It was found that 0.1 mol L^-1 of ammonia took 200 s to be decomposed…

The correct answer is Option (1) → \(2.5 × 10^{-4}\, mol\, L^{-1}\, s^{-1}\)

In a zero order reaction the rate does not depend on the concentration, so the concentration falls by the same amount in each unit of time. The integrated rate equation is:

    \( \displaystyle [\mathrm{R}] = [\mathrm{R}]_0 - kt \)

At the half-life, \( [\mathrm{R}] = [\mathrm{R}]_0/2 \):

    \( \displaystyle \frac{[\mathrm{R}]_0}{2} = [\mathrm{R}]_0 - k\,t_{1/2} \)

    \( \displaystyle k = \frac{[\mathrm{R}]_0}{2\,t_{1/2}} \)

Here \( [\mathrm{R}]_0 = 0.1\ \mathrm{mol\,L^{-1}} \) and \( t_{1/2} = 200\ \mathrm{s} \):

    \( \displaystyle k = \frac{0.1\ \mathrm{mol\,L^{-1}}}{2 \times 200\ \mathrm{s}} \)

    \( \displaystyle k = 2.5 \times 10^{-4}\ \mathrm{mol\,L^{-1}\,s^{-1}} \)

So the rate constant is 2.5 × 10⁻⁴ mol L⁻¹ s⁻¹; the unit mol L⁻¹ s⁻¹ is the unit of rate itself, as expected for zero order.