A gaseous second order reaction obeys $- d p_{A} / d t = k_{p} p_{A}^{2}$ with $k_{p} = 2.0 \times 1 0^{- 3} at m^{- 1} s^{- 1}$ at $400 K$ . Taking the gas as ideal, with $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ , find $k_{c}$ in $- d c_{A} / d t = k_{c} c_{A}^{2}$ , in units of $1 0^{- 2} L mo l^{- 1} s^{- 1}$ (two decimals).

Numerical answer type, answer in 10^-2 L mol^-1 s^-1, tolerance 0.02.

Answer: 6.568 10^-2 L mol^-1 s^-1

Substitute

p_{A} = c_{A} RT

- RT d c_{A} / d t = k_{p} (c_{A} RT)^{2}

, so

- d c_{A} / d t = k_{p} RT c_{A}^{2}

and

k_{c} = k_{p} RT

. Numerically

k_{c} = 2.0 \times 1 0^{- 3} \times 0.0821 \times 400 = 0.06568 L mo l^{- 1} s^{- 1}

, i.e.

6.57 \times 1 0^{- 2}