Question

Which of the following is true for $H_{2}O\left(l\right)\rightleftharpoons H_{2}O\left(v\right)$ at 1 atm and ${100}^{\circ }C?$

• A. $\Delta H=\Delta U$
• B. $\Delta H=0$
• C. $\Delta S=0$
• D. $\Delta H=T\Delta S$

Answer 1

The correct option is D

$\Delta H=T\Delta S$

At equilibrium $\Delta G=0$, use $\Delta G=\Delta H-T\Delta S$

$\Delta H$ increases - think latent heat!

$\Delta S$ increases as well - water is vaporising.

Work done is non-zero - gaseous water is produce. Internal energy change cannot be equal to the change in enthalpy.