Practice Third Hour Exam for 5.111 FALL 2014 _________________________________________________________________________________________________________________ Write your name and your TA's name below. Do not open the exam until the start of the exam is announced. The exam is closed notes and closed book. 1. Read each part of each problem carefully and thoroughly. 2. Read all parts of each problem. MANY OF THE LATTER PARTS OF A PROBLEM CAN BE SOLVED WITHOUT HAVING SOLVED EARLIER PARTS. However, if you need a numerical result that you were not successful in obtaining for the computation of a latter part, make a physically reasonable approximation for that quantity (and indicate it as such) and use it to solve the latter parts. 3. A problem that requests you to “calculate” implies that several steps may be necessary for the problem’s solution. You must show these steps clearly and indicate all values, including physical constants used to obtain your quantitative result. Significant figure and unit usage must be correct. 4. If you don’t understand what the problem is requesting, raise your hand and a proctor will come to your desk. 5. Physical constants, formulas and a periodic table are given on the last page. You may detach this page once the exam has started. _________________________________________________________________________________________________________________ Thermodynamics 1a, 1b (18 points) page 2 ____________ Thermodynamics and Chemical Equilibrium 1c, 1d, 2a (18 points) page 3 ____________ Chemical Equilibrium 2b (20 points) page 4 ____________ Solubility 3a, 3b (12 points) page 5 ____________ Acid/Base 4a (12 points) page 6 ____________ 4b, 4c (8 points) page 7 ____________ 4d (12 points) page 8 ____________ Total (100 points) _________________ Name ___________________________________ TA ___________________________________ 1. (30 points) Thermodynamics (a) Consider the following reaction: ClF(g) + F2(g) → ClF3(g) (i) (8 points) Given the following data, calculate the ΔH° for the reaction (ΔHr°) above. Report your answer in units of kJ per mol of ClF3 formed. 2ClF(g) + O2(g) → Cl20(g) + F2O(g) 2ClF3(g) + 2O2(g) → Cl20(g) + 3F2O(g) 2F2(g) + O2(g) → 2F2O(g) ΔHr° (in kJ for each reaction as written) 167.4 341.4 -‐43.4 We can apply Hess’ Law to calculate ΔH°. ½[2F2(g) + O2(g) → 2F2O(g)] ½ [-‐43.4 kJ] = -‐21.7 kJ ½[2ClF(g) + O2(g) → Cl20(g) + F2O(g)] ½ [167.4 kJ] = ½[ Cl20(g) + 3F2O(g) → 2ClF3(g) + 2O2(g)] _________________________________________________ -‐½ [341.4 kJ] = -‐170.7 kJ __________________________________ ClF(g) + F2(g) → ClF3(g) 83.7 kJ -‐108.7 kJ ΔH = -‐108.7 kJ per mol of ClF3(g) formed (ii) (4 points) Would you expect ΔS° for the reaction in part (a) to be positive, negative, or zero. Briefly justify your choice. ΔS° would be negative. The disorder is decreasing (from 2 mol to 1 mol of gas). (b) (6 points) Fill in the signs as (-‐) or (+) for ΔH° and ΔS° to correctly complete the table below. ΔH + -‐ + ΔS -‐ -‐ + Spontaneity of reaction never spontaneous spontaneous only at low temperatures spontaneous only at high temperatures (c ) (8 points) Calculate the temperature at which it is thermodynamically possible for carbon to reduce iron (III) oxide to iron under standard conditions by the following reaction: 2 Fe2O3 (s) + 