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For second order reaction

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$K = \frac{2 .303}{t (b - a)} \log \frac{a (a - x)}{b (b - x)}$

$K = \frac{2 .303}{t (a - b)} \log \frac{a (a - x)}{b (b - x)}$

$K = \frac{2 .303}{t} \log \frac{b (a - x)}{a (b - x)}$

$K = \frac{2 .303}{t (a - b)} \log \frac{b (a - x)}{a (b - x)}$

answer is B.

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Detailed Solution

For a second order reaction when both ractants are differnet and having differnt concentrations then rate constant

K = $\frac{2.303}{t (a - b)} \log \frac{b (a - x)}{a (b - x)}$

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