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The half - life period of a first order chemical reaction is $6.93$ minutes. The time required for the completion of $99\%$ of the chemical reaction will be $(\log 2 =0.301)$

$(A)\;230.3\;minutes \\ (B)\;23.03\;minutes \\(C)\;46.06\;minutes \\(D)\;460.6\;minutes $

1 Answer

$K= \large\frac{0.693}{t_{Y2}}$
$\quad= \large\frac{0.693}{6.93}$
$\quad= minute^{-1}$
$K= \large\frac{2.303}{t} $$ \log \large\frac{100}{100-99}$
[if $a= 100, x=99\;and a-x=1]$
$\therefore \large\frac{0.693}{6.93}= \frac{2.303}{t} $$\log \large\frac{100}{1}$
$t= 46.06\;minute$
Hence C is the correct answer
answered Apr 3, 2014 by meena.p
 

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