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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 A,B & C at the end of 50% reaction is:

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Answer: 5
After 50% reaction is complete,half of moles of A remain in solution,so, $n(A)=2$
$n(B)=\large\frac{3k_1}{k_1+k_2}$$([A_o]-[A]_t)$ $= \large\frac{3}{2}$
$n(C)=\large\frac{k_2}{k_1+k_2}$$([A_0]-[A]_t)$ $= \large\frac{3}{2}$
Total moles =n(A)+n(B)+n(C) $= 5$
answered Dec 16, 2013 by sreemathi.v
edited Jul 26, 2014 by balaji.thirumalai

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