Question #c81c4

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Question #c81c4

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Answer 1 · 2015-11-03T12:51:59

$t = 248791.2 sec = 69.1 t$ $t=248791.2 sec = 69.1 t$

Explanation:

$r a t e = - \frac{d A}{d t} = k {[A]}^{2}$ $rate=-(dA)/(dt)=k[A]^2$

Integrated rate Law:
$\frac{1}{{[A]}_{t}} = k \cdot t + \frac{1}{{[A]}_{0}}$ $1/([A]_t)=k*t+1/([A]_0)$

0,249 M of A is used = $95.77 %$ $95.77%$
0.011 M of A is not used = $4.23 %$ $4.23%$

fraction of A not used: $\frac{4.23}{100} = \frac{1}{23.64}$ $4.23/100=1/23.64$

${[A]}_{t} = \frac{{[A]}_{0}}{23.64}$ $[A]_t=([A]_0)/23.64$

and $t = t_{\frac{1}{23.64}}$ $t=t_(1/23.64)$

$\frac{1}{\frac{{[A]}_{0}}{23.64}} = k \cdot t_{\frac{1}{23.64}} + \frac{1}{{[A]}_{0}}$ $1/(([A]_0)/23.64)=k*t_(1/23.64)+1/[A]_0$

$t_{\frac{1}{23.64}} = \frac{1}{k} (\frac{23.64}{{[A]}_{0}} - \frac{1}{{[A]}_{0}}) = \frac{1}{3.5 \cdot 10^{- 4}} (\frac{22.64}{0.260})$ $t_(1/23.64)=1/k(23.64/[A]_0-1/[A]_0)=1/(3.5*10^(-4))(22.64/0.260)$

$t_{\frac{1}{23.64}} = 248791.2 sec = 69.1 t$ $t_(1/23.64)=248791.2 sec = 69.1 t$

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Question #c81c4

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