Unit 5 – Part 1: Thermodynamics

1. Unit 5 – Part 1: Thermodynamics • Entropy and the Second Law of Thermodynamics • Gibbs Free Energy • Free Energy and Equilibrium Constants

2. Entropy and the 2nd Law of Thermodynamics • Industrial chemists are responsible for designing cost-effective manufacturing processes. 2 NH3 (g) + CO2 (g) NH2CONH2 (aq) + H2O (l) urea • Commercial uses of urea: • nitrogen fertilizer for plants • used to manufacture certain plastics and adhesives

3. Entropy and the 2nd Law of Thermodynamics • Some of the questions a chemist must consider: • Does the reaction need to be heated? • How much product will be present at equilibrium? • Does the reaction naturally proceed in this direction?

4. Entropy and the 2nd Law of Thermodynamics • In order to determine if a reaction proceeds naturally in the direction written, we need to know if it is spontaneous. • Capable of proceeding in the direction written without needing to be driven by an outside source of energy

5. Entropy and the 2nd Law of Thermodynamics • Examples of spontaneous processes: • an egg breaking when dropped • a ball rolling down a hill • ice  water at room temperature • Examples of non-spontaneous processes: • a ball rolling up a hill • water  ice at room temperature

6. Entropy and the 2nd Law of Thermodynamics • Examples of spontaneous reactions: • CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g) • 2 N (g)  N2 (g) • Examples of non-spontaneous reactions: • 2 H2O (g) 2 H2(g) + O2(g) • O2 (g)  2 O (g)

7. Entropy and the 2nd Law of Thermodynamics • Many but not all spontaneous processes are exothermic. • Enthalpy (DH) alone cannot be used to predict whether or not a reaction is spontaneous. • To predict whether a reaction is spontaneous, we need to use the second law of thermodynamics and a thermodynamic quantity called entropy.

8. Entropy and the 2nd Law of Thermodynamics • Entropy (S): • a thermodynamic quantity related to the disorder or randomness of a system. • The more disordered or random the system is, the larger its entropy is. • A state function • not path dependent

9. Entropy and the 2nd Law of Thermodynamics • Every chemical has an entropy associated with it that depends on its physical state, temperature, and pressure: • H2O (l) 69.91 J/mol.K at 25oC/1 atm • H2O (g) 188.83 J/mol.K at 100oC/1 atm • Appendix C: • table of thermodynamic properties including S

10. Entropy and the 2nd Law of Thermodynamics • The entropy change(DS) can be calculated for any process: • DS = Sfinal - Sinitial • DS = Sproducts - Sreactants • Sign conventions for DS: • DS = positive  more disordered • DS = negative  less disordered

11. Entropy and the 2nd Law of Thermodynamics Example: The following process occurs at 0oC and 1 atm pressure. Does the system become more ordered or more disordered?

12. Entropy and the 2nd Law of Thermodynamics • The sign of DS can be predicted: • In general, any change that increases the overall disorder or randomness will result in a positive value for DS. • In general, the overall entropy increases when: • a molecule (or anything else) is broken into two or more smaller molecules • there is an increase in the number of moles of a gas • a solid changes to a liquid or gas • a liquid changes to a gas

13. Entropy and the 2nd Law of Thermodynamics Example: Without doing any calculations, predict whether DS will be positive or negative. Breaking an egg N2 (g) + 3 H2(g) 2 NH3 (g) 2NH3(g) + CO2(g) NH2CONH2(aq) + H2O (l)

14. Entropy and the 2nd Law of Thermodynamics • The entropy change (DS) for a reaction or process can be calculated using the following equation: DSo = S n Soproducts - S m Soreactants where So = the standard molar entropy • Note: This is similar to the method used to calculate DHo for a reaction!

15. Entropy and the 2nd Law of Thermodynamics • Standard molar entropy (So) : • the entropy value for one mole of a chemical species in its standard state • 1 atm pressure • 1 M (for those in solution) • NOTE: Unlike DHof, the standard molar entropy of a pure element is NOT zero.

16. Entropy and the 2nd Law of Thermodynamics Example: Predict whether the entropy will increase or decrease for the following reaction. Calculate DSo. C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g)

17. Entropy and the 2nd Law of Thermodynamics • The Second Law of Thermodynamics can be used to predict whether a reaction will occur spontaneously. • The total entropy of a system and its surroundings always increases for a spontaneous process. • How does the change in entropy relate to the spontaneity of a chemical reaction or process?

18. Gibbs Free Energy • Simply looking at the sign of DS for a chemical reaction or process does not tell you if the reaction is spontaneous. • Spontaneous reactions involve an overall increase in the entropy of the universe. • Reactions that have a large, negative DH tend to be spontaneous:

19. Gibbs Free Energy • The Gibbs free energy (G) is used to relate both the enthalpy change and the entropy change of a reaction to its spontaneity. G = H - TS where G = Gibbs free energy (“free energy”) H = enthalpy T = temperature (K) S = entropy

20. Gibbs Free Energy • Free energy is a state function. • The change in free energy (DG) of a system can be used to determine the spontaneity of a process or reaction. • For a process occurring at constant temperature: DG = DH - TDS

21. Gibbs Free Energy • For a reaction occurring at constant temperature and pressure, the sign of DG can be used to determine if a reaction is spontaneous in the direction written: • DG = negative • reaction is spontaneous in the forward direction • DG = zero • reaction is at equilibrium

22. Gibbs Free Energy • The sign of DG can be used to determine if a reaction is spontaneous in the direction written (cont): • DG = positive • reaction is not spontaneous in the direction written • work must be supplied by the surroundings to make the reaction occur in the direction written • reaction is spontaneous in the reverse direction

23. Gibbs Free Energy Example:Using the definition of DG, calculate the DG for the following reaction at 35oC: 2 H+(aq) + S2-(aq) H2S(g) DH = -61.9 kJ DS = + 183.6 J/K

24. Gibbs Free Energy • The standard free energy of formation(DGof) has been tabulated for many different substances. (see Appendix C) • the change in free energy associated with the formation of 1 mole of a substance from its elements under standard conditions • pure solid • pure liquid • gas at 1 atm pressure • solution with 1 M concentration

25. Gibbs Free Energy • There is not a standard temperature for determining DGof. • 25oC is often used for tables of data • values can be calculated at other temperatures as well • DGof for an element in its standard state is zero.

26. Gibbs Free Energy • The standard free energy change for a chemical process can be calculated using the following expression: DGo = S n DGof(products) - S m DGof(reactants) • Note: This is similar to the way we calculated DHo and DSo

27. Gibbs Free Energy Example: Calculate DGo for the following reaction using the standard free energies of formation. 2 KClO3 (s)  2 KCl (s) + 3 O2 (g)