The ion-product constant for water is \(1.0 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\) and \(3.8 \times 10^{-14}\) at \(40^{\circ} \mathrm{C}\). Is the process $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ endothermic or exothermic?

Short Answer

Expert verified
The given process of water disassociating into ions is endothermic.

Step by step solution

01

Comprehend the information

The first step is understanding the given information. The ion-product constant for water, \(K_w\), at 25 Celsius is \(1.0 \times 10^{-14}\), and at 40 Celsius it is \(3.8 \times 10^{-14}\). It is noticed that \(K_w\) is increasing with the increase in temperature.
02

Apply Le Chatelier's Principle

Next, apply Le Chatelier's Principle which states that when a reaction at equilibrium is subjected to a change in condition, it strives to achieve a new equilibrium state by favouring either the forward or the backward reaction. Here, we see, that as temperature increases, the ion-product constant, \(K_w\), also increases. Following Le Chatelier's Principle, this means the equilibrium has shifted towards the right or in other words, the forward reaction is being favoured.
03

Determine endothermic or exothermic reaction

This shift towards the forward reaction with increasing temperature suggests that the reaction is endothermic. This is because, according to Le Chatelier's Principle, an increase in temperature favours the endothermic reaction. In an endothermic reaction, heat is effectively a reactant, because it is absorbed from the surroundings.
04

Conclusion

Based on the direction of the shift in equilibrium with the increase in temperature, it can be concluded that the disassociation of water into ions is an endothermic process.

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