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A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min. respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ______ min.
(Round off to the nearest integer)
    Correct answer is '108'. Can you explain this answer?
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    A and B decompose via first order kinetics with half-lives 54.0 min an...
    Concept: First-order kinetics, Half-life, Equimolar mixture

    Solution:

    Let the initial concentration of A and B be 'x' mol/L each.

    After time 't', the concentrations of A and B can be given as:

    [A] = x.e^(-kt) (1)

    [B] = x.e^(-kt/3) (2)

    where 'k' is the rate constant.

    Given, half-life of A = 54.0 min

    Hence, k = ln(2)/54.0 = 0.0128 min^(-1)

    Similarly, half-life of B = 18.0 min

    Hence, k = ln(2)/18.0 = 0.0386 min^(-1)

    Now, we need to find the time taken for [A] to become 16 times that of [B].

    i.e., [A]/[B] = 16

    Substituting equations (1) and (2) in the above equation, we get:

    e^(kt/3) = 16e^(kt)

    Taking natural logarithm on both sides, we get:

    kt/3 = ln(16) + kt

    Solving for 't', we get:

    t = 108.2 min

    Rounding off to the nearest integer, the time taken for [A] to become 16 times that of [B] is 108 min.

    Therefore, the answer is 108.
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    A and B decompose via first order kinetics with half-lives 54.0 min and 18.0 min. respectively. Starting from an equimolar non reactive mixture of A and B, the time taken for the concentration of A to become 16 times that of B is ______ min.(Round off to the nearest integer)Correct answer is '108'. Can you explain this answer?
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