B]/2n2
2(n1 – n2) = 22
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.....................
23 days ago
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