Equal initial concentrations ( $0.20 mol L^{- 1}$ each) of $A$ and $B$ react as $A + B P$ with rate $= k [A] [B]$ , $k = 0.30 L mo l^{- 1} mi n^{- 1}$ . The percentage of $A$ converted in $10 min$ is

Answer: C

Because

[A] = [B]

at all times,

1/ [A] = 1/ a + k t

, which gives the fraction reacted

x / a = ka t / (1 + ka t)

. Here

ka t = 0.30 \times 0.20 \times 10 = 0.60

, so

x / a = 0.60/1.60 = 0.375

, i.e. 37.5%. The value 62.5% is the fraction remaining.