Chemical Kinetics and Nuclear ChemistryMedium
Question
When the concentration of reactant in the reaction A → B is increased by 8 times, the rate increases only by 2 times. The order of reaction is
Options
A.3
B.1/3
C.2
D.1/2
Solution
r = K.[A]n
$\therefore\frac{r_{2}}{r_{1}} = \left( \frac{\left\lbrack A_{2} \right\rbrack}{\left\lbrack A_{1} \right\rbrack} \right)^{n} \Rightarrow 2 = (8)^{n} \Rightarrow n = \frac{1}{3}$
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