The thermal decomposition of HI on gold surface is an example of
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Zero order reaction
In a zero order reaction the rate does not depend on the concentration of the reactant: Rate \( = k[\mathrm{R}]^{0} = k \).
On a gold surface at high pressure the surface is saturated with HI molecules, so the rate stays constant:
\( \displaystyle \ce{2HI(g)} \xrightarrow{\ce{Au}\text{ surface}} \ce{H2(g) + I2(g)} \qquad \text{Rate} = k \)
So the thermal decomposition of HI on gold is a zero order reaction, like the decomposition of ammonia on hot platinum.
Why the other options are incorrect
Option (1) First order reaction — Incorrect. The rate would be proportional to [HI], as in the decomposition of \( \ce{N2O5} \) (Rate \( = k[\ce{N2O5}] \)); on a saturated surface it is not.
Option (3) Second order reaction — Incorrect. The rate would be proportional to [HI]2; this is the rate law of HI decomposing in the gas phase without the gold surface, not on it.
Option (4) Pseudo first order reaction — Incorrect. This is a reaction in which one reactant is in such large excess that its concentration hardly changes, as in the acid hydrolysis of ethyl acetate in water: Rate \( = k[\ce{CH3COOC2H5}] \).
NCERT (2026-27 edition):
Zero Order Reactions – …
