Question

If in the fermentation of sugar in an enzymatic solution that is $0.12M$, the concentration of the sugar is reduced to $0.06M$ in $10h$ and $0.03M$ in $20h$, what is the order of the reaction:

• A. 1
• B. 2
• C. 3
• D. 0

Answer 1

The correct option is A 1

Given, Concentration is reduced from 0.12 M to 0.06 M in 10 h.

So, $t_{1/2}=10h$

Conc reduced from 0.12 M to 0.03 M in 20hr.

So, $t_{3/4}=20h$

$t_{3/4}=2\times t_{1/2}$

This is valid only for 1st Order Reaction.

So,Order is 1