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:
One mole of ice is converted to liquid at 273 K, `H_(2)O(s)` and `H_(2)O(l)` have entropies 38.20 and 60.03 J `"mol"^(-1) K^(-1)`. Enthalpy change in the conversion will be:

Options

A. 59.59 J/mol
B. 593.95 J/mol
C. 5959.5 J/mol
D. 59595 J/mol

Answer

To solve the problem of finding the enthalpy change (ΔH) when one mole of ice is converted to liquid water at 273 K, we can follow these steps:

### Step 1: Understand the relationship between ΔG, ΔH, and ΔS
We know from thermodynamics that:
\[ \Delta G = \Delta H - T \Delta S \]
At equilibrium, the change in Gibbs free energy (ΔG) is zero:
\[ 0 = \Delta H - T \Delta S \]
This implies:
\[ \Delta H = T \Delta S \]

### Step 2: Calculate the change in entropy (ΔS)
The change in entropy (ΔS) for the reaction can be calculated using the formula:
\[ \Delta S = S_{\text{products}} - S_{\text{reactants}} \]

From the problem, we have:
- Entropy of ice (H2O(s)): \( S_{\text{ice}} = 38.20 \, \text{J mol}^{-1} \text{K}^{-1} \)
- Entropy of liquid water (H2O(l)): \( S_{\text{water}} = 60.03 \, \text{J mol}^{-1} \text{K}^{-1} \)

Now, we can calculate ΔS:
\[ \Delta S = S_{\text{water}} - S_{\text{ice}} \]
\[ \Delta S = 60.03 \, \text{J mol}^{-1} \text{K}^{-1} - 38.20 \, \text{J mol}^{-1} \text{K}^{-1} \]
\[ \Delta S = 21.83 \, \text{J mol}^{-1} \text{K}^{-1} \]

### Step 3: Calculate ΔH using the temperature
Now that we have ΔS, we can use the temperature (T = 273 K) to find ΔH:
\[ \Delta H = T \Delta S \]
\[ \Delta H = 273 \, \text{K} \times 21.83 \, \text{J mol}^{-1} \text{K}^{-1} \]
\[ \Delta H = 5959.59 \, \text{J} \]

### Step 4: Convert to kJ (if necessary)
Since enthalpy changes are often expressed in kJ:
\[ \Delta H = 5.96 \, \text{kJ} \]

### Final Answer
Thus, the enthalpy change (ΔH) for the conversion of one mole of ice to liquid water at 273 K is approximately:
\[ \Delta H \approx 5.96 \, \text{kJ} \]