For $H X_{2} O (l) H X_{2} O (g)$ at $298 K$ , take $Δ H^{\circ} = 44.0 kJ mo l^{- 1}$ and $Δ S^{\circ} = 118.8 J mo l^{- 1} K^{- 1}$ (standard pressure $1 bar$ , $1 bar = 750 mmHg$ , $R = 8.314 J mo l^{- 1} K^{- 1}$ ). Find the equilibrium vapour pressure of water in mmHg (one decimal).

Numerical answer type, answer in mmHg, tolerance 0.2.

Answer: 23.3 mmHg

Δ G^{\circ} = 44000 - 298 \times 118.8 = 8597.6 J mo l^{- 1}

. For this equilibrium

K = P_{H X_{2} O} / P^{\circ}

, so

ln K = - 8597.6/ (8.314 \times 298) = - 3.4702

and

K = 0.03111

. Hence the vapour pressure is

\approx 0.03111 bar \times 750 = 23.3 mmHg