A first order reaction has a half-life of $30 min$ . Find the time for $99%$ completion in minutes (nearest integer; use $ln 2 = 0.693$ and $ln 10 = 2.303$ ).

Numerical answer type, answer in min, tolerance 1.

Answer: 199.3939 min

k = 0.693/30 = 0.0231 mi n^{- 1}

. For

99%

completion

[A]_{0} / [A] = 100

, so

t = (2.303/ k) lo g 100 = 2.303 \times 2/0.0231 = 199.4

, i.e. about

199 min