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Given the following parallel reactions, where $\large\frac{k_1}{k_2}=\frac{1}{3}$. Initially only 4 moles of A are present. The total number of moles of C at the end of 50% reaction is:

$\begin{array}{1 1} 2 \\ 1 \\ 1.5 \\ 0.666 \end{array}$

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Answer: 1.5
$n(C)=\large\frac{k_2}{k_1+k_2}$$([A_0]-[A]_t)$ $= \large\frac{3}{2}$ (Given $\large\frac{k_1}{k_2}=\frac{1}{3}$)
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai
 

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