Rate of a reaction is of 2nd order w.r.t. A and Ist order w.r.t. B. If concentration of A is halved and concentration of B becomes 1/3rd of initial concentration. The rate of reaction becomes:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1/12 times
The rate law shows how the rate depends on each concentration raised to its order:
\( \displaystyle \text{Rate} = k[\mathrm{A}]^2[\mathrm{B}]^1 \)
New concentrations: [A] → [A]/2 and [B] → [B]/3.
\( \displaystyle \text{New rate} = k\left(\frac{[\mathrm{A}]}{2}\right)^2\left(\frac{[\mathrm{B}]}{3}\right) = \frac{1}{4}\times\frac{1}{3}\times k[\mathrm{A}]^2[\mathrm{B}] \)
\( \displaystyle \text{New rate} = \frac{1}{12}\times\text{original rate} \)
So the rate becomes 1/12 times the original rate.
NCERT (2026-27 edition):
Order of a Reaction – “In the rate equation (3.4)
Rate = k [A]x [B]y
x and y indicate how sensitive the rate is to the change in concentration of A and B. Sum of these exponents, i.e., x + y in (3.4) gives the overall order of a reaction whereas x and y represent the order with respect to the reactants A and B respectively.”