The decomposition of NH3 on the platinum surface is zero order reaction. If k = 2.5 × 10−4 mol L−1 s−1 the rate of production of H2 is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 7.5 × 10−4 mol L−1 s−1
On a platinum surface ammonia decomposes as
\( \ce{2NH3(g)} \xrightarrow{\text{Pt}} \ce{N2(g) + 3H2(g)} \)
For a zero order reaction the rate does not depend on concentration: rate = k[NH3]0 = k = 2.5 × 10−4 mol L−1 s−1. The rate of the reaction is related to the rate of formation of each product through its stoichiometric coefficient:
\( \displaystyle \text{Rate} = -\frac{1}{2}\frac{d[\ce{NH3}]}{dt} = \frac{d[\ce{N2}]}{dt} = \frac{1}{3}\frac{d[\ce{H2}]}{dt} = k \)
\( \displaystyle \frac{d[\ce{H2}]}{dt} = 3k = 3 \times 2.5 \times 10^{-4} = 7.5 \times 10^{-4}\ \text{mol L}^{-1}\,\text{s}^{-1} \)
NCERT (2026-27 edition):
Zero Order Reactions – “… Zero order reactions are relatively uncommon but they occur under special conditions. Some enzyme catalysed reactions and reactions which occur on metal surfaces are a few examples of zero order reactions. The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure.
\( \ce{2NH3(g)} \xrightarrow[\text{Pt catalyst}]{\text{1130 K}} \ce{N2(g) + 3H2(g)} \)
\( \text{Rate} = k[\ce{NH3}]^0 = k \)
In this reaction, platinum metal acts as a catalyst. At high pressure, the metal surface gets saturated with gas molecules. So, a further change in reaction conditions is unable to alter the amount of ammonia on the surface of the catalyst making rate of the reaction independent of its concentration.”