How does temperature affect the dissociation of acids?

[nymbler_064]

Hi everyone,

There is one major concept I am struggling with, and that is why acids dissociate more at higher temperatures. How can this be explained in terms of molecular energy levels becoming accessible at higher temperature, and the fact that the spacing between adjacent energy levels id closer for the products (more molecules). That is, the products do not have as much kinetic energy as the reactants, but can promote particles to high levels more easily.
This is sort of explained better here:http://www.chem1.com/acad/webtext/thermeq/TE1.html

What I am struggling with is how this relates to the Gibbs free energy equation? The aforementioned facts would suggest the the entropy changes, but people say that the entropy is assumed to stay the same for Gibbs, with only temperature changing (thus attributing for increased dissociation).

I would really appreciate any feedback or clarification on any of these issues,
Many thanks.

 

[Mapes]

Hi nymbler_064, welcome to PF!

What I am struggling with is how this relates to the Gibbs free energy equation? The aforementioned facts would suggest the the entropy changes, but people say that the entropy is assumed to stay the same for Gibbs, with only temperature changing (thus attributing for increased dissociation).

Where did you hear this? The Gibbs free energy G = U +PV -TS is useful for analyzing systems at equilibrium at constant temperature (and pressure). Higher temperatures generally favor configurations with higher entropy, because they minimize G. And more dissociation means higher entropy. Does this make sense?

 

[nymbler_064]

I sort of understand the basics in terms of T increasing the value of the TdeltaS term in the free energy equation, hence decreasing the value of deltaG. However, I don't understand how this relates to the statistical (i think?) thermodynamics issue of more free energy levels becoming available at higher temperatures (ie. products gain entropy faster than reactants, due to smaller spacing between states, but have lower kinetic energy to start with due to endothermic dissociation)? The link in my first post shows what sort of thing I am talking about.

In my mind, it seems to suggest that the rate at which free energy decreases increases at higher temperature, so ka increases more significantly at higher temperatures.

 

[Mapes]

(ie. products gain entropy faster than reactants, due to smaller spacing between states, but have lower kinetic energy to start with due to endothermic dissociation)? The link in my first post shows what sort of thing I am talking about.

I don't quite like the way that web page illustrates those energy levels. The levels aren't spaced evenly that way for two monatomic hydrogen atoms, as far as I know. Rather, there is duplication of energy levels, so that the overall density increases. In other words, (sufficiently spaced) H atoms shouldn't have energy levels 3 for one, 4 for the other, 5 for one, 6 for the other; but rather 3 for one, 3 for the other, 5 for one, 5 for the other, for example.

Also, I wouldn't say that dissociated products gain entropy faster than the reactant, but that they have a higher total entropy than the reactant because the two products (the solvated proton and its conjugate) can now attain decoupled degrees of freedom. The original acid always had a lower energy U than the dissociated components. But the larger entropy of the components makes their total free energy G = U + PV -TS lower when T grows large, and this switch drives dissociation.

 

[nymbler_064]

Thanks so much for your reply, Mapes.

However, I am still unsure how to explain the fact the extent of dissociation/ka of an acid increases at increasing temperatures. I know that because the TdeltaS term increases, the free energy change increases, but where do the energy levels come into it? Would delta S increase as temperature increases? Is there a way to link those energy level diagrams into such an explanation for increasing ka?

Many thanks again.

Also, I just have a question regarding the website illustration. Would you say that the product molecules have a lesser quantity of kinetic energy available to them than the reactants at a given temperature (thus lowering entropy), due to the energy required to break the bond? I am a little confused about the vertical displacement part.

 

[Mapes]

However, I am still unsure how to explain the fact the extent of dissociation/ka of an acid increases at increasing temperatures.

Well, the easiest way is through the relationship $\Delta G=-RT\ln K_a$, but I realize you want to develop the intuition behind this equation.

where do the energy levels come into it?

The products (the solvated proton and its conjugate) have a higher density of energy levels in part because they are two species. However, the energy levels are higher than the energy levels of the reactant (the acid) because the products don't get the benefit of being chemically bonded. At low temperatures, therefore, the only occupied energy levels are those of the acid. At higher temperatures, the energy levels of the reactants will fill up faster than those of the acid, and thus it is far more likely that we'll find the components in the dissociated state.

Would delta S increase as temperature increases?

It's not important in this discussion; at this level, it's often assumed that the entropy of a component is temperature-independent. The important part is the entropy difference between the products and the reactants.

Also, I just have a question regarding the website illustration. Would you say that the product molecules have a lesser quantity of kinetic energy available to them than the reactants at a given temperature (thus lowering entropy), due to the energy required to break the bond?

No; the products are in thermal equilibrium with the environment, so they will have the same average kinetic energy whether they just dissociated or whether they were added to the solution independently and separately. Kinetic energy and entropy are not coupled in this way.

 

[nymbler_064]

Thankyou so much - I think I am finally starting to understand!

Just one last question, what do you mean when you say that the entropy difference between products and reactants is important but dS is Independent of temperature? If dS=Sproducts-Sreactants, wouldn't the entropy change increase as temperature increases.

