Question

Show that the half-life of zero order reaction is `t_(1/2) = ([A]_0)/(2k)`.

Answer 1

Since we have, k = `([A]_0 - [A]_t)/t`   ... (i)

At `t = t_(1/2), [A]_t = ([A]_0)/2`

Substituting in equation (i), we get

∴ k = `([A]_0 - ([A]_0)/2)/t_(1/2)`

∴ k = `((2[A]_0 - [A]_0)/2)/t_(1/2)`

= `[A]_0/(2.t_(1/2))`

∴ `t_(1/2) = [A]_0/(2k)`