Solubility of a gas as a function of temperature. The van\'t Hoff equation relat

Question

Solubility of a gas as a function of temperature. The van't Hoff equation relates the equilibrium constant of a reaction to change in temperature assuming that the enthalpy of a reaction (delta HR degree) 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. The van't Hoff equation can be written: where KT is the equilibrium constant at temperature T. KTO is the equilibrium constant at temperature To. and DELTAHRDEGREE is the standard enthalpy of the reaction. Using the van't Hoff equation, calculate the standard enthalpy of reaction for the solubility of O2 in water using its equilibrium solubility at 10degreeC (from Problem #2) and the equilibrium value at 25degreeC (which you will need to look up).

Explanation / Answer

we are not having the data of question2 and also the equillibrium values of Kt and Kto..

ln(Kt/Kto) = -delta Hr/ 8.314(1/283 - 1/298 )

temperature should be in kelvin..

use kt and kto values solve the above equation

otherwise the post the data of Kt and kto values i will solve for you

please rate me

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