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Consider the following first order completing reactions : $X\;\overset{k_1}{\longrightarrow}\;A+B$ and $Y\;\overset{k_2}{\longrightarrow}\;C+D$ if $50\%$ of the reaction of $X$ was completed when $87.5\%$ of the reaction of $Y$ was completed , the ratio of their constants $(k_1/k_2)$ is

$(a)\;1\qquad(b)\;4.06\qquad(c)\;3\qquad(d)\;4.65$

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Answer: 0.333
Given 50% of the reaction of X was completed when 87.5% of the reaction of Y was completed, the time is the same in both cases..
$\Rightarrow$ Time $t = \large\frac{1}{k_1}$$ \ln $$\; \large\frac{100}{50}$$ = \large\frac{1}{k_2}$$ \ln $$\; \large\frac{100}{12.5}$
$\Rightarrow \large\frac{k_1}{k_2}$$ = \large\frac{\ln \large\frac{100}{50}}{\ln \large\frac{100}{12.5}}$$ = 0.333$
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai
 

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