The way I see it (and I need you to correct me if I'm wrong), increasing temperature increases both T and dS in the TdS term. So higher temperatures magnify the dS, which is greater at higher temperatures.

 

[Mapes]

If [$\Delta$]S=Sproducts-Sreactants, wouldn't the entropy change increase as temperature increases.

Why?

 

[nymbler_064]

Because as the temperature increases, the number of energy levels/microstates available to products increases more substantially than those available to the reactants, so the difference will be greater...?

 

[Mapes]

Because as the temperature increases, the number of energy levels/microstates available to products increases more substantially than those available to the reactants, so the difference will be greater...?

That's a characteristic of higher entropy, not increasing entropy.

 

[nymbler_064]

And that is where I am getting confused...
What is the relationship between temperature and entropy change. The term TdS in the free energy equation suggests that this term only increases due to the temperature. Is this because more microstates are becoming accessible? Why does the entropy change stay the same?

 

[Mapes]

What is the relationship between temperature and entropy change.

Here's how we can find the entropy dependence on temperature. Heat capacity at constant pressure is defined as

$$C_P=T\left(\frac{\partial S}{\partial T}\right)_P.$$

So

$$dS=\frac{C_P}{T}dT.$$

Integrating gives us

$$S=S_0+C_P\ln\left(\frac{T}{T_0}\right).$$

Let's say that we have a entropy difference $\Delta S=S_{0,\mathrm{P}}-S_{0,\mathrm{R}}$ between reactants R and products P. If the heat capacities are similar, the difference will be little changed by changes in temperature (to prove this to yourself, try calculating $\Delta S$ at $T=T_0+10^\circ\mathrm{C}$). Even if the heat capacities are quite different, the logarithmic dependence means that a 10°C increase, for example, makes little difference (again, prove this to yourself by doing the calculations). In contrast, the amount of acid dissociation depends exponentially on temperature and on the difference in energy levels between the products and reactants.

Is this because more microstates are becoming accessible?

The number of energy levels doesn't change much with temperature. The key idea is that at higher temperature, many more of them are populated.

 

[nymbler_064]

So, as the temperature increases, more energy levels can be populated by the products than the reactants... and this increases the value of the TdS term?

And you say that the extent of acid dissociation at different temperatures is exponential? This is what I thought, but most places seem to show it as linear...

I've just realized where I think I'm getting confused --> what is the difference between entropy change and entropy difference?

THanks so much for persevering with me. I really appreciate it.

 

[Mapes]

So, as the temperature increases, more energy levels can be populated by the products than the reactants

Yes.

And you say that the extent of acid dissociation at different temperatures is exponential? This is what I thought, but most places seem to show it as linear...

Yes, $K_\mathrm{a}\propto\exp(-1/T)$.

 

[nymbler_064]

Thanks so much.

Is entropy change different to entropy change? (last question, I promise!)

 

[Mapes]

Is entropy change different to entropy change?

?

 

[nymbler_064]

You have been saying that the difference between the number of energy levels is significant, but that dS is not, so i am getting a bit confused on what the difference is?

 

[Mapes]

With reactions we have to be precise about which dS or $\Delta S$ we're talking about: (1) the difference in entropy values between reactants and products, or (2) the difference in entropy values for a single material at different temperatures. The first has a strong (exponential) influence on the direction of a reaction. The second generally has a much smaller influence, for two reasons: (1) the entropy values of the reactants vs. products may increase at similar rates with temperature, so that the difference between them remains approximately constant, and/or (2) temperature has a relatively small (logarithmic) influence on entropy for a single material. I hope this helps clarify?

 

[nymbler_064]

Yes, definitely - you are absolutely amazing!

So just to clarify, can I discuss the dS for the dissociation as the entropy difference between the products and reactants (which should be exponential)?
And so then the value of ka could increase exponentially with temperature (but only slightly)?

 

[Mapes]

So just to clarify, can I discuss the dS for the dissociation as the entropy difference between the products and reactants (which should be exponential)?
And so then the value of ka could increase exponentially with temperature (but only slightly)?

The entropy difference ΔS between an acid and its components (specifically, the larger total entropy of the components) is generally responsible for acid dissociation. The acid dissociation constant $K_\mathrm{a}$ depends exponentially on the entropy difference ΔS, the energy difference ΔU between the acid and its components, and the temperature T:

$$K_\mathrm{a}\propto e^{-\Delta G/RT}\propto e^{\Delta S/R} e^{-\Delta U/RT}$$

 

[nymbler_064]

Thanks so much for all your help!

Just to link the exponential trend, could you say that the change in the value of ka is greater at higher temperatures. That is, if you increased the temperature of a system from 50C to 70C, you would see a greater change in the extent of dissociation than from 5C to 25C?

 

[nymbler_064]

Does anyone know if this would be correct? I just need this one last piece of information!

Thanks so much.

 

[SpectraCat]

Does anyone know if this would be correct? I just need this one last piece of information!

Thanks so much.

You have the equations available .. why not plot them yourself and see if you can answer your own question?

 

[nymbler_064]

I've already done plotted the results of the experiment, and did find an exponential trend. I just wanted to check it could be backed up by theory. :)