When concentrations of the reactants are increased sixteen times, the rate becomes two times. The reaction is of:
Answer & explanation
Correct answer: option 1
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)
\(\frac{2r}{r} = \frac{k[16conc]^n}{k[conc]^n}\)
\(2 = 16^n\)
\(2^1 = 2^{4n}\)
\(4n = 1\)
\(n = \frac{1}{4}\)