# K_w and auto-ionization of water

1. Dec 3, 2013

### Zondrina

$K_w$ and auto-ionization of water

1. The problem statement, all variables and given/known data

The $K_w$ value decreases as the temperature decreases. Is the auto-ionization of water an exothermic or endothermic reaction in the forward direction?

2. Relevant equations

3. The attempt at a solution

Water is able to auto-ionize into $H_3O^{+1}$ and $OH^{-1}$ in an equilibrium reaction:

$H_2O_{(L)} + H_2O_{(L)} \leftrightharpoons H_3O^{+1}_{(aq)} + OH^{-1}_{(aq)}$

The equilibrium constant for this reaction is defined as $K_w = [H_3O^{+1}][OH^{-1}]$. It is known that the $K_w$ value increases as the temperature increases and decreases as the temperature decreases.

If $K_w$ is decreasing because temperature is decreasing, it means that the concentration of products is decreasing and so the reaction must move forward to produce more products and restore equilibrium. This means that heat must be a by-product of the reaction and therefore the overall reaction must be exothermic so that the equilibrium will shift right as the temperature decreases.

Is this reasonable?

Last edited: Dec 3, 2013
2. Dec 3, 2013

### Keipi

The autoionization of water is an endothermic reaction.

If we increases the temperature Kw increases and the equilibrium shifts towards the products. Because an increase in temperature favors the endothermic side (the products, in this case) we can conclude that the forward reaction is endothermic and not exothermic.

3. Dec 3, 2013

### Staff: Mentor

But it doesn't shift right.

Think in terms of LeChatelier's principle.

4. Dec 3, 2013

### Zondrina

Ah so I should think of it the other way around.

If $K_w$ is decreasing because temperature is decreasing, it means that the concentration of products is decreasing BECAUSE the reaction is shifting to the left to compensate for the heat loss. While this shift occurs, it is clear that products become reactants.

Since we added heat and the equilibrium shifted left, we can conclude the overall reaction is endothermic.

Thanks guys :)