For $A + B P$ with rate = $k [A] [B]$ and $k = 0.45 L mo l^{- 1} mi n^{- 1}$ , both reactants start at $0.10 mol L^{- 1}$ . Find $[A]$ after $20 min$ in units of $1 0^{- 2} mol L^{- 1}$ (two decimals).

Numerical answer type, answer in 10^-2 mol/L, tolerance 0.02.

Answer: 5.26 10^-2 mol/L

With equal initial concentrations,

[A] = [B]

throughout, so the rate law reduces to second order:

1/ [A] = 1/ [A]_{0} + k t = 10 + 0.45 \times 20 = 19.0 L mo l^{- 1}

. Hence

[A] = 5.26 \times 1 0^{- 2} mol L^{- 1}