Question:
For the reaction $H_2O (l) \rightleftharpoons H_2O(g)$ at 373 K and 1 atm pressure
For the reaction $H_2O (l) \rightleftharpoons H_2O(g)$ at 373 K and 1 atm pressure
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At equilibrium: $\Delta G = 0 \Rightarrow \Delta H = T\Delta S$.
Updated On: Apr 10, 2026
- $\Delta H=0$
- $\Delta E=0$
- $\Delta H=T\Delta S$
- $\Delta H=\Delta E$
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The Correct Option is C
Solution and Explanation
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)
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)
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