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

`\boxed{\text{23.4 mmHg}}`

Step-by-step 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 * 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} * \frac{\text{750.1 mmHg}}{\text{1 bar}} = \textbf{23.4 mmHg}\\\\\text{The vapour pressure of water at $25 ^(\circ)\text{C}$ is $\boxed{\textbf{23.4 mmHg}}$}`