A flask contains a mixture of compounds A and B. Both compounds decompose by first - order kinetics. The half lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is: (Use ln 2 = 0.693 )
A flask contains a mixture of compounds A and B. Both compounds decompose by first - order kinetics. The half lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is: (Use ln 2 = 0.693 )
Option 1 -
120
Option 2 -
300
Option 3 -
900
Option 4 -
180
Asked by Shiksha User
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1 Answer
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Correct Option - 3
Detailed Solution:For A
0.693/300 = (2.303/t) log (A? /A? )For B
0.693/180 = (2.303/t) log (B? /B? )Given A? = B? & A? = 4B?
Substituting & solving we get t = 900 s
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