3 C (s) à 4 Fe (s) + 3 CO2 (g) ∆Sr° = +558.4 J K-‐1 Note that the ∆Hf° of Fe2O3 (s) is -‐824.2 kJ/mol and the ∆Hf° for of CO2 is -‐393.5 kJ/mol ∆Hr˚ = ∆Hf˚(products) -‐ ∆Hf˚(reactants) = [4 mol(0 kJ/mol) + 3 mol(-‐393.5 kJ/mol)] – [2 mol(-‐824.2 kJ/mol) + 3 mol(0 kJ/mol)] = 467.9 kJ ∆Gr˚ = ∆Hr˚ -‐ T∆Sr˚ = 0 (at equilibrium) T∆Sr˚ = ∆Hr˚ T = ∆Hr˚/∆Sr˚ = 467,900 J/558.4 J = 837.9 K (d) (4 points) Given that CO2 has a ∆Gf° of -‐394.4 kJ. Predict whether CO2 is more stable or less stable relative to its elements. Briefly explain your answer. CO2 is more stable with respect to its elements. Since the ∆Gf˚ is negative, the formation of CO2 is stable, therefore the decomposition in to its elements is disfavored. 2. (26 points) Chemical Equilibrium (a) (6 points) The compound 1,3-‐di-‐t-‐butylcyclohexane exists in two conformations that are known as the chair and the boat conformations, because their structures resemble those objects. An equilibrium exists between these two forms, represented by the equation: chair boat At 580 K, 6.42% of the molecules are in the chair form. Calculate the equilibrium constant for the preceding reaction as written at 580 K. If 6.42% of the compound is in the chair conformation, then 93.58% of the compound is in the boat conformation. The percentages are proportional to the concentrations of the specific conformation. K= [boat]/[chair] = 93.58/6.42 = 14.6 at 580 K (b) Consider the equilibrium 2 P2 (g) P4 (g) State the direction (toward product, toward reactant, no change) that the equilibrium will shift when each of the following changes occurs at constant temperature. Briefly explain your answers. (i) (5 points) P2 (g) is added Toward products. When excess reactant is added, Q < K, and forward direction will be spontaneous (or) When excess reactant is added, the system will shift toward products until the excess reactant is consumed. (ii) (5 points) P4 (g) is removed Toward products. When product is removed, Q < K, and forward direction will be spontaneous (or) When product is removed, the system will shift toward products until equilibrium is reached again. (iii) (5 points) the total pressure is increased by addition of N2(g) No Change. If the total pressure is increased by addition of an inert gas, then the volume didn’t change. If volume doesn’t change, then the partial pressures of gases didn’t change. (iv) (5 points) the volume is increased Toward Reactant. When the volume increases, the pressure decreases, so the system responds in such a way to increase the total pressure. Thus the equilibrium shifts from fewer moles (1) to more moles (2) (toward reactant) to increase the total pressure. ‘ 3. (12 points) Solubility Hydrocarbons can have limited solubility in water partially due to entropic effects. Thus researchers were surprised when a particular hydrocarbon was found to be soluble in water, especially since solvent cage formation was known to accompany the dissolving process, decreasing the entropy of the system. (a) (6 points) Given that this hydrocarbon is soluble in water, explain what must be true about the sign and magnitude of the enthalpy of solution for this hydrocarbon. If the hydrocarbon is soluble, ∆G must be negative. For ∆G to be negative, the sign of ∆Hsol must be negative and the magnitude must be larger than T∆S (∆G = ∆H -‐ T∆S). (b) (6 points) Predict the effect, if any, that increasing temperature will play in the spontaneity of the dissolving process for this hydrocarbon. Briefly explain your answer. Increasing temperature is likely to decrease spontaneity (or) decrease dissolution (or) decrease solubility. Since ∆S is negative, increasing temperature will increase the magnitude of the T∆S term, which will make ∆G a smaller negative number or even a positive number. 