The integrated rate law for a first-order reaction is as follows:

$\overline{){\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{t}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{+}}{\mathbf{ln}}{\mathbf{}}{{\mathbf{\left[}}{\mathbf{A}}{\mathbf{\right]}}}_{{\mathbf{0}}}}$

where:

**[A] _{t}** = concentration at time t

** **

**Calculate rate constant, k:**

let [A]_{0} = 100%

[A]_{t} = 100% - 75% = 25% (left of the reactant)

$\overline{){\mathbf{ln}}{\mathbf{}}{\mathbf{\left[}\mathbf{A}\mathbf{\right]}}_{{\mathbf{t}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{kt}}{\mathbf{+}}{\mathbf{ln}}{\mathbf{}}{{\mathbf{\left[}}{\mathbf{A}}{\mathbf{\right]}}}_{{\mathbf{0}}}}\phantom{\rule{0ex}{0ex}}\mathbf{ln}\mathbf{}\mathbf{(}\mathbf{25}\mathbf{\%}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{kt}\mathbf{+}\mathbf{ln}\mathbf{}\mathbf{(}\mathbf{100}\mathbf{\%}\mathbf{)}\phantom{\rule{0ex}{0ex}}\mathbf{3}\mathbf{.}\mathbf{2188}\mathbf{=}\mathbf{-}\mathbf{k}\mathbf{(}\mathbf{320}\mathbf{}\mathbf{s}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{.}\mathbf{6051}\phantom{\rule{0ex}{0ex}}\mathbf{3}\mathbf{.}\mathbf{2188}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{6051}\mathbf{=}\mathbf{-}\mathbf{k}\mathbf{(}\mathbf{320}\mathbf{}\mathbf{s}\mathbf{)}\phantom{\rule{0ex}{0ex}}\frac{\overline{)\mathbf{-}}\mathbf{1}\mathbf{.}\mathbf{3863}}{\overline{)\mathbf{-}}\mathbf{320}}\mathbf{=}\frac{\overline{)\mathbf{-}}\mathbf{k}\mathbf{\left(}\overline{)\mathbf{320}\mathbf{}\mathbf{s}}\mathbf{\right)}}{\overline{)\mathbf{-}\mathbf{320}\mathbf{}\mathbf{s}}}$

**k = 4.3321x10 ^{-3} s^{-1}**

**Solving for ****time t****:**

let [A]_{0} = 100%

[A]_{t} = 100% - 90% = 10% (left of the reactant)

A first-order reaction is 75.0% complete in 320. s.

b. How long does it take for 90.0% completion?

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