Auto Ionization of Water: Understanding the Equilibrium and Heat Transfer

[Evangeline101]

Homework Statement

Homework Equations

The Attempt at a Solution

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I'm confused with this question... first off isn't heat being removed from the product side since the Kw value is on the product side?

If I assume heat is being removed from the product side, wouldn't the reaction shift to the right to restore equilibrium causing the reaction to be exothermic in the forward direction?

A lot of people are saying it is endothermic but I am not sure how? wouldn't that mean the heat is being removed from the reactant side? Thanks.

 

[vbrasic]

I'm confused with this question... first off isn't heat being removed from the product side since the Kw value is on the product side?

What do you mean the $k_w$ is on the products side? The $k_w$ is just the rate of reaction, it doesn't belong to the products side or reactants side. (It is associated with a direction though.)

If I assume heat is being removed from the product side, wouldn't the reaction shift to the right to restore equilibrium causing the reaction to be exothermic in the forward direction?

A lot of people are saying it is endothermic but I am not sure how? wouldn't that mean the heat is being removed from the reactant side?

The people saying endothermic are correct. If temperature decreases, we would expect the reaction to proceed more vigorously in the direction of heat release to restore equilibrium. However, in your case, the rate decreases (meaning that the reverse reaction is becoming more favorable). This suggests that the reverse reaction is actually the exothermic reaction, and that the forward reaction is endothermic as suggested by your peers.

 

[Evangeline101]

The people saying endothermic are correct. If temperature decreases, we would expect the reaction to proceed more vigorously in the direction of heat release to restore equilibrium. However, in your case, the rate decreases (meaning that the reverse reaction is becoming more favorable). This suggests that the reverse reaction is actually the exothermic reaction, and that the forward reaction is endothermic as suggested by your peers.

So does this mean heat is being removed from the reactant side?

 

[vbrasic]

So does this mean heat is being removed from the reactant side?

No it means that heat is being added to the reactant side when the reaction proceeds in the forward direction. If a system is in equilibrium and we decrease the temperature (loosely related to the total heat present) we expect that the reaction will proceed in whichever direction restores the equilibrium temperature. If the forward reaction is exothermic (meaning the reactants are releasing heat; i.e. heat is being removed), then we expect that in order to restore the lost temperature, the reaction will "speed up" in the forward direction. However, in your case, the temperature decrease causes the reverse reaction to be favored (that is, forward rate is decreased). This tells us, that when we proceed in the forward direction, that more heat must be being added to the reactants, taking away from the heat available in the environment, moving the system further away from the equilibrium environmental temperature. We are talking about heat being taken away from/put into the surroundings.

In an endothermic reaction, we are taking heat from our surroundings and adding it to the energy of the products causing the surroundings to cool down.

 

[Borek]

The kw is just the rate of reaction

No, it is an equilibrium constant.

Sure, it looks like the rate equation for the neutralization, but it is not.

 

[vbrasic]

No, it is an equilibrium constant.

Sure, it looks like the rate equation for the neutralization, but it is not.

Ah, yes, you are correct. Sorry about that. My answer is still correct, though not for the aforementioned reasons. I will revise it here:

The equilibrium constant for auto-ionization of water is given by, $$k_{w}=\frac{[OH^-][H_3O^+]}{[H_2O]}.$$ Granted that $[H_2O]=1$ always, we have that $k_w=[OH^-][H_3O^+]$. The greater the concentrations of hydroxide, (OH^-) and hydronium, (H₃O⁺), the greater this value. Now, your question tells you that the equilibrium constant is decreasing, meaning that the concentrations are getting smaller (i.e. the forward reaction is less favorable). We can use the exact same reasoning as before. If temperature decreases, the reaction will proceed more vigorously in the direction which releases heat to restore equilibrium temperature. This implies then, that the forward reaction is not exothermic (as it is not being favored), and so, the forward reaction is then endothermic.

Sorry again about that mix-up. For whatever reason I was thinking of $k$ as in the rate constant.

 

[Borek]

Granted that $[H_2O]=1$ always

That's again not true - [H₂O]=1000/18=55.6 M. What is important here is that this concentration in typical situation can be considered to be constant and as such safely ignored.

Doesn't mean you aren't right about the reaction being endothermic.

 

[Ygggdrasil]

Homework Statement

View attachment 114243

Homework Equations

The Attempt at a Solution

[/B]
I'm confused with this question... first off isn't heat being removed from the product side since the Kw value is on the product side?

If I assume heat is being removed from the product side, wouldn't the reaction shift to the right to restore equilibrium causing the reaction to be exothermic in the forward direction?

A lot of people are saying it is endothermic but I am not sure how? wouldn't that mean the heat is being removed from the reactant side?Thanks.

It would be helpful to consider a few equation. The Kw is an equilibrium constant, which means it tells you about change in free energy (ΔG) of water ionization. Do you know the equation that connects ΔG to Kw. If Kw increases, how does ΔG change? If Kw decreases, how does ΔG change?

ΔG itself is influenced by the change in entropy (ΔS) and change in enthalpy (ΔH) during the reaction. Do you know the equation that links ΔG to ΔH and ΔS? What can you say about ΔS of the reaction?