4. (32 points) Acid/Base 20.00 mL of a 0.1000 M solution of the weak base ammonia (NH3) is titrated with a 0.1000 M solution of the strong acid HCl at 25°C. The base ionization constant (Kb) for NH3 is 1.8 x 10-‐5 at 25°C. (a) (12 points) Calculate the pH of the solution after addition of 5.00 mL of HCl titrant. You can assume that [OH-‐] at equilibrium satisfies the 5% rule without checking. THIS IS LIKE A BUFFER PROBLEM Option 1: 20.00 x 10-‐3 L x 0.1000 M = 0.002000 mol NH3 (Initial) 5.00 x 10-‐3 L x 0.1000 M = 0.000500 mol HCl added 0.002000 mol NH3 -‐ 0.000500 mol HCl added = 0.001500 mol NH3 left 0.000500 mol NH4+ formed Volume = 25.00 x 10-‐3 L [NH3] = 0.001500 mol/25.00 x 10-‐3 L = 0.06000 M [NH4+] = 0.000500 mol/25.00 x 10-‐3 L = 0.0200 M Initial Change at equilibrium NH3 + H2O 0.06000 –x 0.06000 – x Kb = 1.8 x 10-‐5 = (0.0200 + x )(+x) ≈ (0.0200)(+x) (0.06000 – x ) 0.06000 NH4+ + 0.0200 +x 0.0200 + x x = 5.4 x 10-‐5 pOH = -‐ log [OH-‐] = -‐ log (5.4 x 10-‐5) = 4.267 pH = 14.00 – pOH = 14.00 – 4.267 = 9.73 Option 2: 20.00 x 10-‐3 L x 0.1000 M = 0.002000 mol NH3 (Initial) 5.00 x 10-‐3 L x 0.1000 M = 0.000500 mol HCl added 0.002000 mol NH3 -‐ 0.000500 mol HCl added = 0.001500 mol NH3 left 0.000500 mol NH4+ formed Volume = 20.00 x 10-‐3 L + 5.00 x 10-‐3 L = 25.00 x 10-‐3 L [NH3] = 0.001500 mol/25.00 x 10-‐3 L = 0.06000 M [NH4+] = 0.000500 mol/25.00 x 10-‐3 L = 0.0200 M pH ≈ pKa – log ([HA]/[A-‐]) pH = 9.255 – log (0.0200/0.06000) pH = 9.255 – log (0.3333) pH = 9.255 – (-‐0.4771) = 9.732 = 9.73 Ka = Kw/Kb = 1.0 x 10-‐14/1.8 x 10-‐5 = 5.56 x 10-‐10 pKa = -‐log Ka = -‐log 5.56 x 10-‐10 = 9.255 OH-‐ +x +x (b) (5 points) Calculate the pH at the half-‐equivalence (half-‐stoichiometric) point. pH ≈ pKa at half-‐equivalence point Ka = Kw/Kb = 1.0 x 10-‐14/1.8 x 10-‐5 = 5.56 x 10-‐10 pKa = -‐log Ka = -‐log 5.56 x 10-‐10 = 9.255 pH = 9.26 or 9.25 (c ) (3 points) Calculate the volume of strong acid (HCl) titrant needed to reach the equivalence (stoichiometric) point. 20.00 mL x 0.1000 M = (VOL)(0.1000 M) VOL = 20.00 mL (d ) (12 points) Calculate the pH to 2 decimal places at the equivalence (stoichiometric) point. Check any assumption. THIS IS LIKE A WEAK ACID IN WATER PROBLEM. 20.00 x 10-‐3 L x 0.1000 M = 0.002000 mol NH3 (Initial) At equivalence point, all NH3 has been converted to NH4+ Volume = 20.00 x 10-‐3 L + 20.00 x 10-‐3 = 40.00 x 10-‐3 L [NH4+] = 0.002000 mol/40.00 x 10-‐3 L = 0.05000 M Initial Change at equilibrium NH4+ + H2O 0.05000 –x 0.05000 – x NH3 Ka = Kw/Kb = 1.0 x 10-‐14/1.8 x 10-‐5 = 5.56 x 10-‐10 Ka = 5.56 x 10-‐10 = (x2) (0.05000 – x ) ≈ (x2) 0.05000 x = 5.27 x 10-‐6 Assumption is valid: 5.27 x 10-‐6/0.05000 x 100 % = 0. 0105% (<5%) pH = -‐ log [H3O+] = -‐ log (5.27 x 10-‐6) = 5.28 +x +x + H3O+ +x +x

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