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Standard XII

Physics

KE & PE Graph

For a second ...

Question

For a second order kinetics, a straight line is obtained by plotting dx/dt against:

a -x

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a+x

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log C

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1 / (a - x)^{2}

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Solution

The correct option is A $1 / (a - x)^{2}$
$\frac{1}{[A]_{t}} = - k t + \frac{1}{[A o]}$
$\frac{1}{a - x} = - k t + \frac{1}{a} \to b y \frac{d t}{d x} b o t h s i n e$
$- 1 \frac{d t}{d x} {[\frac{1}{a - x}]}^{2} = t k$
$\frac{d x}{d t} = \frac{1}{k} {[\frac{1}{a - x}]}^{2}$

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For a second order kinetics, a straight line is obtained by plotting dx/dt against:

The correct option is A 1/(a−x)21[A]t=−kt+1[Ao]1a−x=−kt+1a→by dtdxboth sine−1dtdx[1a−x]2=tkdxdt=1k[1a−x]2

The correct option is A $1 / (a - x)^{2}$
$\frac{1}{[A]_{t}} = - k t + \frac{1}{[A o]}$
$\frac{1}{a - x} = - k t + \frac{1}{a} \to b y \frac{d t}{d x} b o t h s i n e$
$- 1 \frac{d t}{d x} {[\frac{1}{a - x}]}^{2} = t k$
$\frac{d x}{d t} = \frac{1}{k} {[\frac{1}{a - x}]}^{2}$