Pure $PH X_{3}$ decomposes in a rigid vessel at constant temperature: $4 PH X_{3} (g) P X_{4} (g) + 6 H X_{2} (g)$ (all species gaseous, ideal). Initially the total pressure rises at $18.0 Torr mi n^{- 1}$ . Find the initial rate of fall of $p_{PH X_{3}}$ in $Torr mi n^{- 1}$ (one decimal place).

Numerical answer type, answer in Torr/min, tolerance 0.1.

Answer: 24 Torr/min

Per unit extent of reaction,

4

mol of gas become

7

mol, so

Δ n = + 3

and

d p_{tot} / d t = 3 \dot{ξ}

. Also

- d p_{PH X_{3}} / d t = 4 \dot{ξ}

. Hence

\dot{ξ} = 18.0/3 = 6.0

and

- d p_{PH X_{3}} / d t = 4 \times 6.0 = 24.0 Torr mi n^{- 1}