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

Answer 2

In the present case if

[A]₀ = 100,

[A] = 100 − 30 = 70

∴ k = `2.303/t log_10  [A]_0/([A])`

= `2.303/40 log_10  100/70`

= 0.0575 log₁₀ (1.4285)

= 0.0575 × 0.1549

= 0.0089

= 8.9 × 10⁻³ min⁻¹

t_1/2 = `0.693/k`

= `0.693/(8.9 xx 10^-3)`

= 77.86 min.