If $a$ is the initial concentration and $a - x$ is the concentration of reactant at time ' $t$ ', then the rate constant equation for second order is:

K = \frac{2.303}{t} \frac{a}{a - x}

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K = \frac{1}{t} \frac{a}{a - x}

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K = \frac{1}{t} \frac{x}{a (a - x)}

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K = \frac{1}{t^{2}} \frac{a}{(a - x)}

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Solution

The correct option is C $K = \frac{1}{t} \frac{x}{a (a - x)}$
For a second-order reaction, rate equation is $\frac{- d x}{d t} = K x^{2} .$
The rate constant for a second-order reaction is expressed as $K = \frac{1}{t} [\frac{1}{A_{t}} - \frac{1}{A_{0}}] .$
Now putting,
$A_{0} = a$
$A_{t} = a - x$
We get,
$K = \frac{1}{t} [\frac{1}{a - x} - \frac{1}{a}]$
$K = \frac{1}{t} \frac{x}{a (a - x)}$

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If a is the initial concentration and a−x is the concentration of reactant at time 't', then the rate constant equation for second order is:

The correct option is C K=1txa(a−x)For a second-order reaction, rate equation is −dxdt=Kx2.The rate constant for a second-order reaction is expressed as K=1t[1At−1A0].Now putting, A0=aAt=a−xWe get,K=1t[1a−x−1a]K=1txa(a−x)

If $a$ is the initial concentration and $a - x$ is the concentration of reactant at time ' $t$ ', then the rate constant equation for second order is: