Question

Consider the reaction at 25 °C. H2O(l) ↔ H2O(g) ΔG° = 8.6 kJ/mol Calculate the pressure of water at 25 °C (Hint: Get K eq)

Answer 1

Answer:

$\boxed{\text{23.4 mmHg}}$

Explanation:

H₂O(ℓ) ⟶ H₂O(g)

$K_{\text{p}} = p_{\text{H2O}}$

$\text{The relationship between \Delta G^{\circ} and K_{\text{ p}} is}\\\Delta G^{\circ} = -RT \ln K_{\text{p}}$

Data:  

T = 25 °C

ΔG° = 8.6 kJ·mol⁻¹

Calculations:

T = (25 + 273.15) K = 298.15 K

$\begin{array}{rcl}8600 & = & -8.314 \times 298.15 \ln K \\8600 & = & -2478.8 \ln K\\-3.47 & = & \ln K\\K&=&e^{-3.47}\\& = & 0.0311\end{array}$

Standard pressure is 1 bar.

$p_{\text{H2O}} = \text{0.0311 bar} \times \dfrac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\\\\\text{The vapour pressure of water at 25 ^{\circ}\text{C} is \boxed{\textbf{23.4 mmHg}} }$