Question

A first order reaction is 40% complete in 20 minutes. Calculate the time required for 90% completion of the same reaction.

Answer 1

For the first order reaction is:

k = `2.303/t log  ([A_0])/([A])`

Assume, [A₀] = 100%, t = 20 minutes

Therefore, [A] = 100 – 40 = 60

k = `(2.303/20) log (100/60)`

= 0.11515 × log (1.667)

= 0.11515 × 0.2219

= 0.02555 min⁻¹

Hence, the value of the rate constant is 0.02555 min⁻¹

Using the same rate constant, we find the time t when 90% is complete (x = 90)

[A] = 100 – 90 = 10

k = `(2.303/0.02555) log (100/10)`

= 90.136 × 1

= 90.136 min