For a second order reaction $\frac{d x}{d t} = k (a - x)^{2} .$ 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 $\frac{1}{a . k}$
In general for $n^{t h}$ order reaction,

$t_{1 / 2 (n)} = \frac{2^{n - 1} - 1}{(n - 1) k_{n} (a)^{n - 1}} (n \geq 2)$

$\Rightarrow t_{1 / 2 (n)} \propto \frac{1}{a^{(n - 1)}}$

$\Rightarrow t_{1 / 2 (n)} \propto a^{(1 - n)}$

For second order half life will be $\frac{1}{a . k}$

Hence, the correct option is $A$

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Similar questions

Q. For a second order reaction dx/dt =

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 dxdt=k(a−x)2. Its half life period is:

The correct option is A 1a.kIn general for nth order reaction,t1/2(n)=2n−1−1(n−1)kn(a)n−1(n≥2)⇒t1/2(n)∝1a(n−1)⇒t1/2(n)∝a(1−n)For second order half life will be 1a.kHence, the correct option is A

For a second order reaction $\frac{d x}{d t} = k (a - x)^{2} .$ Its half life period is: