Rate of a reaction changes from \(2.48 × 10^{-3}\, \ mol^{-1}s^{-1}\) to \(4.96 × 10^{-3}\, \ mol^{-1}s^{-1}\) when concentration of reactant, is changed from \(0.6\, \ M\) to \(2.4\, \ M\) respectively, the order of reaction is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.5
Write the rate law as rate = k[R]n and divide the two rates:
\( \displaystyle \frac{r_2}{r_1} = \left(\frac{[R]_2}{[R]_1}\right)^n \;\Rightarrow\; \frac{4.96 \times 10^{-3}}{2.48 \times 10^{-3}} = \left(\frac{2.4}{0.6}\right)^n \)
\( \displaystyle 2 = 4^n = 2^{2n} \;\Rightarrow\; 2n = 1 \;\Rightarrow\; n = 0.5 \)
Quadrupling the concentration only doubles the rate, so the reaction is of order one-half. The order is found from experiment and can be a fraction.
Note: It seems that there is a mistake in the original NTA paper, which prints the unit of rate as mol−1 s−1 when it should read mol L−1 s−1.
NCERT (2026-27 edition):
Order of a Reaction
