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14.42: QRT Butane (B) and 2-methypropane (MP) can convert from one isomer to the other. At 25oC, KC = 2.5 for this reaction. If 0.10 mole of each isomer were put in a 1.0 L container at 25oC, what would the MP concentration be at equilibrium? Report [MP] = A) 0.10 M B) 2.5 M C) 0.043 M D) 0.14 M KC = [MP] / [B] = 2.5 1 To solve these kinds of problems, construct a table of amounts below the reaction equation. Initial conc conc chge eq conc 0.10 M -x 0.10-x 0.10 M +x 0.10+x Q = 1, K > 1 so reaction will produce more product K = (0.10+x) / (0.10-x) = 2.5 x = 0.0429 [MP] = 0.10 + 0.0429 = 0.14 M 2 Le Chatelier's Principle (not covered in lecture) MSJ text, pp. 695 703. Foundations of Chemistry #31, pp. 215 220. Ch. 14 QRT Homework 55, 57, 59, 67, 71, 73 3 What determines the magnitude of an equilibrium constant, K We found that K was determined by a ratio of rate constants involving forward and reverse reactions. We also found that K is related to two energy factors for a reaction: the enthalpy change and the entropy change. - RT ln(K) = Go = Ho T So We connected these two ideas by using the Arrhenius equation, where Ea is the activation energy (i.e. free energy of activation: Ea = Goa. k = A e-Ea/RT 4 RT ln( K ) G o G o RT H o 1 S o ln( K ) R T R ln( K ) G = H - TS Equation for a straight line! The enthalpy effect says that the reaction favors the species (reactants or products) that have the lower energy. The entropy effect says that the reaction favors the species (reactants or products) that disperse the energy over the larger number of states. The importance of the entropy effect increases as the temperature increases. 5 For each reaction predict whether K >1 (product favored) or K < 1 (reactant favored). H o 1 S o ln( K ) R T R 2NO + O2 2NO2 K>1 K<1 Can't tell exothermic Enthalpy favors products, entropy favors reactants Expect reactants to be favored at high T. 6 For each reaction predict whether K >1 (product favored) or K < 1 (reactant favored). H o 1 S o ln( K ) R T R 2O3(g) 3O2(g) exothermic A) K >1 B) K<1 C) Can't tell N2(g) + 3Cl2(g) 2NCl3(g) endothermic A) K >1 B) K<1 C) Can't tell 7 How do plants and people get the nitrogen they need to produce proteins and nucleic acids? Nitrogen (N2) is abundant in the atmosphere. But N2 is very stable and unreactive. Ammonia (NH3) is more reactive and therefore a more useful form of nitrogen. How can nitrogen be converted to ammonia? N2 + 3H2 2NH3 The industrial Haber-Bosch process requires high pressures (200 atm) and high temperatures (500oC). Why? Legumes (plants like peas, beans, peanuts, soybeans) do this conversion in your garden (1 atm, 22oC). How? 8 World Production of Ammonia Industrial Haber Bosch process Invented during World War 1 to produce nitrate explosives 120,000,000,000 kg per year (30% of total) That's 120 billion! Legumes / Rhizobia / Nitrogenase 240,000,000,000 kg per year (60%) Lightning Combustion and 40,000,000,000 kg per year (10%) 9 Why use 200 atm and 500oC? For the reaction N2(g) + 3H2(g) 2NH3(g): Ho = - 92 kJ/mol, So = - 199 J/mol K, Go= - 32 kJ/mol Calculate values for the equilibrium constant at 300 K and at 500 K. R = 8.314 J/mol K G o RT H o 1 S o ln( K ) R T R ln( K ) What did you get? A) 300 K = 3.7 x 105 and 500 K = 2.2 x 103 B) 12.8 and 7.70 C) 12 x 10-3 and 7.7 x 10-3 D) 2.55 and 7.70 Want K to be large, raising the temperature reduces K! 10 Why use 200 atm and 500oC? For the reaction N2(g) + 3H2(g) 2NH3(g): Ho = - 92 kJ/mol, So = - 199 J/mol K, Go= - 32 kJ/mol G o RT H o 1 S o ln ( K ) R T R ln ( K ) 300 K = 3.7 x 105 and 500 K = 2.2 x 103 Reaction is exothermic, that's good! Entropy effect favors reactants, so want low T! But at low T the reaction rate is slow because the activation energy is high! k = A e-Ea/RT It requires a lot of energy to break the triple bond in N2. (Bond energy = 945 kJ/mol) 11 Why use 200 atm and 500oC? N2(g) + 3H2(g) 2NH3(g) So the following reaction conditions are needed. (Haber - Bosch Process). High T to compensate for the high Ea and slow rate. High gas pressures to drive the reaction to the right: LeChatelier's principle! A flow system to continuously removes the ammonia. Use catalysts to lower the activation energy. But the catalysts are not as good as nature's! 12 Nature Does Better! The enzyme, nitrogenase, catalyzes the reaction N2 to NH3. This enzyme is only found in certain bacteria, called Rhizobia. Rhizobia are found in root nodules of legumes. Farmers rotate soy beans and corn from one year to the next to take advantage of this free source of fertilizer. 13 Nitrogenase Nitrogenase is a catalyst that lowers the activation energy for converting N2 into NH3 so it can be done at garden temperatures. Despite intense efforts the mechanism of N2 fixation by nitrogenase is not completely understood so it has not been possible to duplicate it for industrial use. The structure of nitrogenase was determined by two former Stony Brook professors, Caroline Kisker and Herman Schindelin, when they were post docs at Caltech. 14 Active Site of Nitrogenase Fe7MoS9 Fe = orange S = yellow Mo = dark blue N O = red N = light blue H = black NH3 15 Can cars run on coal, manure, and other biomass? C(s) + O2(g) CO2(g) Ho = -394 kJ/mol, So = 3 J/mol K, Go = -394 kJ/mol Yes, but there are problems. Pollutants like Hg and S are released. Not easily transported and burned. But Coal is abundant. Gasification and liquefaction Removes pollutants Resolves issues with transportation, storage, and burning. 16 What is required to improve the utilization of coal and biomass as an energy source? Gasification Liquefaction Reforming for use in fuel cells Oxidation in fuel cells Key Questions What temperatures are needed for these reactions? How much free energy can be extracted from them? Is it feasible?

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