For a second order reaction dx/dt = $k (a - x)^{2}$ . Its half life period is:

\frac{1}{a . k}

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\frac{0.693}{k}

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\frac{a}{k}

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\frac{0.693}{a k}

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Solution

The correct option is A $\frac{1}{a . k}$
Half-life Method
Relation between half-life period of a reaction and initial concentration is as $: t^{1 / 2} α \frac{1}{a^{n - 1}}$
For first order reaction (Half life $α a^{0})$
For second order reaction (Half life $α 1 / a$ )
For third order reaction (Half life $α 1 / a^{2}$ )

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Similar questions

Q. For a second order reaction

\frac{d x}{d t} = k (a - x)^{2} .

Its half life period is:

Q. If the half life period of a reaction is

0.693 k

, its order will be

Match the following
Order of reaction Half-life period
1)1st order reaction a) $(k a_{0})^{- 1}$
2) Second order reaction b) $\frac{0.693}{k}$
3) Zero order reaction c) $\frac{a_{0}}{2 k}$

Match the following
$\begin{matrix} Order of reaction & Half-life period 1)1st order reaction & a) (k a_{0})^{- 1} 2) Second order reaction & b) \frac{0.693}{k} 3) Zero order reaction & c) \frac{a_{0}}{2 k} \end{matrix}$

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For a second order reaction dx/dt = k(a−x)2. Its half life period is:

The correct option is A 1a.kHalf-life MethodRelation between half-life period of a reaction and initial concentration is as :t1/2α1an−1For first order reaction (Half life αa0)For second order reaction (Half lifeα1/a)For third order reaction (Half life α1/a2)

For a second order reaction dx/dt = $k (a - x)^{2}$ . Its half life period is: