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For the reaction AA \rightarrow products, successive half-lives are observed to be 10.0, 20.0, and 40.0 min for an experiment in which [A]0=0.10 M.[ \mathrm { A } ] _ { 0 } = 0.10\ \mathrm { M } . Calculate the concentration of A at the following times. 80.0 min

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Answered 2 years ago
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Judging by the fact that half-life times are rising it is clear this is a second-order reaction\textbf{second-order reaction}. We can, therefore, determine the reaction rate constant:

t1/2=1k[A]0\text{t}_{1/2} = \dfrac{1}{\text{k}\left[\text{A}\right]_0}

k=1t1/2[A]0=110 min×0.10 M\text{k} = \dfrac{1}{\text{t}_{1/2}\left[\text{A}\right]_0} = \dfrac{1}{10 \ \text{min} \times 0.10 \ \text{M}}

k=1 M1 s1\boxed{\text{k}= 1 \ \text{M}^{-1} \ \text{s}^{-1}}

Therefore:

1[A]=kt+1[A]0=1 M1 s1×80 min+10.1 M=90 M1\dfrac{1}{\left[\text{A}\right]} = \text{kt} + \dfrac{1}{\left[\text{A}\right]_0} = 1 \ \text{M}^{-1} \ \text{s}^{-1} \times 80 \ \text{min} + \dfrac{1}{0.1 \ \text{M}} = 90 \ \text{M}^{-1}

[A]=0.01 M\boxed{\left[\text{A}\right] = 0.01 \ \text{M}}

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