The van’t Hoff equation relates the equilibrium constant of a reaction to change

Question

The van’t Hoff equation relates the equilibrium constant of a reaction to change in temperature
assuming that the enthalpy of a reaction (?HR°) is constant as a function of temperature. The
assumption of constant enthalpy is generally valid over a short temperature range. Derivation
and discussion of the van’t Hoff equation is given in Krauskopf and Bird, Chapter 8, pg. 203-
210.
Using the van't Hoff equation, calculate the standard enthalpy of reaction for the solubility of O2 in water using its equilibrium solubility at 10°C (from Problem #1) and the equilibrium value at 25°C (which you will need to look up).

Thank you.

Explanation / Answer

Applying Henry's Law constant:

ln(588 atm-L/mol/792.6442 atm-L/mol) = -deltaHRo /(8.314 J/mol K) [1/(283) - 1/(298)]

-0.2986 = -deltaHRo /(8.314 J/mol K) [1.7786 x 10-4 1/K]

1679.0759 = deltaHRo /(8.314 J/mol)

deltaHRo = 13959.837 J/mol

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