In a particular reaction `t_(1//2)` was found to increase 16 times when initial concentration of the reactant was reduced to one fourth What is the order of the reaction ?

Options

A. Zero
B. One
C. Two
D. Three

Answer

D

Explanation

As time taken to reduce the concentration 1/2 is 5 hours and time taken to reduce the concentration to 1/4 to 10 hours. The reaction is of first order.
`t_(1//2)prop(1)/(a^(n-1))`
`((t_(1//2))_(2))/((t_(1//2))_(1))=[(a_(1))/(a_(2))]^(n-1)`
Here `((t_(1//2))_(2))/((t_(1//2))_(1))=16 and(a_(1))/(a_(2))=4`
`:. 16=4^(n-1)`
`4^(2)=4^(n-1)`
`n=3`.