Derive integrated rate law for a zero order reaction `A rarr` product.

Explanation

A reaction in which the rate is independent of the concentration of the reactant over a wide range of concentration is called as zero order reaction. Let us consider the following by hypothetical zero order reaction.

A plot of [A] Vs time for a zero order reaction `A rarr` product with initial concentration of [A] = 0.5 M and `k = 1.5xx10^(-2)mol^(-1)L^(-1)"min"^(-1)` The rate law can be written as
Rate `= k[A]^(0) " "` ...(1)
`(-d[A])/(dt)=k(1) " " therefore ([A]^(0)=1)`
`rArr -d[A]=kdt`
Integrate the above equation between the limits of `[A_(0)]` at zero time and [A] at some leter time 't',
`-int_([A_(0)])^([A])d[A]=k int_(0)^(t) dt`
`-([A])_([A_(0)])^([A])=k(t)_(0)^(t)`
`[A_(0)]-[A]=kt " "` ....(2)
`k=([A_(0)]-[A])/(t)`
Equation (2) is in the form of a straight line
`y=mx + c`
i.e., `[A] = -kt+[A_(0)]`
`rArr y=c + mx`
A plot of [A] vs time gives a straight line with a slope of -k and y - intercept of `[A_(0)]`.