When the concentrations of the reactants are increased sixteen times, the rate of the reaction becomes two times. The order of the reaction is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 1/4
The order n tells how the rate depends on concentration. With all the reactant concentrations scaled together:
\( \displaystyle \text{Rate}=k\,[\text{concentration}]^{n} \)
Making the concentrations 16 times larger makes the rate 2 times larger:
\( \displaystyle \frac{\text{Rate}_2}{\text{Rate}_1}=16^{n}=2 \)
\( \displaystyle (2^{4})^{n}=2^{1}\;\Rightarrow\;4n=1 \)
\( \displaystyle n=\frac{1}{4} \)