The decomposition of ammonia on platinum surface is a zero order reaction. How much time it will take for \(1 × 10^{-4} \text{mol L}^{-1}\) of ammonia to reduce into half of its concentration?
(\(k = 0.5 × 10^{-4}\text{ mol L}^{-1}s^{-1}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1 s
For a zero order reaction the rate does not depend on concentration, so the concentration falls linearly with time:
\( [R] = [R]_0 - kt \)
At the half-life [R] = [R]0/2, which gives
\( \displaystyle t_{1/2} = \frac{[R]_0}{2k} \)
With [R]0 = 1 × 10−4 mol L−1 and k = 0.5 × 10−4 mol L−1 s−1:
\( \displaystyle t_{1/2} = \frac{1 \times 10^{-4}}{2 \times 0.5 \times 10^{-4}} = 1\ \text{s} \)
On platinum the surface is saturated with \( \ce{NH3} \) at high pressure, so the rate stays constant:
\( \ce{2NH3(g)} \xrightarrow[\text{Pt catalyst}]{\text{1130 K}} \ce{N2(g) + 3H2(g)} \); Rate = k[NH3]0 = k
NCERT (2026-27 edition):
Zero Order Reactions – …
