For the autoionization of water, ΔH° = 5.58 x 104 J. 2H2O (l) ⇌ H3O + (aq) + OH - (aq) Kw = 1.0 x 10-14 at 25 °C Assuming that ΔH° is constant over the temperature range 25-100°C, calculate KW at 53 °C. (Hint: Units!) (for values in scientific notation use e to represent x10^)

1.9 x 10⁻¹⁴

Explanation:

The Gibbs-Helmholtz equation which gives us the dependendence of the equilibrium constant, k, at the different temperatures, T₁ and T₂, will be used to solve this question:

ln (k₂/k₁) = - ΔHº/R (1/T₂ - 1/T₁)

Here we are to assume ΔHº is constant over the temperature range 25-100 ºC.

We have all the data input required, so lets substitute and solve for k₂

T₁ = (25 + 273) K = 298 K

T₂ = (53 + 273) K = 326 K

R = 8.314 J

ln (k₂/1.0 x 10⁻¹⁴) = - 5.58 x 10⁴ J/K x (8.314 J/K (1/326 K - 1/298 K))

ln (k₂/1.0 x 10⁻¹⁴) = 1.9

Notice the units cancel each other as we would expect it to be since k₂/k₁ is unitless.

Now take inverse ln to both sides of the equation:

k₂ / 1.0 x 10⁻¹⁴ = 1.93 ⇒ k₂ = 1.9 x 10⁻¹⁴

This result is logical because the reaction is endothermic given ΔH° is positive at 25 ºC, so at 53 ºC we would expect k₂ greater than k₁.