2H_2O rightleftharpoonsH_3O^(+) + HO^-
As you have correctly interpreted, this is a bond-breaking reaction, which requires substantial energy input to break the strong O-H bond, and the small equilibrium constant, 10^-14 (at 298*K), reflects this strength. Now this reaction was exhaustively measured at a precise temperature of 298*K. At higher temperatures, given a bond breaking reaction, we would expect that the equilibrium should lie farther over to the RHS, the product side, and thus pH should DECREASE at higher temperatures.
And thus at 100 ""^@C, pH=6.14, as anticipated (this is almost a tenfold increase in the extent of reaction!). See [the lower part of here.](http://chemguide.co.uk/physical/acidbaseeqia/kw.html). Of course under this scenario, pOH=6.14 necessarily.
Note that typically any equilibrium is a function of temperature. A standard temperature of 298*K is assumed for simplicity.
