Question

Which of the following is true for the reaction?

$H_{2}O\left(l\right)\leftrightharpoons H_{2}O\left(g\right)$ at 100${}^{\circ }C$ and 1 atmosphere.

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

Answer 1

The correct option is D $\Delta H=T\Delta S$

Solution:- (D) $\Delta H=T\Delta S$

${H_{2}O}_{\left(l\right)}\leftrightharpoons {H_{2}O}_{\left(g\right)}$

As we know that, at equilibrium,

$\Delta G=0$

$\therefore \Delta H-T\Delta S=0$

$\Rightarrow \Delta H=T\Delta S$