Chapter 13, Problem 100FP

Interpretation Introduction

(a)

Interpretation:

Δ_rG⁰ at 298 K for the vaporization ${\text{H}}_2\text{O (l, 1 bar) }\rightleftharpoons {\text{ H}}_2\text{O (g, 1 bar)}$ should be determined in two different ways using the data in the table given.

Concept introduction:

$\Delta G=\Delta H-T.\Delta S$

$\begin{array}{l}{\Delta }_r{H}^0={\displaystyle \sum {\Delta }_f{H}^0(\text{products})-{\displaystyle \sum {\Delta }_f{H}^0(\text{reactants})}}\\ {\Delta }_r{S}^0={\displaystyle \sum \Delta S^0(\text{products})-{\displaystyle \sum \Delta S^0(\text{reactants})}}\\ {\Delta }_r{G}^0={\displaystyle \sum {\Delta }_f{G}^0(\text{products})-{\displaystyle \sum {\Delta }_f{G}^0(\text{reactants})}}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The value of K for the vaporization and vapor pressure of water at 298.15 K should be determined.

Concept introduction:

${\Delta }_r{G}^0=-RT\ln K$

${\text{H}}_2\text{O (l, 1 bar) }\rightleftharpoons {\text{ H}}_2\text{O (g, 1 bar)}$

Interpretation Introduction

(c)

Interpretation:

The vapor pressure should be converted to mmHg

Concept introduction:

$1\text{ bar = 750}\text{.062 mmHg}$

Interpretation Introduction

(d)

Interpretation:

Calculations in part (b) and (c) should be repeated starting ${\Delta }_r{G}^{{}^0}=8.590\text{ kJ/mol}$.

Concept introduction:

${\Delta }_r{G}^0=-RT\ln K$

${\text{H}}_2\text{O (l, 1 bar) }\rightleftharpoons {\text{ H}}_2\text{O (g, 1 bar)}$