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In the following first order reactions : $A+ Regent \quad \underrightarrow{K_1} \quad product; \qquad B+ Reagent \quad \underrightarrow {K_2} \quad Product$ The ratio of $\large\frac{K_1}{K_2}$ when only $50 \%$ of B reacts in a given time when $94 \%$ of a has been reacted is :

$(a)\;4.06 \\ (b)\;0.246 \\(c)\;2.06 \\(d)\;0.06 $
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1 Answer

$K_2= \large\frac{2.303}{t_2}$$ \log \large\frac{100}{50}$ for $50 \%\;B $ reacted.
$K_1= \large\frac{2.303}{t_1}$$ \log \large\frac{100}{6}$ for $94 \%\;A $ reacted.
$\therefore \large\frac{K_2}{K_1} =\frac{t_1}{t_2} \times \frac{0.3010}{1.2218}$
Since $t_2=t_1$ because $50\%\;B$ has reacted when $94 \%$ A has reacted.
$\therefore \large\frac{K_2}{K_1}=\frac{0.3010}{1.2218}$$=0.246$
and $\large\frac{K_1}{K_2}$$=4.006$
Hence c is the correct answer.
answered Apr 2, 2014 by meena.p

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