Arrange the following rate expressions in decreasing order of their overall order of reaction:
A. Rate = k[A]3/2[B]1/2
B. Rate = k[A]−1[B]3/2
C. Rate = k[A]1/2[B]1/2
D. Rate = k[A]1/2[B]
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → A, D, C, B
The order of a reaction is read from its rate law. For Rate = k[A]x[B]y, x is the order with respect to A and y the order with respect to B. The order of the whole reaction is the sum of the powers, x + y. The powers are found by experiment, so they can be fractions, zero or negative.
Adding the powers for each rate law:
\( \displaystyle \text{A: Rate}=k[\mathrm{A}]^{3/2}[\mathrm{B}]^{1/2},\quad \text{order}=\frac{3}{2}+\frac{1}{2}=2 \)
\( \displaystyle \text{B: Rate}=k[\mathrm{A}]^{-1}[\mathrm{B}]^{3/2},\quad \text{order}=-1+\frac{3}{2}=\frac{1}{2} \)
\( \displaystyle \text{C: Rate}=k[\mathrm{A}]^{1/2}[\mathrm{B}]^{1/2},\quad \text{order}=\frac{1}{2}+\frac{1}{2}=1 \)
\( \displaystyle \text{D: Rate}=k[\mathrm{A}]^{1/2}[\mathrm{B}],\quad \text{order}=\frac{1}{2}+1=\frac{3}{2} \)
In decreasing order:
\( \displaystyle 2>\frac{3}{2}>1>\frac{1}{2} \)
\( \displaystyle \mathrm{A}>\mathrm{D}>\mathrm{C}>\mathrm{B} \)
Shortcut: A has the highest order and B the lowest. Only Option (4) starts with A and ends with B.
NCERT (2026-27 edition):
Order of a Reaction – …
