Ch. 17: Thermodynamics: Entropy and Free Energy Entropy I. Entropy, S, is the measure of the disorder of a system. A. This, like enthalpy, cannot be measured. B. Thus, only the change in disorder (∆S) can be measured. II. A reaction is spontaneous (more on this later) if: A. heat decreases (H, enthalpy) B. disorder increases (S, entropy) III. Spontaneity depends on BOTH enthalpy AND entropy. IV. Why does entropy happen? Probability! A. It’s harder to keep things in order (look at my room). B. Therefore, it takes less energy to allow things to become disorderly.
V. Entropy of states of matter: A. Ssolid< Sliquid<<< Sgas VI. Entropy increases generally when: A. Liquids/solids become gases B. Solids dissolve C. # mols of gases increases D. # mols increases VII. Standard entropies (Sº) tend to increase with increasing molar mass.
The Laws of Thermodynamics I. First Law A. Energy is neither created nor destroyed B. a.k.a. Conservation of Energy C. Gives rise to Hess’s Law (products - reactants) D. Side note: heat flows from hot (high temps) to cold (low temps). II. Second Law A. Systems have an overall tendency to increase in entropy (S) B. disorder increases (+S) III. Third Law A. Entropy of a pure crystalline substance at absolute zero is zero (S=0 @ 0K)
Ch. 17: Thermodynamics: Entropy and Free Energy Spontaneity I. Many chemical reactions spontaneously happen. A. A reaction is spontaneous if it happens without any outside intervention. B. This has nothing to do with rate (speed). Even if a reaction happens slowly, it is considered spontaneous. II. A reaction is spontaneous if: A. heat decreases (H, enthalpy) B. disorder increases (S, entropy) III. Thus, spontaneity depends on BOTH enthalpy AND entropy. IV. Processes that are spontaneous at one temp. can be non-spontaneous at another temp.
Gibbs Free Energy (G) I. Definition A. Energy available to do work B. Dependent on enthalpy (H) and entropy (S) C. Equation: ∆G = ∆H - T∆S D. For standard (298K and 1 atm) states, ∆G̊ = ∆H̊ - T∆S̊ 1. This shows that the effect of entropy on spontaneity is dependent on temperature E. T is temperature in K. Units for G and H are usually kJ and S is usually in J so be careful! F. Determines the spontaneity of a reaction.
II. The Significance of the Signs Property ∆H ∆S ∆G
Positive (+)
Negative (-)
Ch. 17: Thermodynamics: Entropy and Free Energy III. A Special Case! A. When ∆G = 0, the system is at equilibrium. B. The equation changes: ∆H = T∆S or ∆H/∆S = T 1. You can find the temperature at a given point if the system is at equilibrium (i.e. boiling point of a reaction) 2. Ex: What is the boiling point of a solution where the ∆Hvap=23.5kJ/mol and ∆S=34.5J/Kmol? a. At boiling, there is equilibrium between liquid and gas phase.
b.
IV. Spontaneous or not? Use ∆G = ∆H - T∆S ∆G
∆H
∆S
Spontaneous?
V. Example: 2 C4H10(g) + 13 O2(g) ------> 8 CO2(g) + 10 H2O(l) The reaction represented above is spontaneous at 25 °C. Assume that all reactants and products are in their standard states. (a) Predict the sign of ΔS° for the reaction and justify your prediction. (b) What is the sign of ΔG° for the reaction? How would the sign and magnitude of ΔG° be affected by an increase in temperature to 50 °C? Explain your answer.
Ch. 17: Thermodynamics: Entropy and Free Energy (c) What must be the sign of ΔH° for the reaction at 25°C? How does the total bond energy of the reactants compare to that of the products?
(d) When the reactants are placed together in a container, no change is observed even though the reaction is known to be spontaneous. Explain this observation.
Thermo. Calculations I. Calculating ∆H (four ways): Units usually kJ or kJ/mol A. Calorimetry (q=mc∆T) B. Hess’s Law 1. v. 1: Combining ∆H of a set of reactions. 2. v. 2: ∑∆Hºf, products - ∑∆Hºf, reactants (Remember: ∆Hºf is 0 for an element in its standard state) C. Bond enthalpies (energies): bonds broken-bonds formed. II. Calculating ∆S: Units usually J or J/mol A. ∑Sºproducts - ∑Sºreactants B. When calculating with S AND G or H, remember to convert so that the units match! C. Unlike ∆H, ∆Sº for an element in standard state ≠ 0. D. Ex. Calculate ∆Sº for the rxn: 2NiS(s) + 3O2(g) --> 2SO2(g) + 2NiO(s) E. Ex. Calculate ∆Sº for the rxn: Al2O3(s) + 3H2(g) --> 2Al(s) + 3H2O(g) III. Calculating ∆G (five ways) A. ∆G = ∆H - T∆S or ∆G̊ = ∆H̊ - T∆S̊ B. ∆Gº = -nF∈º (more on this in this next chapter) C. ∑∆Gºf, products - ∑∆Gºf, reactants 1. Just like ∆Hºf, an element in its standard state has a ∆Gºf =0 2. Ex: Calculate ∆Gº for the reaction: 2CH3OH + 3O2 --> 2CO2 + H2O D. ∆G = ∆Gº+RTlnQ 1. Used when substances are not at standard state (not at 1 atm). 2. Usually used for gas problems. E. ∆Gº = -RTlnK (K=equilibrium constant; R=8.3145J/molK; T=temp. in K) IV. Practice: A. Calculate ΔH, ΔS, and ΔG for the following reaction: Use the appropriate appendices in your book to find individual values: 2SO2 (g) + O2(g) ⟶ 2SO3 (g)
Ch. 17: Thermodynamics: Entropy and Free Energy B. Consider the ammonia synthesis rxn: N2(g) + 3H2(g) ↔ 2NH3 (g) Where ΔGº = -33.3 kJ/mol of N2 consumed at 25º C. Calculate the value for the equilibrium constant C. Example 2: Calculate the value of ΔGº for the following reaction at 389K where at equilibrium, [NH3]=2.0M, [H2]=1.25M, and [N2]=3.01M. N2(g) + 3H2(g) ↔ 2NH3 (g)
V. AP Practice: C2H2(g) + 2 H2(g) ---> C2H6(g) Information about the substances Substance
S° (J/mol K)
C2H2(g) H2(g) C2H6(g)
200.9 130.7 --------
ΔH°f (kJ/mol) Bond 226.7 0 -84.7
C-C C=C C-H H-H
Bond Energy (kJ/mol) 347 611 414 436
(a) If the value of the standard entropy change, ΔS°, for the reaction is -232.7 joules per mole Kelvin, calculate the standard molar entropy, S°, of C2H6 gas.
(b) Calculate the value of the standard free-energy change, ΔG°, for the reaction. What does the sign of ΔG° indicate about the reaction above?
(c) Calculate the value of the equilibrium constant, K, for the reaction at 298 K.
(d) Calculate the value of the C≡C bond energy in C2H2 in kilojoules per mole.
Free Energy and Equilibrium I. When a reaction is at equilibrium, A. ∆G = 0 because neither the forward or reverse reaction will spontaneously occur. B. Q=K because the concentrations present are the equilibrium concentrations. II. When the reaction is at equilibrium, ∆Gº is directly related to K. When... A. ∆Gº = 0, K = 1 B. ∆Gº < 0, K > 1 C. ∆Gº > 0, K < 1
