A reaction $A P$ has a half-life of $40 min$ for $[A]_{0} = 0.50 mol L^{- 1}$ and $20 min$ for $[A]_{0} = 1.00 mol L^{- 1}$ . Starting from $[A]_{0} = 0.25 mol L^{- 1}$ , calculate the time in minutes (nearest integer) for 80% of $A$ to be consumed.

Numerical answer type, answer in min, tolerance 0.5.

Answer: 320 min

Halving the concentration doubles the half-life, so the order is

n = 1 + lo g (40/20) / lo g (1.00/0.50) = 2.0

(second order). Then

k = 1/ (a t_{1/2}) = 1/ (0.50 \times 40) = 0.05 L mo l^{- 1} mi n^{- 1}

. For 80% consumption at 0.25 M,

[A] = 0.050

M and

t = (1/0.050 - 1/0.25) /0.05 = 320 min