For $A + B P$ , rate $= k [A] [B]$ with $k = 0.0462 L mo l^{- 1} mi n^{- 1}$ . A mixture is prepared with $[A]_{0} = 0.010 mol L^{- 1}$ and $[B]_{0} = 1.0 mol L^{- 1}$ . Using $ln 2 = 0.693$ , the time for 87.5% of $A$ to react is

Answer: A

Even at 87.5% conversion only

0.00875

M of

B

is used (under 1% of 1.0 M), so

[B]

is effectively constant and

k^{'} = k [B]_{0} = 0.0462 mi n^{- 1}

. Then

t_{1/2} = 0.693/0.0462 = 15 min

. Since 87.5% means

1 - (1/2)^{3}

, three half-lives are needed:

t = 45 min