# The reaction2 A → Bis second order. The initial concentration ofA is 1.258 M and its half-life is 69 sec.  Howlong will it take [A] to drop to one ninthof its initial value?Answer in units of s Hint: Use the correct integrated rate law equation for a second order reaction Before we can use the integrated rate law we must first find the rate constant for the reaction.  We can use the half life equation for a second order reaction:

t = 1/(k[A'])

where t is the half life, [A'] is the initial concentration, and k is the rate constant.  Plugging in the given values:

69 sec = 1/(k(1.258 M))

k = 0.0115 L/mol*sec

Now we can use the integrated rate law for a second order reaction:

1/[A] = kt + 1/[A']

Since the initial concentration of A is 1.258 M then 1/9 of that value is 1.258/9 = 0.14 M.  We can now input the various values and solve for t:

1/0.14 = 0.0115t + 1/1.258

7.143 = 0.0115t + 0.795

0.0115t = 6.348

t = 552 seconds

The answer is 552 seconds.

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