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0 votes
21.2k views
in Chemical kinetics by (79.8k points)
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Two substances A(t1/2 = 5 mins) and B(t1/2 = 15 mins) are taken is such a way that initially [A] = 4[B]. The time after which the conentration of both will be equal is 

(A) 5 min 

(B) 15 min 

(C) 20 min 

(D) concentration can never be equal 

2 Answers

+1 vote
by (18.1k points)
selected by
 
Best answer

Given t1/2 of (A) = 5 min

t1/2 of (B) = 15 min.

and initially = [A] = 4[B]

using hit and trial method -

Let after 15 min the concentration of [A]5 = [B]15

using formula -

N = N0 (1/2)n

where,

N = concentration after n half life time

N0 = initial concentration

n = Number of half life time.

Therefore,

concentration of A after 15 min -

[A]15 = [A]0 (1/2)3 where [B]0 = initial concentration

[A]15 = [A]0/8

concentration of B after 15 min -

[B]15 = [B]0 (1/2)1

[B]15 = [B]0/2

∴ [A]15 = [B]15

⇒ [A]0/8 = [B]0/2

∵ Given [A] = 4[B]

∴ 4[B]0/8 = [B]0/2

⇒ [B]0/2 = [B]0/2

Hence, after 15 min. the concentration of A and B become equal.

+2 votes
by (76.1k points)

Correct Option (B) 15 min  

Explanation:

t1/2 of A is 5 min

∴ in 15 mins it will become 1/8 of initial and t1/2 of B is 15 mins 

 ∴ in 15 mins it will become 1/2 of initial 

∴ ratio of [A] : [B] after 15 min is 4 : 1 

But given [A] = 4[B] 

 [A] = [B] after 15 min

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