Question

A first order decomposition reaction takes 40 minutes for 30% decomposition. Calculate its `"t"_(1//2)` value.

Answer 1

Here t = 40 min,

`"t"_(1//2)` = ?

Let a = 100

∴ x = 30% of 100 = 30

Using the formula,

t = `2.303/"k" log  "a"/("a" - "x")`

⇒ 40 = `2.303/"k" log  100/(100 - 30)`

40 = `2.303/"k" log  100/70`

⇒ 40 = `2.303/"k" (log 10 - log 7)`

40 = `2.303/"k" (1 - 0.8451)`

⇒ 40 = `2.303/"k" xx 0.1549`

⇒ k = `0.357/40`

= 0.0089 min⁻¹

∴ `"t"_(1//2) = 0.693/"k" = 0.693/0.0089` = 77.86 min