The rate of a gaseous reaction becomes half when the volume of the vessel is doubled. The order of the reaction is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → One
Let rate = k[A]n. When the volume of the vessel is doubled, the amount of gas is unchanged, so its concentration becomes [A]/2:
\( \displaystyle \text{new rate} = k\left(\frac{[\text{A}]}{2}\right)^n = \frac{1}{2^n} \times \text{rate} \)
The rate becomes half, so
\( \displaystyle \frac{1}{2^n} = \frac{1}{2} \;\Rightarrow\; n = 1 \)
The reaction is of the first order.
Why the other options are incorrect
Option (1) Zero — Incorrect. The rate of a zero order reaction does not depend on concentration, so it would not change at all.
Option (3) Two — Incorrect. The rate would fall to (1/2)2 = one fourth.
Option (4) Three — Incorrect. The rate would fall to (1/2)3 = one eighth.