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Exam 4-solutions

# Exam 4-solutions - Version 385 Exam 4 mccord(50970 This...

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Version 385 – Exam 4 – mccord – (50970) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The enthalpy of fusion of methanol (CH 3 OH) is 3.16 kJ/mol. How much heat would be absorbed or released upon freezing 25.6 grams of methanol? 1. 3.95 kJ absorbed 2. 0.253 kJ absorbed 3. 2.52 kJ released correct 4. 0.253 kJ released 5. 3.95 kJ released 6. 2.52 kJ absorbed Explanation: MW CH 3 OH = 32.042g ΔH(fusion) = - ΔH(freezing) q = 25 . 6 g × 1mol 32 . 042g × - 3 . 16 kJ mol = - 2 . 525 kJ 002 10.0 points The standard molar enthalpy of formation of ammonium chloride, NH 4 Cl (s), is - 314 . 43 kJ/mol. What is the standard molar internal energy of formation of ammonium chloride? 1. - 316 . 91 kJ/mol 2. +311 . 95 kJ/mol 3. - 314 . 43 kJ/mol 4. - 321 . 87 kJ/mol 5. - 309 . 47 kJ/mol 6. - 306 . 99 kJ/mol correct 7. - 319 . 39 kJ/mol Explanation: Δ H = - 314 . 43 kJ/mol T = 298 . 15 K (standard temperature) 1 2 N 2 (g) + 2 H 2 (g) + 1 2 Cl 2 (g) NH 4 Cl(s) n i = ( . 5 + 2 + . 5) mol = 3 mol n f = 0 mol Δ n = (0 - 3) mol = - 3 mol Δ E = q + w, q = Δ H and w = - P Δ V = - Δ nRT = - ( - 3 mol)(8 . 314 J / mol · K) × (298 . 15 K) parenleftbigg kJ 1000 J parenrightbigg = 7 . 43646 kJ Work is per 1 mole of N 2 H 4 formed, so Δ E = Δ H + w = - 314 . 43 kJ / mol + 7 . 43646 kJ / mol = - 306 . 994 kJ / mol 003 10.0 points Consider the dissolution of CaCl 2 : CaCl 2 (s) Ca 2+ (aq) + 2Cl (aq) Δ H = - 81 . 5 kJ A 13.1-g sample of CaCl 2 is dissolved in 130 g of water, with both substances at 25.0 C. Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of 4.18 J/g C. 1. 39.2 C 2. 41.1 C correct 3. 33.7 C 4. 57.4 C 5. 43.2 C

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Version 385 – Exam 4 – mccord – (50970) 2 Explanation: CaCl 2 = 111. g/mol 13.1/111. = 0.118 mol of CaCl2 which will release (.118)(81.5)= 9.62 kJ of heat when it dissolves. mass of solution = 13.1 + 130 = 143.1 g q = m · C s · Δ T Δ T = q/ ( m · C s ) Δ T = 9620 / (143 . 1 × 4 . 18) = 16 . 1 C T final = 25 . 0 + 16 . 1 = 41 . 1 C 004 10.0 points The sublimation of solid carbon dioxide is a spontaneous process. Predict the sign (+, - , or 0) of Δ G r , Δ H r , and Δ S r , respectively. 1. - , +, + correct 2. - , - , - 3. 0, +, + 4. - , 0, + 5. - , +, - Explanation: Δ G is negative for a spontaneous reaction. Sublimation requires energy to facilitate the solid becoming a gas, so the process is en- dothermic (Δ H is positive). Finally, the en- tropy of a gas is more than that of a solid, so disorder increases (Δ S is positive). 005 10.0 points Calculate Δ H for the formation of 2 . 72 moles of MgCl 2 at 298 K and 1 atm by the reaction Mg(OH) 2 (s) + 2 HCl(aq) MgCl 2 (s) + 2 H 2 O( ) . 1. 115.25 2. 125.392 3. 90.356 4. 33.192 5. 36.419 6. 56.703 7. 25.816 8. 99.576 9. 104.647 10. 120.321 Correct answer: 125 . 392 kJ. Explanation: Reactants: Δ H f Mg(OH) 2 (s) = - 924 . 7 kJ/mol Δ H f HCl(aq) = - 167 . 4 kJ/mol Products: Δ H f MgCl 2 (s) = - 641 . 8 kJ/mol Δ H f H 2 O(g) = - 285 . 8 kJ/mol Δ H 0 rxn = summationdisplay n Δ H 0 f prod - summationdisplay n Δ H 0 f rct = [2 ( - 285 . 8 kJ / mol) + ( - 641 . 8 kJ / mol)] - [2 ( - 167 . 4 kJ / mol) + ( - 924 . 7 kJ / mol)] = parenleftBig 46 . 1 kJ mol rxn parenrightBigparenleftBig 1 mol rxn 1 mol MgCl 2 parenrightBig × (2 . 72 moles MgCl 2 ) = 125 . 392 kJ 006 10.0 points Using the provided bond enthalpy data, cal- culate the change in enthalpy for the following reaction: CH 4 (g) + O 2 (g) ←→ CH 2 O(g) + H 2 O(g) 1. 577 kJ · mol 1 2. - 577 kJ · mol 1 3. - 349 kJ
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