Chapter 19 Part 3: Free Energy.

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Presentation transcript:

Chapter 19 Part 3: Free Energy

Effects of Temperature on Spontaneity. H2O (l)  H2O (g) What happens to the entropy? One mole of water has a volume of 18 mL. One mole of water vapor at 100 °C has a volume of 31 liters. ΔS system is + in this reaction.

What is happening to ΔS universe? In H2O (l)  H2O (g), the reaction is endothermic, requiring energy from the surroundings. This draw of energy reduced the entropy of the surroundings. Thus ΔS surround is negative.

Temperature Since ΔS system and ΔS surround are in opposition for the vaporization of water… Temperature must have an effect on the relative importance of the two terms.

Heat Flow Entropy changes in the surroundings are primarily determined by the heat flow. Exothermic Rxns  the entropy of the surroundings. Endothermic Rxns  the entropy of the surrounding.

Heat as a driving Force. Heat as a driving force depends on the temperature at which the process occurs. If the temperature is high a small heat addition will make little change. If the temperature is low a small heat addition will make a significant change.

Expression of heat and entropy ΔS surround = ΔH T The sign of ΔS depends on the direction of heat flow The magnitude of ΔS depends on Temp.

Problems Calculate the ΔS surround for each of these reactions at 25 °C and 1 atm. Sb2S3 (s) + 3Fe (s)  2Sb (s) + 3FeS(s) ΔH = 125 kj Sb4O6 (s) + 6C(s)  4Sb (s) + 6CO (g) ΔH = 778 kj

Use ΔS surround = ΔH T T = 25 + 273 = 298 K

Answers ΔS surround= (125kj)/298K=0.419kj/K

Free Energy Spontaneity is a function of: ΔH, T & ΔS This function is called Gibbs Free Energy and is defined by: G=H-TS H is enthalpy, T is Temperature in Kelvins and S is entropy

For a Process at a constant temperature ΔG= ΔH –TΔS At a constant Temperature and Pressure, a process is spontaneous in the direction for which free energy decreases… or ΔG is negative.

Gibb's Free energy