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For the reaction:
The rate is given by
$Rate = k_{1} [RX] [{OH}^{-}] + k_{2} [RX]$
Calculate % of $RX$ which reacts by 2nd order mechanism when $[OH]^{-}$ is 0.1 M.

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$\frac{k_{1} \times 100}{k_{1} + k_{2}}$

$\frac{k_{2} \times 100}{k_{1} + k_{2}}$

$\frac{10 k_{1}}{k_{1} + k_{2}}$

$\frac{100 k_{1}}{k_{1} + 10 k_{2}}$

answer is D.

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Detailed Solution

$\begin{aligned} % by 2 nd order = \frac{k_{1} [R X] [{OH}^{-}]}{k_{1} [R X] [{OH}^{-}] + k_{2} [R X]} \times 100 \\ = \frac{k_{1} \times 0.1}{k_{1} \times 0.1 + k_{2}} \times 100 = \frac{100 k_{1}}{k_{1} + 10 k_{2}} \end{aligned}$

Hence, the correct option is D.

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