How many half-lives are needed to complete the zeroth order reaction ?

Options

A. Two
B. Four
C. Infinite
D. Eight

Answer

A

Explanation

The half-life of a reaction, symbolized by `t_(1//2)` , is the time required for the reactant concentration to drop to one half of its intial value.
Suppose a zero order reaction of the from
`a0` initial concentration
`ArarrB`
`(a-x)x` concentration at time `t`
Integrated rate law is
`K_(0)=(x)/(t)`
where `K_(0)` is the rate constantl of a zero order reaction. When `t=t_(1//2)` , half of the reatant is converted into product. thus `x=a//2` . Using integrated rate law, we can write
`t_(1//2)=(a)/(2K)`
Multiplying both sides by `2` , we get
`2t_(1//2)=(a)/(K)=t_(oo)`
where `t_(oo)` is the time taken for the completion of the reaction. Thus, two half lives are required to complete the zero order reaction.
For first order reaction, the integrated rate law is
`t=(2.303)/(k)log((a)/(a-x))`
Substituting the value of `K` in terms of half-life we get
`t=(2.303)/(0.693//t_(1//2))log((a)/(a-x))`
When reaction is complete, `t=t_(oo)` and `x=a` . Therefore
`t_(oo)=(2.303t_(1//2))/(0.693)log((a)/(a-a))`
`=(2.303)/(0.693)t_(1//2)log((a)/(0))`
`=(2.303)/(0.693)t_(1//2)(oo)`
Therefore, infinite half lives are needed to complete a first order reaction.