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

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

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?

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

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

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)

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.

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

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

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

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?

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

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)

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!

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.

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)

Entropy and the 2nd Law of Thermodynamics How does the change in entropy relate to the spontaneity of a chemical reaction or process? 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.

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:

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

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

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

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

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

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

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.

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

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)