Question

For the reaction $H_2O (l) \rightleftharpoons H_2O(g)$ at 373 K and 1 atm pressure

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

Hint:

At equilibrium: $\Delta G = 0 \Rightarrow \Delta H = T\Delta S$.

Answer 1

The Correct Option is C

Step 1: Thermodynamics Formula
Use the Gibbs free energy equation: $\Delta G = \Delta H - T\Delta S$.
Step 2: Equilibrium Condition
At 373 K and 1 atm, water and steam are in equilibrium. For any system at equilibrium, $\Delta G = 0$.
Step 3: Result
Substituting $\Delta G = 0$ into the equation gives $0 = \Delta H - T\Delta S$, which rearranges to $\Delta H = T\Delta S$.
Final Answer: (c)