Question

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 49 °C

Answer 1

The calculation of the autoionization constant (Kw) of water at 49 °C requires using the van't Hoff equation and the provided ΔH° and Kw at 25 °C, accounting for the endothermic nature of the reaction, and expressing temperatures in Kelvin.

Step-by-step explanation:

To calculate the autoionization constant (Kw) of water at 49 °C, we can use the van't Hoff equation which relates the equilibrium constant to temperature changes. Since we are given that the ΔH° for the reaction is 5.58 x 104 J and that the Kw at 25 °C is 1.0 x 10−14, we can assume the following van't Hoff equation form:

ln(Kw2/Kw1) = (ΔH°/R) ((1/T1) - (1/T2))

Where Kw1 is the equilibrium constant at T1 (25°C or 298K), Kw2 is the equilibrium constant at T2 (49 °C or 322K), ΔH° is the enthalpy change of the reaction, and R is the universal gas constant (8.314 J/mol·K). Solving for Kw2 using the provided values and converting the temperatures to Kelvin, we can calculate the new equilibrium constant at 49 °C.

It is important to remember that temperature must be in Kelvin when using the van't Hoff equation and that since the autoionization of water is endothermic, the value of Kw will increase with temperature.