Can the Second Law of Thermodynamics Explain Freezing Water at -5°C?

[Aeon]

Homework Statement

Show that the second law of thermodynamics predicts the spontaneous freezing of liquid water at -5$$^{o}$$C under 1 bar of constant pressure. Assume that Cp is temperature independent.

Standard water fusion enthalpy = 6.008 kJ/mol @ 273.15 K

$$\geq$$ has to be interpreted as greater than, not greater or equal, in this example.

Homework Equations

A: $$dS$$ $$\geq$$ $$\frac{\partial q}{T}$$

B: $$\Delta$$$$S_{transition}$$ = $$\frac{\Delta H_{transition}}{T_{transition}}$$

C: $$\Delta S = \int \frac{Cp}{T} dT$$

D: $$\Delta H = q$$ in an isobaric process

The Attempt at a Solution

After having wasted an hour or so on this presumably easy problem, I cannot get the second law to predict the freezing.

$$Cp\ ln(T) \geq \frac{\Delta H_{transition}}{T_{transition}}$$ ?

Some advice would be appreciated. I know the proof is simple. I'm having some brain fog atm though.

Thanks!

 

[Mapes]

Hi Aeon, welcome to PF. What thermodynamic potential stays constant for phase changes at constant temperature and pressure? And using this, can you find the entropy of freezing from the enthalpy of freezing?

 

[Aeon]

I have been reading on the forums for some time. I never had to ask a question yet though.
Thanks!

On a phase transition diagram, Gibbs free energy of one phase is equal to G of another phase when you follow the phase transition curves.

Is the enthalpy of freezing opposite of the enthalpy of fusion?

 

[Mapes]

On a phase transition diagram, Gibbs free energy of one phase is equal to G of another phase when you follow the phase transition curves.

Exactly. So by using the definition of G, the entropy of fusion can be determined. Now you can calculate the difference in G between the phases - and a spontaneous process will tend to minimize G.

Is the enthalpy of freezing opposite of the enthalpy of fusion?

Yes, and that was a typo; I meant fusion! Enthalpy and entropy of freezing would have the opposite signs.

 

[DeeNos]

First, we need to understand the second law of thermodynamics, which states that the total entropy of a closed system will always increase over time, or remain constant in ideal cases where the system is in equilibrium or reversible. This means that for a process to occur spontaneously, it must result in an increase in the system's entropy.

In the case of liquid water freezing at -5^{o}C, we can use the equations given to show how the second law predicts this to be a spontaneous process.

First, we can use equation D to calculate the heat released (q) during the freezing process. Since we are assuming constant pressure, q is equal to the change in enthalpy (ΔH) of the system.

q = ΔH = 6.008 kJ/mol

Next, we can use equation C to calculate the change in entropy (ΔS) of the system during the freezing process. Since Cp is temperature independent, we can take it out of the integral.

ΔS = \int \frac{Cp}{T} dT = Cp \ln(\frac{T_{final}}{T_{initial}})

For water freezing at -5^{o}C, we can assume an initial temperature of 0^{o}C and a final temperature of -5^{o}C.

ΔS = Cp \ln(\frac{268.15}{273.15}) = -0.018 kJ/mol-K

Finally, we can use equation B to calculate the change in entropy during the transition from liquid to solid (ΔS_{transition}).

ΔS_{transition} = \frac{\Delta H_{transition}}{T_{transition}} = \frac{6.008 kJ/mol}{268.15 K} = 0.022 kJ/mol-K

Since ΔS_{transition} is positive and greater than ΔS, we can see that the freezing process will result in an increase in entropy, making it a spontaneous process according to the second law of thermodynamics.

Therefore, the second law predicts that liquid water will spontaneously freeze at -5^{o}C under 1 bar of constant pressure.