Question

Evaluation of hydrogen bond and evaporation model

Answer 1

In this challenge we will be evaluating a simple 25-bead model of how water evaporate, where each bead represent a water molecule \(\sf [H_{2}O]_{25}\). The dashed lines represent hydrogen bonds. Each hydrogen bond has a strength of 20 kJ/mol, which means that the enthalpy at 0 K is decreasing with 20 kJ when 1 mol hydrogen bonds has been made.
We may assume that the pressure is constant at 1 bar at all temperatures and ignore all rotational and vibrational contributions. The entropy for a water molecule is 145.04 J/mol*K at is assumed to be independent of the temperature.

a) Calculate the Gibbs free energy for the reaction at 25 \(^{\circ}\)C

b) If the model is an attempt to model liquid water, is it most likely for the state to be liquid or gas at 25 \(^{\circ}\)C?

c) Assuming that the changes in the enthalpy and entropy are independent of the temperature, estimate the boiling point from the model.

d) Using the experimental free energy difference, between gaseous and liquid water at 25\(^{\circ}\)C, which is 9.56 kJ/mol (per mol water), and the corresponding value for the enthalpy is 44.01 kJ/mol. What should the boiling point be assuming the statefunctions to be independent of temperature.

e) The boiling points derived from our 25-bead model of water is much too low compared to that computed experimental values:
\[\Large \frac{ T _{b}^{\sf model} }{ T _{b}^{\exp} }<1\]
Is this because of the value of entropy, enthalpy, or both predicted by the model?

Answer 2

the model is found in the attachment.

Answer 3

Lets start with a)....what have you tried?

Answer 4

I already got the answers, this is more of a challenge for others.

Answer 5

ok

Answer 6

So according to the model, in the 25 molecules, there are 50 H-bonds

a) \(\Delta G=\Delta H-T\Delta S\)

\(\Delta G=50*\underbrace{[\dfrac{(20000 ~J/mol)}{N_A}-(25+273.15)K\dfrac{(145.04~ J/mol*K)}{N_A}]}_{1~~H-bond}\)

\(\Delta G=1.93*10^{-18}\)

b) \(H_2O_{(l)} \rightleftharpoons H_2O_{(g)}\)

the equilibrium would lie on the liquid side because \(\Delta G<<1\)

c) BP is when \(\Delta G=0\)

\(0=[\dfrac{(20000 ~J/mol)}{N_A}-T\dfrac{(145.04~ J/mol*K)}{N_A}]\)

\(T=\dfrac{\dfrac{(20000 ~J/mol)}{N_A}}{\dfrac{(145.04~ J/mol*K)}{N_A}}=\dfrac{(20000 ~J/mol)}{(145.04~ J/mol*K)}=137.9~K\)


d) \(9560~ J/mol =(44010~ J/mol)-T*(145.04~ J/mol*K)\)

\(T=237.5~K\)

e) \(\dfrac{T_B~model}{T_B~exp}=0.58\)

the values for BP i calculated are really far off, i can't really comment on what parameter is incorrect. Any hints on where i went wrong?

Answer 7

@aaronq
Okay I gotta admit I've put out a number of traps for people to find, which we usually ignore when doing our simplified thermodynamics :) but this time we can't ignore it all! :D My first trap is in the evaluation of the enthalpy:
I write that we may ignore all rotational and vibrational contributions, but that leaves behind the translational contribution!
The translation contribution from statical thermodynamics can be evaluated to
\[\large \left[ \Delta H(T)-\Delta H(0) \right]_{trans}=\frac{ 5 }{ 2 }RT\]
This is a super evil contribution and has to be taken account for. The enthalpy at the temperature T for the model must be:
\[\large \Delta H(T)=\Delta H(0)+ 25\left( \frac{ 5 }{ 2 }RT \right)-\frac{ 5 }{ 2 }RT\]
Also I only count 40 hydrogen bounds, but when looking past all that, you manage to calculate \(\Delta H(0)\).

My next trap is in entropy and is the one I expected must trouble with:
The entropy is, believe it or not, mass dependent of the particles. This contribution we too must take into account!
So for our model we have the following reaction and thereby following entropy:
\[\Large \sf \left( H _{2}O \right)_{25} (l) \rightarrow 25~H _{2}O (g)\]
\[\Large \Delta S=25~ S_{H_{2}O}-S _{\left( H_{2}O \right)_{25}}\]
The mass dependency for the entropy can be written as:
\[\large S(m _{2})=S(m_{1})+\frac{ 3 }{ 2 }R \ln \left( \frac{ m _{2} }{ m _{1} } \right)\]
After all those considerations, it is good old
\[\Large \Delta G=\Delta H-T \Delta S\]

Answer 8

About the last one I give a fast hint cause I know you super fast see it if I write in in terms of enthalpy and entropy:
We define the boiling point as the temperature in which the liquid phase and the gaseous is in equilibrium, that is \(\Delta G=0\), rearrange the state functions and you got a full overview of all the errors.

