## Estimating the boiling point of water

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water_boilingpoint

## Estimating the boiling point of water

John Kitchin

I got distracted looking for Shomate parameters for ethane today, and came across this website on predicting the boiling point of water using the Shomate equations. The basic idea is to find the temperature where the Gibbs energy of water as a vapor is equal to the Gibbs energy of the liquid.

```clear all; close all
```

## Liquid water

```Hf_liq = -285.830; % kJ/mol
S_liq = 0.06995;   % kJ/mol/K
shomateL = [-203.6060
1523.290
-3196.413
2474.455
3.855326
-256.5478
-488.7163
-285.8304];
```

## Gas phase water

Interestingly, these parameters are listed as valid only above 500K. That means we have to extrapolate the values down to 298K. That is risky for polynomial models, as they can deviate substantially outside the region they were fitted to.

```Hf_gas = -241.826; % kJ/mol
S_gas = 0.188835;  % kJ/mol/K

shomateG = [30.09200
6.832514
6.793435
-2.534480
0.082139
-250.8810
223.3967
-241.8264];
```

## Compute G over a range of temperatures.

```T = linspace(0,200)' + 273.15; % temperature range from 0 to 200 degC

t = T/1000;

% normalize sTT by 1/1000  so entropies are in kJ/mol/K
sTT = [log(t) t t.^2/2 t.^3/3 -1./(2*t.^2) 0*t.^0 t.^0 0*t.^0]/1000;
hTT = [t t.^2/2 t.^3/3 t.^4/4 -1./t 1*t.^0 0*t.^0 -1*t.^0];

Gliq = Hf_liq + hTT*shomateL - T.*(sTT*shomateL);
Ggas = Hf_gas + hTT*shomateG - T.*(sTT*shomateG);
```

## Interpolate the difference between the two vectors to find the boiling point

The boiling point is where Gliq = Ggas, so we solve Gliq(T)-Ggas(T)=0 to find the point where the two free energies are equal.

```f = @(t) interp1(T,Gliq-Ggas,t);
bp = fzero(f,373)
```
```bp =

373.2045

```

## Plot the two free energies

You can see the intersection occurs at approximately 100 degC.

```plot(T-273.15,Gliq,T-273.15,Ggas)
line([bp bp]-273.15,[min(Gliq) max(Gliq)])
legend('liquid water','steam')
xlabel('Temperature \circC')
ylabel('\DeltaG (kJ/mol)')
title(sprintf('The boiling point is approximately %1.2f \\circC', bp-273.15))
``` ## Summary

The answer we get us 0.05 K too high, which is not bad considering we estimated it using parameters that were fitted to thermodynamic data and that had finite precision and extrapolated the steam properties below the region the parameters were stated to be valid for.

```% tags: thermodynamics
```