It was found that $0.1\, mol\, L^{-1}$ of ammonia took 200 s to be decomposed to half the initial concentration. If the decomposition follows zero order reaction, then the rate constant is

$2.5 × 10^{-4}\, mol\, L^{-1}\, s^{-1}$

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

For a zero-order reaction, the half-life is given by:

$t_{1/2} = \frac{[A]_0}{2k}$​​

Given:

• Initial concentration, $[A]_0 = 0.1 \, \text{mol L}^{-1}$
• Half-life, $t_{1/2} = 200 \, \text{s}$

Rearranging the formula:

$k = \frac{[A]_0}{2t_{1/2}}$

$k = \frac{0.1}{2 \times 200} = \frac{0.1}{400} = 2.5 \times 10^{-4} \, \text{mol L}^{-1}\text{s}^{-1}$