Does the initial temperature affect the change in temp?

[Hen Sharir]

Say I dissolve NaOH in water - a very exothermic reaction.
Does it matter if I dissolve it in water that is at room temperature or at a different temperature?

Will changing the initial temperature of the water affect the dH of the reaction (hence affect the change in temperature at the end of the reaction)?

 

[dRic2]

Yes, temperature plays a big role. But it is more important the change of Gibbs free energy according to the Van't Hoff equation:
https://en.wikipedia.org/wiki/Van_'t_Hoff_equation

Enthalpy of reaction also changes with temperature but its variation is "usually" negligible (it depends, of course).

 

[Hen Sharir]

Yes, temperature plays a big role. But it is more important the change of Gibbs free energy according to the Van't Hoff equation:
https://en.wikipedia.org/wiki/Van_'t_Hoff_equation

Enthalpy of reaction also changes with temperature but its variation is "usually" negligible (it depends, of course).

ok so let me know if I understand right (from what I read in the link you shared).
In exothermic reactions - as the initial temperature increases, dH/R becomes smaller hence dH being a smaller value?

so if I apply this to the example I gave with NaOH - as I increase the initial temperature of the water, dH of the reaction decreases hence the change in temperature is smaller?

 

[dRic2]

No. Gibbs free energy is related to the equilibrium constant and to the enthalpy of reaction (which is not to be confused with the total $ΔH$) . Van't Hoff equation tells us that exothermic reactions are less favored at higher temperature. This means that you get less products if you work at higher temperature.

If you have less product the total $ΔH$ decreases and so you may have a smaller change in temperature

 

[Ygggdrasil]

No. Gibbs free energy is related to the equilibrium constant and to the enthalpy of reaction (which is not to be confused with the total $ΔH$) . Van't Hoff equation tells us that exothermic reactions are less favored at higher temperature. This means that you get less products if work at higher temperature.

If you have less product the total $ΔH$ decreases and so you may have a smaller change in temperature

Solids are generally more soluble at higher temperatures. Is this not true for NaOH and other solutes that dissolve exothermically?

 

[dRic2]

Solids are generally more soluble at higher temperatures. Is this not true for NaOH and other solutes that dissolve exothermically?

I don't know. Mass transfer processes (like diffusivity) and kinetics should have a big role here. I was just reasoning about the thermodynamics of the reaction. I think this problem is more complex than I thought. I just wanted to make some stuff clear about exothermic reaction, I don't know what happens in particular if you dissolve NaOH in water.

 

[dRic2]

@Ygggdrasil I checked on my thermodynamic book and I found out that solubility for solids (in liquids) can be expressed as:

$x_{i} = exp \left ( \frac {ΔH_{i, fus}} {R} \left ( \frac 1 {T_{i, fus}} - \frac 1 T \right) \right )$

So if the temperature increases then also the solubility increases. But this is true for non-reagent systems and so this should not be true for $NaOH$ because the reaction

$NaOH → OH^-_{(aq)} + Na^+_{(aq)}$

takes place.

Moreover $OH^-_{(aq)}$ and $Na^+_{(aq)}$ diffusivity should increase with temperature so they would move to places where their concentration is lower with higher speed. This means that the layer where the reaction takes place has a lower concentration of products and so...

To sum it up... I have no idea of what should happen

 

[Borek]

If nothing else helps - check tables.

 

[dRic2]

If nothing else helps - check tables.

It seems from these tables that solubility of NaOH increases with temperature so I guess mass transfer is more important here than chemical equilibrium

 

[Hen Sharir]

It seems from these tables that solubility of NaOH increases with temperature so I guess mass transfer is more important here than chemical equilibrium

does it matter that solubility increases if in both cases all of the NaOH dissolves?

edit: or well not ALL of it but in both cases, it reaches the equilibrium

 

[Ygggdrasil]

I don't know. Mass transfer processes (like diffusivity) and kinetics should have a big role here. I was just reasoning about the thermodynamics of the reaction. I think this problem is more complex than I thought. I just wanted to make some stuff clear about exothermic reaction, I don't know what happens in particular if you dissolve NaOH in water.

While mass transfer/diffusion and kinetics will affect the speed at which the reaction reaches equilibrium, it shouldn't affect the position of the equilibrium.

 

[dRic2]

While mass transfer/diffusion and kinetics will affect the speed at which the reaction reaches equilibrium, it shouldn't affect the position of the equilibrium.

That's true but

Moreover OH−(aq)OH(aq)−OH^-_{(aq)} and Na+(aq)Na(aq)+Na^+_{(aq)} diffusivity should increase with temperature so they would move to places where their concentration is lower with higher speed. This means that the layer where the reaction takes place has a lower concentration of products and so...

Otherwise how can NaOH solubility increase if exothermic reaction are less favored at higher T?EDIT: I though of a way to explain it. NaOH in water should give the reaction $NaOH → Na^+ + OH^-$, so we can write

$NaOH_{s} ↔ NaOH_{l}$ // Solid-Liquid equilibrium
$NaOH_{l} ↔ Na^+ + OH^-$

If we consider the solubility of NaOH as the concentration of $NaOH_{l} + Na^+ + OH^-$ we can say this:

The reaction $NaOH_{l} ↔ Na^+ + OH^-$ gives less products, but this only means that there will be less NaOH dissociated. In fact NaOH is still present in the liquid solution but in its non-dissociated form. That being said we can focus on the first equation $NaOH_{s} ↔ NaOH_{l}$. This is a simple non-reagent equilibrium so it can be described with formula I mentioned earlier:

$x_i = exp \left ( \frac {ΔH_{NaOH, fus}} R \left ( \frac 1 {T_{NaOH, fus}} - \frac 1 T \right ) \right )$

This explain why solubility still increase with Temperature. Do you think this sounds reasonable?

 

[Hen Sharir]

That's true but
Otherwise how can NaOH solubility increase if exothermic reaction are less favored at higher T?EDIT: I though of a way to explain it. NaOH in water should give the reaction $NaOH → Na^+ + OH^-$, so we can write

$NaOH_{s} ↔ NaOH_{l}$ // Solid-Liquid equilibrium
$NaOH_{l} ↔ Na^+ + OH^-$

If we consider the solubility of NaOH as the concentration of $NaOH_{l} + Na^+ + OH^-$ we can say this:

The reaction $NaOH_{l} ↔ Na^+ + OH^-$ gives less products, but this only means that there will be less NaOH dissociated. In fact NaOH is still present in the liquid solution but in its non-dissociated form. That being said we can focus on the first equation $NaOH_{s} ↔ NaOH_{l}$. This is a simple non-reagent equilibrium so it can be described with formula I mentioned earlier:

$x_i = exp \left ( \frac {ΔH_{NaOH, fus}} R \left ( \frac 1 {T_{NaOH, fus}} - \frac 1 T \right ) \right )$

This explain why solubility still increase with Temperature. Do you think this sounds reasonable?

I spoke to my Chemistry teacher about it and she said that since NaOH is already VERY soluble in water the increase in temperature shouldn't have a noticeable effect on its equilibrium.

 

[dRic2]

Yeah, changes are small, but they exist (according to wikipedia)