The reaction 2X B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When…

For zero order: $t_{1/2} = \frac{[\mathrm{X}]_0}{2k}$

From given data: $6 = \frac{0.2}{2k} \Rightarrow k = \frac{0.2}{12} = \frac{1}{60}\,\mathrm{M\,h^{-1}}$

Time to go from 0.5 M to 0.2 M: $t = \frac{[\mathrm{X}]_0 - [\mathrm{X}]}{k} = \frac{0.5 - 0.2}{1/60} = 0.3 \times 60 = 18\,\mathrm{h}$

Rate $= k$ (zero order), $-\frac{d[\mathrm{X}]}{dt} = 2k$ (since 2 mol X consumed per step).

$t_{1/2}$ for X: $\frac{[\mathrm{X}]_0}{2 \times 2k} = 6 \Rightarrow 2k = \frac{0.2}{12} = \frac{1}{60}$, so effective rate $= \frac{1}{60}\,\mathrm{M\,h^{-1}}$.

$t = \frac{0.5 - 0.2}{1/60} = 18\,\mathrm{h}$. Still 18.

With $k_{eff} = \frac{[X]_0}{2 t_{1/2}} = \frac{0.2}{12}$: $t = \frac{0.3}{0.2/12} = \frac{0.3 \times 12}{0.2} = 18$. Answer: Option (3)