In a second order reaction, the time needed for the initial concentration of the reactant to reduce to half that value is:

independent of the initial concentration

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proportional to the initial concentration

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inversely proportional to initial concentration

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proportional to the square of the initial concentration

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Solution

The correct option is C inversely proportional to initial concentration
We know that for a $n^{t h}$ order reaction,
$h a l f l i f e \propto \frac{1}{a^{n - 1}}$
Hence for a $2^{n d}$ order reaction; $t_{1 / 2} \propto \frac{1}{a^{n - 1}} \propto \frac{1}{a}$ (where $a$ is initial concentration).

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A + B_{2} \to A B + B

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In a second order reaction, the time needed for the initial concentration of the reactant to reduce to half that value is:

The correct option is C inversely proportional to initial concentrationWe know that for a nth order reaction, halflife∝1an−1Hence for a 2nd order reaction; t1/2∝1an−1∝1a (where a is initial concentration).

The correct option is C inversely proportional to initial concentration
We know that for a $n^{t h}$ order reaction,
$h a l f l i f e \propto \frac{1}{a^{n - 1}}$
Hence for a $2^{n d}$ order reaction; $t_{1 / 2} \propto \frac{1}{a^{n - 1}} \propto \frac{1}{a}$ (where $a$ is initial concentration).