Answer 9

ohh there are 40 bonds lol i had too many xmas drinks prior to answering this. sooo statistical thermodynamics, I'm gonna have to come back and try this later on.

Very nice question though!

Answer 10

Just a fast note for my self and for @aaronq (cause I know he enjoy thermodynamics) so we can justify the translational enthalpy contribution:
the translational partition is given by:
\[\Large q ^{T}=\frac{ V }{ \Lambda^3 }=\left( \frac{ 2 \pi m k_b T }{ h^2 } \right)^{3/2}V\]
Where \(\Lambda\) I think is called the thermo radiation (I need a source)
Equation after Atkins and McQuarrie

Answer 11

Found it! \(\Lambda\) is called the "thermal wavelength" or "the thermal de Broglie wavelength" my bad!

Answer 12

haha this question keeps getting more daunting with every post!

Answer 13

Meh pronblem solved here:

Answer 14

Problem*

Answer 15

Here is what I got @aaronq . What do you think?
a)
\[\Large \Delta G _{model}^{\Theta}=\Delta _{r}H-T ~ \Delta _{r}H\]
\[\large \Delta _{r} H _{T=0~K}=40 \times 20 \frac{ kJ }{ mol }=800 \frac{ kJ }{ mol }\]
The transnational contribution:
\[\large \Delta H(T)=\Delta H(0)+ 25\left( \frac{ 5 }{ 2 }RT \right)-\frac{ 5 }{ 2 }RT\]
\[\large \Delta _{r}H=800 \frac{ kJ }{ mol }+25\left( \frac{ 5 }{ 2 }R ~ 298.15~K \right)-\frac{ 5 }{ 2 }R~298.15 ~K=948.7\frac{ kJ }{ mol }\]
The entropy for the model must be:
\[ \Large \Delta S=25~S_{H_{2}O}-S _{\left( H _{2}O \right)_{25}}\]
As we got a mass change the entropy is calculated from:
\[\large S(m _{2})=S(m _{1})+\frac{ 3 }{ 2 } \ln \left( \frac{ m _{2} }{ m _{1} } \right)\]
The ratio between the two compounds is 25, so the entropy is:
\[\large \Delta_r S=25 \times 0.145\frac{ kJ }{ mol~K }-\left( 0.145\frac{ kJ }{ mol~K }+\frac{ 3 }{ 2 }R~\ln(25) \right)=3.44 \frac{ kJ }{ mol~K }\]
\[\large \Delta G _{model}^{\Theta}=948.7 \frac{ kJ }{ mol }-298.15~K \times 3.44 \frac{ kJ }{ mol~K }=-76.3 \frac{ kJ }{ mol }\]

b)
\(\Delta G<0\), so the equilibrium goes spontaneous towards the produces, that is the gas form.

c)
For a phase shift \(\Delta G=0\) so we get boiling point:
\[\large T_b ^{model}=\frac{ \Delta_rH }{ \Delta_r S }\]
\[\large T _{b}^{model}=\frac{ 948.7 \frac{ kJ }{ mol } }{ 3.44 \frac{ kJ }{ mol~K } }=275.79~K=2.69~ ^{\circ}C\]
d)
\[\Large T _{b}^{\exp}=\frac{ \Delta H }{ \Delta S }\]
Finding the change in entropy:
\[\large \Delta S=\frac{ -\left( \Delta G-\Delta H \right) }{ T }\]
\[\large \Delta S=\frac{ -\left( 8.56 \frac{ kJ }{ mol }-44.01 \frac{ kJ }{ mol } \right) }{ 298.15 ~K }=0.1189 \frac{ kJ }{ mol ~K }\]
\[\large T _{b}^{\exp}=\frac{ 44.01 \frac{ kJ }{ mol } }{ 0.1189 \frac{ kJ }{ mol~K } }=370.14~K=97 ~^{\circ}C\]
e)
We look at the fraction:
\[\Large \frac{ T _{b}^{model} }{ T _{b}^{\exp} }=\frac{ \frac{ \Delta H _{model}^{\Theta} }{ \Delta S _{model}^{\Theta} } }{ \frac{ \Delta H _{\exp}^{\Theta} }{ \Delta S _{\exp}^{\Theta }} }=\frac{ \Delta H _{model}^{\Theta} }{ \Delta H _{\exp}^{\Theta} }\frac{ \Delta S _{\exp}^{\Theta} }{ \Delta S _{model}^{\Theta} }\]
\[\large 0.75=0.86 \times 0.86\]
This means that the boiling point for the model is to low, cause the change in enthalpy is to low and the change in entropy is to high.

It has to be noted that the experimental values are per mol \(H_{2}O\), whereas the model values are per mol \((H_{2}O)_{25}\), so the model values has to be divided with 25 with the use of same units.