What is the change in entropy of the water?

Free Energy

Gibb's Free Energy \(G\) Spontaneity Equilibrium view all

Worksheets

From a strictly by-the-formulas point of view, entropy change is heat transfer divided by the temperature over which the heat transfer occurs. The heat transfer is clearly the same for both volumes but positive for the cold volume and negative for the hot volume (heat flowed out of the hot volume and into the cold volume) but the average temperature over which it occurs is lower for the cold volume (it went from cold to equilibrium) than it is for the hot volume (which went from hot to equilibrium), so the positive heat transfer is divided by a smaller number than the negative heat transfer, therefore the total entropy change is positive.

From a statistical mechanics point of view, there are clearly more configurations of the water molecules resembling the final state than there are for the initial state. In the initial state, molecules of different average energy are bound in separate volumes, while in the final state, all molecules have the same average energy and they are free to move anywhere within the total volume.

Intuitively, consider that it would be extremely unlikely for the final state to spontaneously evolve to the initial state.

And in the most general sense, entropy is a measure of how close a system is to equilibrium. The initial state is not at equilibrium and the final state is.

Phase changes

When a system receives an amount of energy q at a constant temperature, T, the entropy increase DS is defined by the following equation. Hence, the magnitude of ΔS for a reversible process such as a phase change is calculated

\[ \Delta S = \dfrac{q_{rev}}{T}\]

with the temperature in Kelvin. Since entropy changes are much smaller than enthalpy changes, they are usually reported in J K–1 mol–1.

Examples of reversible processes are

Boiling: As temperature is constant, ΔS = ΔHvap/T
Melting: As temperature is constant, ΔS = ΔHfus/T

For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth’s atmosphere). As a result, \(q_{surr}\) is a good approximation of \(q_{rev}\), and the second law may be stated as the following:

\[ΔS_\ce{univ}=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \label{4}\]

We may use this equation to predict the spontaneity of a process as illustrated in Example \(\PageIndex{1}\).

The entropy change for the process

\[\ce{H2O}(s)⟶\ce{H2O}(l) \nonumber\]

is 22.1 J/K and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at −10.00 °C? Is it spontaneous at +10.00 °C?

We can assess the spontaneity of the process by calculating the entropy change of the universe. If ΔSuniv is positive, then the process is spontaneous. At both temperatures, ΔSsys = 22.1 J/K and qsurr = −6.00 kJ.

At −10.00 °C (263.15 K), the following is true:

\[\begin{align} ΔS_\ce{univ}&=ΔS_\ce{sys}+ΔS_\ce{surr}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T} \nonumber\\ &=\mathrm{22.1\: J/K+\dfrac{−6.00×10^3\:J}{263.15\: K}=−0.7\:J/K} \nonumber

\end{align} \nonumber\]

\(S_{univ} < 0\), so melting is nonspontaneous (not spontaneous) at −10.0 °C.

At 10.00 °C (283.15 K), the following is true:

\[ΔS_\ce{univ}=ΔS_\ce{sys}+\dfrac{q_\ce{surr}}{T}\nonumber\]

\[\mathrm{=22.1\:J/K+\dfrac{−6.00×10^3\:J}{283.15\: K}=+0.9\: J/K}\nonumber\]

\(S_{univ} > 0\), so melting is spontaneous at 10.00 °C.

Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of Suniv?

Answer:

Entropy is a state function, and freezing is the opposite of melting. At −10.00 °C spontaneous, +0.7 J/K; at +10.00 °C nonspontaneous, −0.9 J/K.

Ice melting provides an example in which entropy increases in a small system, a thermodynamic system consisting of the surroundings (the warm room) and the entity of glass container, ice, water which has been allowed to reach thermodynamic equilibrium at the melting temperature of ice. In this system, some heat (δQ) from the warmer surroundings at 298 K (25 °C; 77 °F) transfers to the cooler system of ice and water at its constant temperature (T) of 273 K (0 °C; 32 °F), the melting temperature of ice. The entropy of the system, which is δQ/T, increases by δQ/273K. The heat δQ for this process is the energy required to change water from the solid state to the liquid state, and is called the enthalpy of fusion, i.e. ΔH for ice fusion.

It is important to realize that the entropy of the surrounding room decreases less than the entropy of the ice and water increases: the room temperature of 298 K is larger than 273 K and therefore the ratio, (entropy change), of δQ/298K for the surroundings is smaller than the ratio (entropy change), of δQ/273K for the ice and water system. This is always true in spontaneous events in a thermodynamic system and it shows the predictive importance of entropy: the final net entropy after such an event is always greater than was the initial entropy.

As the temperature of the cool water rises to that of the room and the room further cools imperceptibly, the sum of the δQ/T over the continuous range, “at many increments”, in the initially cool to finally warm water can be found by calculus. The entire miniature ‘universe’, i.e. this thermodynamic system, has increased in entropy. Energy has spontaneously become more dispersed and spread out in that ‘universe’ than when the glass of ice and water was introduced and became a 'system' within it.

Calculate entropy change when 36.0 g of ice melts at 273 K and 1 atm. If you look up the enthalpy of fusion for ice in a table, you would get a molar enthalpy of 6.01 kJ/mol.

dS = (6.01 kJ mol-1)/272 K * (36 g)/(18 g mol-1)

= 1.22 kJ / K