Free energy, G=H-TS, is a state function that includes whether a reaction is spontaneous or non-spontaneous. If you think of TS as the part of the system's energy that is disordered already, then (H-TS) is the part of the system's energy that is still ordered and therefore free to cause spontaneous change by becoming disordered.
Also, `DeltaG=DeltaH-TDeltaS`
To see what this equation for free energy change has to do with spontaneity let us return to relationship.
`DeltaS_("total")=DeltaS_("sys")+DeltaS_("surr") = DeltaS + DeltaS_("surr")`
(It is generally understood that symbols without subscript refer to the system not the surroundings.)
`DeltaS_("surr")=-(DeltaH)/T`, where `DeltaH` is the heat gained by then system at constant pressure.
`DeltaS_("total") = DeltaS -(DeltaH)/T`
`rArr TDeltaH_("total")=DeltaH-TDeltaS`
`rArr -TDeltaS_("total") =DeltaH-TDeltaS`
i.e. `DeltaG=-TDeltaS_("total")`
From second law of thermodynamics, a reaction is spontaneous if `DeltaS_("total")` is positive, non-spontanous if `DeltaS_("total")` is negative and at equilibrium if `DeltaS_("total")` is zero.
Since, `-TDeltaS=DeltaG` and since `DeltaG` and `DeltaS` have opposite signs, we can restate the thermodynamic criterion for the spontaneity of a reaction carried out at constant temperature and pressure.
If `DeltaG lt 0`, the reaction is spontaneous.
If `DeltaG gt 0`, the reaction is non-spontanous.
If `DeltaG=0`, the reaction is at equilibrium.
In the equation, `DeltaG=DeltaH-TDeltaS`, temperature is a weighting factor that determine the relative importance of enthalpy contribution to `DeltaG`.
Read the above paragraph carefully and answer the following questions based on above comprehension:
Which of the following is true for the reaction?
`H_(2)O(l)` `H_(2)O(g)` at `100^(@)` C and 1 atmosphere?

Options

A. `DeltaS=0`
B. `DeltaH=0`
C. `DeltaH=DeltaE`
D. `DeltaH=TDeltaS`

Explanation

To solve the question regarding the reaction \( H_2O(l) \) to \( H_2O(g) \) at \( 100^\circ C \) and 1 atmosphere, we will analyze the situation using the concepts of Gibbs free energy, enthalpy, and entropy.

### Step-by-Step Solution:

1. **Identify the Reaction**: The reaction given is the phase change of water from liquid to gas:
\[ H_2O(l) \rightarrow H_2O(g) \]

2. **Recognize the Conditions**: The reaction occurs at \( 100^\circ C \) (which is the boiling point of water) and at 1 atmosphere of pressure.

3. **Understand Equilibrium**: At the boiling point, the liquid and gas phases of water are in equilibrium. This means that the rate of evaporation of water equals the rate of condensation.

4. **Apply Gibbs Free Energy**: According to the Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, the change in Gibbs free energy \( \Delta G \) is zero:
\[ \Delta G = 0 \implies \Delta H = T \Delta S \]

5. **Interpret the Terms**:
- \( \Delta H \) is the change in enthalpy, which corresponds to the enthalpy of vaporization for water at its boiling point.
- \( T \) is the temperature in Kelvin. At \( 100^\circ C \), this is \( 373.15 \, K \).
- \( \Delta S \) is the change in entropy associated with the phase change from liquid to gas.

6. **Conclusion**: Since the reaction is at equilibrium at the boiling point, we can conclude that:
\[ \Delta H = T \Delta S \]
This indicates that the enthalpy change for the phase transition is equal to the temperature multiplied by the entropy change.

### Final Answer:
The correct statement regarding the reaction \( H_2O(l) \rightarrow H_2O(g) \) at \( 100^\circ C \) and 1 atmosphere is:
\[ \Delta H = T \Delta S \]
This corresponds to option D, which states that at equilibrium, the enthalpy change is equal to the product of temperature and entropy change.