For a reaction A → C+D , the initial concentration of A is 0.010 M. A…

The correct answer is Option (3) → Second Order

Each order has its own integrated rate equation, so the same data give a different value of k for each order. The order is the one whose equation gives the stated numerical value, 9.0. Here [A]₀ = 0.010 M, [A] = 0.001 M and t = 100 s.

Zero order:

    \( \displaystyle k = \frac{[A]_0 - [A]}{t} = \frac{0.010 - 0.001}{100} = 9.0 \times 10^{-5}\,\mathrm{mol\,L^{-1}\,s^{-1}} \)

First order:

    \( \displaystyle k = \frac{2.303}{t}\log\frac{[A]_0}{[A]} = \frac{2.303}{100}\log\frac{0.010}{0.001} = \frac{2.303}{100}\times 1 = 0.0230\,\mathrm{s^{-1}} \)

Second order:

    \( \displaystyle k = \frac{1}{t}\left(\frac{1}{[A]} - \frac{1}{[A]_0}\right) = \frac{1}{100}\left(\frac{1}{0.001} - \frac{1}{0.010}\right) = \frac{1000 - 100}{100} = 9.0\,\mathrm{L\,mol^{-1}\,s^{-1}} \)

Third order:

    \( \displaystyle k = \frac{1}{2t}\left(\frac{1}{[A]^2} - \frac{1}{[A]_0^2}\right) = \frac{1}{200}\left(10^{6} - 10^{4}\right) = 4.95 \times 10^{3}\,\mathrm{L^2\,mol^{-2}\,s^{-1}} \)

Only the second order equation gives k = 9.0, so the reaction is second order.

Shortcut: the concentration falls to one-tenth. For second order, k = (1/[A] − 1/[A]₀)/t = (1000 − 100)/100, which is 9 at a glance; the other orders give a very small (zero, first) or very large (third) number.