The gas-phase reaction $2 A (g) 3 B (g)$ is second order in $A$ . Starting with pure $A$ at $400 mmHg$ (constant volume and temperature), the total pressure is $520 mmHg$ after $10 min$ . Calculate the rate constant $k_{p}$ defined by $- d p_{A} / d t = k_{p} p_{A}^{2}$ , in units of $1 0^{- 4} mmH g^{- 1} mi n^{- 1}$ (two decimals).

Numerical answer type, answer in 10^-4 mmHg^-1 min^-1, tolerance 0.03.

Answer: 3.75 10^-4 mmHg^-1 min^-1

x

is the pressure of

A

reacted,

p_{A} = 400 - x

and

p_{B} = 1.5 x

, so

P_{t} = 400 + 0.5 x

. Then

x = 2 (520 - 400) = 240

and

p_{A} = 160 mmHg

. Second order:

k_{p} t = 1/ p_{A} - 1/ p_{0}

, so

k_{p} = (1/160 - 1/400) /10 = 3.75 \times 1 0^{- 4} mmH g^{- 1} mi n^{- 1}