For $A + B P$ , rate $= k [A] [B]$ . With $[B]_{0} = 0.80 mol L^{- 1}$ (large excess), the half-life of $A$ is $12.0 min$ . In a second run $[B]_{0} = 0.20 mol L^{- 1}$ (still in large excess). Use $ln 2 = 0.693$ and find the time (in minutes, one decimal) for 75% of $A$ to react in the second run.

Numerical answer type, answer in min, tolerance 0.2.

Answer: 96 min

In the first run

k^{'} = 0.693/12.0

, so

k = k^{'} / [B]_{0} = 0.07219 L mo l^{- 1} mi n^{- 1}

. In the second run

k^{''} = k \times 0.20 = 0.01444 mi n^{- 1}

, giving

t_{1/2}^{''} = 48.0 min

. A 75% conversion needs two half-lives, so

t = 96.0 min