A reaction $2 A P$ is second order with $- d [A] / d t = k [A]^{2}$ and $k = 0.032 L mo l^{- 1} s^{- 1}$ . How long (in seconds, nearest integer) does it take for $[A]$ to fall from $0.40 mol L^{- 1}$ to $0.10 mol L^{- 1}$ ?

Numerical answer type, answer in s, tolerance 0.5.

Answer: 234 s

Use

k t = 1/ [A] - 1/ [A]_{0}

. Then

t = (1/0.10 - 1/0.40) /0.032 = (10 - 2.5) /0.032 = 234.38 s

, i.e. about 234 s.