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For a first order $Rx_n$ of rate constant K,$\;t_{ 2/3}$ the time of a reactant to drop to $1/3$) is given by:

Note: $ln \;2=0.693$ & $ln\; 3=1.1$

$\begin{array}{1 1}(a)\;0.693/K&(b)\;1.1/k\\(c)\;0.407/K&(d)\;1.793/K\end{array}$

1 Answer

$\ln\big[\large\frac{[A_o]}{[A_t]}\big]$$=Kt$
$\ln\big[\large\frac{3[A_o]}{[A_o]}\big]$$=Kt_{\large\frac{2}{3}}$
$\Rightarrow \large\frac{1.1}{K}$
Hence (b) is the correct answer.
answered Dec 13, 2013 by sreemathi.v
 

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