The standard enthalpy change upon dissolving one mole of oxygen at 25°C is -11.7 kJ. Use this number and the van't Hoff equation (attached) to calculate the equilibrium (Henry's law) constant for oxygen in water at O°C and at 100°C. Discuss the results briefly. Derive the van't Hoff equation,dIn KΔΗ'dTRT2which gives the dependence of the equilibrium constant on temperature.* HereAH° is the enthalpy change of the reaction, for pure substances in their standardstates (1 bar pressure for gases). Notice that if AH° is positive (loosely speaking,if the reaction requires the absorption of heat), then higher temperature makes thereaction tend more to the right, as you might expect. Often you can neglect thetemperature dependence of AH°; solve the equation in this case to obtainIn K(T3) – In K(T) = G-)AH°11R

Given data: Standard enthalpy of dissolution of oxygen = -11.7 kJ

Answered: Derive the van't Hoff equation, dIn K…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

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The standard enthalpy change upon dissolving one mole of oxygen at 25°C is -11.7 kJ. Use this number and the van't Hoff equation (attached) to calculate the equilibrium (Henry's law) constant for oxygen in water at O°C and at 100°C. Discuss the results briefly.

Derive the van't Hoff equation,
dIn K
ΔΗ'
dT
RT2
which gives the dependence of the equilibrium constant on temperature.* Here
AH° is the enthalpy change of the reaction, for pure substances in their standard
states (1 bar pressure for gases). Notice that if AH° is positive (loosely speaking,
if the reaction requires the absorption of heat), then higher temperature makes the
reaction tend more to the right, as you might expect. Often you can neglect the
temperature dependence of AH°; solve the equation in this case to obtain
In K(T3) – In K(T) = G-)
AH°
1
1
R

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