[Prestomatic]

Here is a problem from a high school contest:

K_w for water is 1.0E-14 at 25 ° C. What is the K_w at 80 ° C, given that the enthalpy of neutralization for a strong acid and strong base is -55.8 kJ/mol?
I understand H20  H + OH is endothermic, thus LeChatlier's principle would move the reaction right, causing K_w to increase; however, I am lost in utilizing the enthalpy value to obtain a numeric answer.

Thanks

[Corribus]

Well I'd encourage you to write out an equation for the neutralization condition, and then see if you know any equations that relate the equilibrium constant of a reaction to temperature and the enthalpy change, which is usually assumed to be constant as a function of temperature.

[Prestomatic]

Well I'd encourage you to write out an equation for the neutralization condition, and then see if you know any equations that relate the equilibrium constant of a reaction to temperature and the enthalpy change, which is usually assumed to be constant as a function of temperature.

Thanks,
I suppose you're indicating the Van't Hoff equation?
ln(K_1/K_2) = -ΔH/R(1/T_2 - 1/T_1)
Plugging in:
K_1 = 1E-14
ΔH = 55.8
R = 8.314
T_2 = 273 + 80
T_1 = 273 + 25
Solving yields K_2 to be 1.0038E-14 (extended extra sig figs), but this answer is incorrect. You mentioned looking at the neutralization reaction (H + OH  H20), but I'm not seeing how this can be used.

[mjc123]

Units!!! ΔH = -55800 J/mol!

[Corribus]

Right. A good lesson in the importance of writing out all your units.

(Writing out the equation for the neutralization helps you know that the dissociation is endothermic. You already figured that out but I wrote it down anyway for the benefit of anyone else who came along and wanted to know that critical part of solving the problem.)

[Prestomatic]

This yielded the correct answer.

Thanks