[rajajidwivedi]

when we consider the rate constant we see that rate does not depend on the initial concentration but when we see the formula for the rate constant for the second order reaction we see that K = x/t.a(a-x) if we assume that we have changed the conc 'm' times then we can see that rate constant will be 1/m times of previous one, how can it be?
thanks in advance
prakash dwivedi

[juanrga]

when we consider the rate constant we see that rate does not depend on the initial concentration but when we see the formula for the rate constant for the second order reaction we see that K = x/t.a(a-x) if we assume that we have changed the conc 'm' times then we can see that rate constant will be 1/m times of previous one, how can it be?
thanks in advance
prakash dwivedi

The rate constant does not depend on the concentration (initial or otherwise)

For second order kinetics

$$ \frac{1}{[A]} - \frac{1}{[A]_0} = kt /$$

which has been obtained from integrating the rate equation for k constant. Thus k in the above expression cannot be a variable and, indeed, it is not

$$ k = \frac{[A]_0-[A]}{t[A][A]_0} = \frac{[A]_0-[A]'}{t'[A]'[A]_0} /$$

[Kadabrium]

if a change will a0  also

[juanrga]

if a change will a0  also

$$[A]_0/$$ is a constant.