Question

Show that `t_(1/2)= 0.693/k` for first reaction.

Answer 1

For first order reaction,

`k = 2.303/t.log_10  [A]_0/[A]_t`

[A]₀ = Initial concentration at time, t = 0

[A]_t = Concentration at time t = t

For half time `t = t_(1/2), [A]_t =[A]_0/2 `

`k=2.303/t_(1/2).log_10  ([[A_0]]/[A]_0)/2`

`=2.303/k.log_10 2`

`=(2.303xx0.3010)/k`

`t_(1/2)=0.693/k`