chem exam 1 - Version 275 Exam 1 Holcombe(52460 This...

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Version 275 – Exam 1 – Holcombe – (52460) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Helpful constants and conversions 1 atm = 760 torr = 760 mm Hg 0 degC = 273.15 K R = 8.314 J/K*mol R = 0.082 L*atm/K*mol 001 10.0 points The observation that the solubility of gases in water decreases with increasing temper- ature means that the enthalpy of solvation H solvation )for gases is 1. exothermic correct 2. endothermic 3. zero 4. greater than the change in entropy Explanation: A high velocity gas molecule must release its excess kinetic energy as heat in order to decelerate and enter the aqueous phase. 002 10.0 points K c = 0 . 040 for the system below at 450 C. PCl 5 (g) PCl 3 (g) + Cl 2 (g) Evaluate K p for the reaction at 450 C. 1. 5 . 2 × 10 2 2. 0.40 3. 6 . 7 × 10 4 4. 0.64 5. 2.4 correct Explanation: R = 0 . 0821 L · atm / mol · K T = 450 C + 273 = 723 K Δ n = n gas products - n gas reactants = 2 - 1 = 1 K p = K c ( RT ) Δ n = (0 . 040)(0 . 0821)(723) = 2 . 37433 003 10.0 points The colligative effects of 1 molal sugar solu- tion would be ? 1 molal sodium chloride solution. 1. greater than 2. the same as 3. less than correct Explanation: Sugar is a non-electrolyte so the concen- tration of particles is the concentration of dissolved sugar. Soduim chloride is an electrolyte which dis- sociates in solution, giving twice as many par- ticles as dissolved: NaCl -→ Na + + Cl Since colligative properties depend on the number of particles in solution, the colligative effect doubles for the NaCl(aq), compared to any identical concentration of sugar. 004 10.0 points What mass of ethylene glycol ((CH 2 OH) 2 with molecular weight 62 g/mol) must be added to 1.00 L of H 2 O (of mass 1 kg) to lower the freezing point to - 5 C? K f H 2 O = 1 . 86 C /m . 1. 25 g 2. 123 g 3. 167 g correct 4. 330 g 5. 66 g
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Version 275 – Exam 1 – Holcombe – (52460) 2 Explanation: MW CH 2 OH = 62 g/mol V H 2 O = 1 L m H 2 O = 1 kg T f = - 5 C T f = T 0 f - Δ T f Δ T f = T 0 f - T f = 0 . 00 C - ( - 5 C) = 5 C Δ T f = K f · m m = Δ T f K f = 5 C 1 . 86 C /m = 2 . 688 m m = n (CH 2 OH) 2 kg water = g (CH 2 OH) 2 MW (CH 2 OH) 2 kg water g (CH 2 OH) 2 = m (MW (CH 2 OH) 2 )(kg water) = 2 . 688 mol (CH 2 OH) 2 kg water × 62 . 0 g (CH 2 OH) 2 mol (CH 2 OH) 2 × (1 kg water) = 167 g (CH 2 OH) 2 005 10.0 points In order for an endothermic reaction to be spontaneous 1. nothing special is required; they are al- ways spontaneous. 2. the entropy increase in the system must equal the entropy decrease in the surround- ings. 3. endothermic reactions are never spontan- teous. 4. heat must be supplied to the system. 5. the entropy increase in the system must be greater than the entropy decrease in the surroundings. correct Explanation: The Second Law of Thermodynamics states that in spontaneous reactions, the universe tends toward a state of greater disorder.
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