On doubling the initial concentration, the half life of a reaction be…

The correct answer is Option (3) → 2

The half-life of a reaction depends on the initial concentration in a way that is fixed by the order of the reaction. From the integrated rate laws:

• zero order: \( t_{1/2} = \frac{[\mathrm{R}]_0}{2k} \), so \( t_{1/2} \propto [\mathrm{R}]_0 \)

• first order: \( t_{1/2} = \frac{0.693}{k} \), independent of \( [\mathrm{R}]_0 \)

• second order: \( t_{1/2} = \frac{1}{k[\mathrm{R}]_0} \), so \( t_{1/2} \propto \frac{1}{[\mathrm{R}]_0} \)

In general, for a reaction of order n:

    \( \displaystyle t_{1/2} \propto \frac{1}{[\mathrm{R}]_0^{\,n-1}} \)

Here doubling the initial concentration halves the half-life:

    \( \displaystyle \frac{t'_{1/2}}{t_{1/2}} = \left(\frac{1}{2}\right)^{n-1} = \frac{1}{2} \)

    \( \displaystyle n - 1 = 1 \Rightarrow n = 2 \)

So the reaction is of second order: the half-life is inversely proportional to the initial concentration.