When concentrations of the reactants are increased sixteen times, the rate becomes two times. The reaction is of:

1/4 order

The correct answer is option 1. 1/4 order

\(R = k[conc]^n\) -----(1)

\(2r = k[16 \text{ conc}]^n\) ------(2)

Dividing (2) by(1)

\(\frac{2r}{r} = \frac{k[16conc]^n}{k[conc]^n}\)

\(2 = 16^n\)

\(2^1 = 2^{4n}\)

\(4n = 1\)

\(n = \frac{1}{4}\)