The decomposition of gaseous ammonia on a hot platinum surface is a zero order reaction at high pressure. Why does the rate of this reaction not depend on the concentration of ammonia?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → At high pressure, the surface of the platinum gets saturated with ammonia molecules.
Platinum acts as a catalyst: ammonia decomposes only on its surface.
\( \displaystyle \ce{2NH3(g)} \xrightarrow[\text{Pt catalyst}]{1130\ \mathrm{K}} \ce{N2(g) + 3H2(g)} \)
\( \displaystyle \mathrm{Rate} = k[\ce{NH3}]^0 = k \)
At high pressure every site on the surface is already covered by ammonia molecules. Adding more ammonia cannot put more molecules on the surface, so the reaction goes on at the same rate whatever the concentration.
Why the other options are incorrect
Option (2) The decomposition is an elementary reaction whose molecularity is zero. — Incorrect. Molecularity is the number of species that collide in an elementary step, so it cannot be zero. It is the order that is zero here.
Option (3) Platinum lowers the activation energy of the decomposition to zero. — Incorrect. A catalyst lowers the activation energy, but not to zero. A lower activation energy changes the value of k, not the way the rate depends on concentration.
Option (4) Ammonia molecules collide with one another more often at high pressure. — Incorrect. More collisions would make the rate rise with concentration; they cannot explain a rate that stays constant.
NCERT (2026-27 edition):
3.3.1 Zero Order Reactions – “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.”