CHEMICAL THERMODYNAMICS. Nature of Energy. ΔE = q + w. w = PΔV
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1 CHEMICAL HERMODYNAMICS Nature of Energy hermodynamics hermochemistry Energy (E) Work (w) Heat (q) Some Definitions Study the transformation of energy from one form to another during physical and chemical processes. Study the relationship between heat and chemical reactions. Capacity to do work or to transfer heat. he application of a force across some distance. Flow of energy from one object to another that causes a change in the temperature of the object. System Surroundings Universe Internal Energy Some hermodynamics erms Substances involved in the chemical and physical changes that we are studying. Everything in the system's environment that can exchange energy with it. System plus its surroundings. Sum of the system's kinetic and potential energies. he first law of thermodynamics = the law of conservation of energy. he total energy of the universe is constant during a chemical or physical change. Energy is neither created nor destroyed. ΔE = q + w Energy is simply transferred between the system and its surroundings. q is positive: heat is absorbed by the system (endothermic process) q is negative: heat is released by the system (exothermic process) w is positive: work is done on the system by the surroundings w is negative: work is done by the system on the surroundings w = PΔV Common type of work associated with a chemical process is the expansion or compression of a gas w is positive: w is negative: compression expansion Page 1
2 1. Calculate the change in energy of the system if 38.9 J of work is done by the system with an associated heat loss of 16.2 J. 2. Calculate the work associated with the contraction of a gas from 75 L to 30 L at a constant external pressure of 6.0 atm. (1 L atm = J) 3. A piston is compressed from a volume of 8.3 L to 2.8 L against a constant pressure of 1.9 atm. In the process, there is a heat gain by the system of 350 J. Calculate the change in energy of the system. (1 L atm = J) Entropy and Spontaneity Spontaneous = occurs without outside help. Spontaneity of any physical or chemical change is favoured: when heat is released during the change (exothermic) when the change causes an increase in entropy Entropy (S) is described as a measure of randomness (disorder) the greater the disorder of a system, the greater its entropy S solid < S liquid < S gas solution formation is favoured by entropy 4. Which of the following pairs is likely to have the higher (positional) entropy per mole at a given temperature? a. solid phosphorous or gaseous phosphorous b. CH 4 (g) or C 3 H 8 (g) c. O 2 (g) at 1 atm or O 2 (g) at atm How to predict whether a reaction will be spontaneous? Answer = he second law of thermodynamics. For a spontaneous process, the entropy of the universe will increase. ΔS univ = ΔS sys + ΔS surr > 0 for a spontaneous process 5. Predict the sign of entropy change, ΔS, for the following processes: a. 2NH 4 NO 3 (s) 2N 2 (g) + 4H 2 O (g) + O 2 (g) b. 2SO 2 (g) + O 2 (g) 2SO 3 (g) c. C 12 H 22 O 11 (aq) C 12 H 22 O 11 (s) Page 2
3 emperature and Spontaneity he impact of the transfer of a given quantity of energy as heat (to or from the surroundings) will be greater at lower temperatures. q rev S= ΔS = change in entropy, q = heat, = temperature in Kelvin rev = reversible process, any process for which a very small change in conditions can reverse its direction. Often, reversible conditions exist during phase changes. q S rev H sys = = sys since q surr = q sys and at constant pressure q sys = ΔH sys H S sys surr = 6. Determine the entropy change (ΔS) when 1 mol H 2 O boils at its normal boiling point of C. he heat of vaporization of water is 44.0 kj/mol. Calculating Entropy Changes in Chemical Reactions Values of entropy are based on some reference point that is common to all compounds. he third law of thermodynamics establishes