2/18/2013. Spontaneity, Entropy & Free Energy Chapter 16. The Dependence of Free Energy on Pressure Sample Exercises

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1 Spontaneity, Entropy & Free Energy Chapter The Dependence of Free Energy on Pressure Why is free energy dependent on pressure? Isn t H, enthalpy independent of pressure at constant pressure? No work? G = H - T S At a constant temperature S large volume > S small volume S low pressure > S high pressure G = G + RT ln(p) G = G + RT ln(q) G is free energy at 1 atm G is free energy at pressure P R is universal gas constant J/K-mol T is temp in Kelvin Q is reaction quotient using partial pressures (p593) The entropy and free energy of an ideal gas depends on its pressure. Remember Q = 0 equilibrium; Q>K shift towards reactants; Q<K shift towards products 16.7 Sample Exercises One method for synthesizing methanol (CH 3 OH) involves reacting carbon monoxide and hydrogen gases: CO (g) + 2H 2 (g) CH 3 OH (l) Calculate the G at 25 C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol. -38 kj/mol G is more negative than G the reaction is more spontaneous at pressures above 1 atm Substance G f kj CH 3 OH (l) 51 H 2 (g) 131 CO (g) 28 1

2 A system will try to achieve the lowest possible free energy Sometime that is equilibrium not a complete reaction REACTANTS PRODUCTS Minimum free energy happens at the completion of products so no equilibrium MINIMUM FREE ENERGY Minimum free energy happens before the completion of products, B, so it is at equilibrium. A is changing into B, but it never fully gets there, so equilibrium (a) At the beginning of the reaction (b) As the reaction is occuring (c) If G reactants = G products then the system is at equilibrium because G = 0 Should happen at minimum free energy A (g) B (g) Start with 1.0 mole of gas A at 2.0 atm Start with 1.0 mole of gas B at 2.0 atm K = P B / P A = 1.5 atm / 0.50 atm = 3.0 At 75% of A has reacted P A =2.0 atm X.25 = 0.5 atm 2

3 At equilibrium G products = G reactants or G = G products G reactants = 0 G = G + RT ln (Q) G = 0 = G + RT ln (Q) G = - RT ln (Q) at equilibrium Q = K G = - RT ln (K) G = G products G reactants = 0 G = G + RT ln (Q) G = - RT ln (K) Case 1: G = 0 Case 2: G < 0 Case 3: G > 0 G products = G reactants K = 1 All pressures = 1 atm The system is at equilibrium. G products < G reactants The system will shift towards products. K > 1 since the products at equilibrium will be greater than the reactants G products > G reactants The system will shift towards reactants. K < 1 since the reactants at equilibrium will be greater than the products Relationship between ΔG and K 3

4 The Temperature Dependence of K G = - RT ln (K) = H - T S ln (K) = H T S -RT -RT ln (K) = - H 1 S + R T R y = m x + b The graph of ln (K) versus 1/T is a straight line graph. With slope = - H /R and intercept S /R This assumes that H and S are independent of temperature. This is a good approximation over a relatively small temperature range. So K is dependent on T because of the G of the system! Sample Exercises Consider the ammonia synthesis reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) where G = kj per mole of N 2 consumed at 25 C. For each of the following mixture of reactants and products at 25 C, predict the direction in which the system will shift to reach equilibrium. a. P NH3 = 1.00 atm; P N2 = 1.47 atm; P H2 = 1.00 x 10-2 atm b. P NH3 = 1.00 atm; P N2 = 1.00 atm; P H2 = 1.00 atm Sample Exercises The overall reaction for the corrosion (rusting) of iron by oxygen is 4Fe (s) + 3O 2 (g) 2Fe 2 O 3 (s) Using the data below, calculate the equilibrium constant for this reaction at 25 C. Substance H f kj/mol S J/K-mol Fe 2 O 3 (s) Fe (s) 0 27 O 2 (g)

5 16.9 Free Energy and Work The maximum possible useful work obtained from a process at constant T & P is equal to the change in free energy. w max = G 100% efficiency Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. Consider a rechargeable battery When current from the battery flows through a circuit it heats the wires. The heat is wasted energy- lost to the surroundings. Not energy doing useful work. When recharging the battery more work is required than the battery discharged. Even though the battery has returned to its original state, the surrounding have not. Second Law of Thermodynamics (restated) In any real cyclic process in the system, work is changed to heat in the surroundings, and the entropy of the universe is increased Cell Potential, Electrical Work, and Free Energy Work is done when electrons are transferred through a a wire Depends on the push behind the electrons which is the emf or potential difference (in volts) 1 volt = 1 J/ 1 C emf = potential difference (V) = work (J) / charge (C) E = -w / q When a cell produces current the potential is positive and the current can be used to do work so the work is negative Rearranging w = q E so w max = - q E max In any real spontaneous process some energy is always wasted the actual work realized is always less than the calculated maximum. Given maximum potential 2.50 V, moles of electrons, Cell Potential, Electrical Work, and Free Energy Given maximum potential 2.50 V, moles of electrons 1.33 mol e-, and actual potential 2.10 V Efficiency can be calculated Calculate q with the Faraday C/ mol e- Calculate the actual work Calculate the efficiency 5

6 Cell Potential, Electrical Work, and Free Energy Now let s include free energy w max = - q E max = G q = nf n = moles of e- G = - nf E max G = - nf E the subscript max is usually left out at standard conditions The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell. Provides an experimental means to obtain G for a reaction Galvanic cells will run in the direction that gives a positive value for E cell and a negative value for G Sample Exercise Using the data in Table 17.1, calculate the G for the reaction Cu 2+ (aq) + Fe (s) Cu (s) + Fe 3+ (aq) Is this reaction spontaneous?l Sample Exercise 17.4 Using the data from Table 17.1, predict whether 1 M HNO 3 will dissolve gold metal to form a 1 M Au 3+ solution. 6

7 Sample Exercise Chlorine dioxide (ClO 2 ), which is produced by the reaction 2NaClO 2 (aq) + Cl 2 (g) 2ClO 2 (g) + 2NaCl (aq) has been tested as a disinfectant for municipal water treatment. Using the data from Table 17.1, calculate the E and the G at 25 C for the production of ClO 2. Table 17.1 BACK 7