this point for entropy. It states that the entropy of a pure perfect crystal at 0 K is zero. o Under standard conditions (298 K, 1 atm), compounds have standard molar entropies ( S 298 units of J/K mol. he standard entropy change, ΔS, for chemical reactions can be determined using: ) in ΔS reaction = Σ ns products Σ ns products n = number of moles 7. Calculate ΔS for the following reactions: a. N 2 O 4 (g) 2NO 2 (g) S [N 2 O 4 (g)] = 304 J/K.mol and S [NO 2 (g)] = 240 J/K.mol b. Fe 2 O 3 (s) + 2Al (s) 2Fe (s) + Al 2 O 3 (s) S [Fe 2 O 3 (s)] = 90 J/K.mol; S [Al (s)] = 28 J/K.mol; S [Fe (s)] = 27 J/K.mol and S [Al 2 O 3 (s)] = 51 J/K.mol Page 3
4 Gibbs Free Energy o determine the spontaneity of a reaction, the change in entropy of the universe must be calculated. A simpler and more useful way is to look at a property called the free energy (G). he Gibbs free energy of a system is defined as G = H S For conditions of constant pressure and temperature, the change in the Gibbs free energy of the system is given by: ΔG = ΔH ΔS Gibbs realized that free energy change is a quantitative indicator of spontaneity for all systems at constant pressure and temperature. ΔG < 0, reaction is spontaneous ΔG > 0, reaction is not spontaneous (spontaneous in reverse direction) ΔG = 0, system is at equilibrium hee methods to calculate standard free energy change: 1. ΔG = ΔH ΔS 2. Manipulation of know equations 3. ΔG reaction = Σ nδg products Σ nδg products 8. Using the data for ΔH and ΔS, calculate ΔG for the following reactions at 25 C and 1 atm. Cr 2 O 3 (s) + 2Al (s) Al 2 O 3 (s) + 2Cr (s) ΔH [Cr 2 O 3 (s)] = 1128 kj/mol; ΔH [Al 2 O 3 (s)] = 1676 kj/mol ΔS [Cr 2 O 3 (s)] = 81 J/K.mol; ΔS [Al (s)] = 28 J/K.mol; ΔS [Al 2 O 3 (s)] = 51 J/K.mol; ΔS [Cr (s)] = 24 J /K.mol 9. Given the following data, calculate ΔG for S (s) + O 2 (g) SO 2 (g) 1. S (s) O 2 (g) SO 3 (g) ΔG = 371 kj/mol 2. 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔG = 142 kj/mol 10. Calculate ΔG for the reaction C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O (g) ΔG f [C 3 H 8 (g)] = 24 kj/mol; ΔG f [CO 2 (g)] = 394 kj/mol; ΔG f [H 2 O (g)] = 299 kj/mol Page 4
5 Free Energy and he Equilibrium Constant When a substance is not in its standard state, its Gibbs free energy depends on its concentration (or its pressure, in the case of gases). he quantitative expression for the Gibbs free energy change for any reaction is: ΔG = ΔG + R lnq ΔG = Gibbs free energy change under non standard state conditions ΔG = standard Gibbs free energy change R = ideal gas constant (8.314 J/K mol), = temperature in Kelvins Q = reaction quotient 11. Calculate ΔG at 700 K for the following reaction: C (s, graphite) + H 2 O (g) CO (g) + H 2 (g) he initial pressures are P[H 2 O (g)] = 0.85 atm; P[CO (g)] = 1.0 x 10 4 atm and P[H 2 (g)] = 2.0 x 10 4 atm. ΔG f [H 2 O (g)] = 229 kj/mol; ΔG f [CO (g)] = 137 kj/mol At equilibrium, ΔG = 0 and the reaction quotient becomes the equilibrium constant (Q = K) he standard Gibbs free energy change relates to temperature and the equilibrium constant by: 0 = ΔG + R lnk or ΔG = R lnk 12. Consider the ammonia synthesis reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔG = 33.3 kj/mol of N 2 consumed at 25 C. At 25 C, P[NH 3 (g)] = 1.00 atm; P[N 2 (g)] = 1.00 atm; P[H 2 (g)] = 1.00 atm. Predict the direction in which the system will shift to reach equilibrium. 13. For the following reaction at 25 C: Cr 2 O 3 (s) + 2Al (s) Al 2 O 3 (s) + 2Cr (s) ΔG = 537 kj. Calculate K. 14. Consider Br 2 (l) Br 2 (g) ΔH = 31.0 kj/mol and ΔS = 93.0 J/K.mol What is the normal boiling point of liquid Br 2? Chemical reactions differ widely in their responses to changing temperature. the direction of spontaneity is determined by the sign of ΔG. the equilibrium constant depends on temperature. Since ΔG = R lnk and ΔG = ΔH ΔS Combining the two equations: lnital K= H R + S R ΔH Page 5
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