Version137–Exam1TThONLY–sutcliffe–(52200) 1 This print-out should have 27 questions. the vapor pressure of octanol is less than 1 Multiple-choice questions may continue on Torr at room temperature while methanol is the next column or page – find all choices more than 100 Torr. before answering. 003 10.0 points 001 10.0 points For a reaction at equilibrium, which of the Which of the species would be expected to following is the best way to think about it? be the most miscible in benzene (C6H6)? 1. The concentration of reactants and prod- 1. silicon dioxide, SiO2 ucts are not related.
2. methanol, CH3OH 2. forward rate = reverse rate correct
3. hydrogen fluoride, HF 3. The concentration of reactants becomes negligible. 4. sugar, C6H12O6 4. The concentration of intermediates be- 5. decane, C10H22 correct comes negligible.
Explanation: 5. concentration of reactants = concentra- “Like dissolves like.” Benzene is highly tion of products non-polar and therefore the most miscible species will be the least polar. In this case, Explanation: that is decane. In chemical systems, equilibrium is dy- namic and both foward and reverse processes 002 10.0 points continue to occur, but at the same rate. Methanol (CH3OH) has a higher vapor pres- sure than octanol (C8H17OH) because 004 10.0 points Why does it take longer to hard boil an egg 1. methanol has weaker dispersion forces in boiling water at higher altitude than at or than octanol correct below sea level?
2. methanol exhibits hydrogen bonding and 1. The boiling point goes down. correct octanol does not 2. There is an increase in atmospheric pres- 3. methanol has stronger dispersion forces sure. than octanol 3. It does not take longer. 4. methanol is non-polar and octanol is po- lar 4. The boiling point increases.
5. octanol exhibits hydrogen bonding and 5. The ambient temperature is lower at methanol does not higher altitude. Explanation: Explanation: Methanol and octanol both are polar and The normal boiling point of a liquid will oc- exhibit hydrogen bonding as a result of the cur when its vapor pressure reaches 760 torr (1 -OH group in the molecule. However, octanol atm). By normal, we mean “normal” condi- is a larger molecule and the long carbon chain tions; i.e., 1 atm pressure. In general, liquids has much larger dispersion forces. As a result will boil when their vapor pressure reaches the Version137–Exam1TThONLY–sutcliffe–(52200) 2 applied (in this case, atmospheric) pressure. soda, which means more gas dissolves in it, so At higher altitudes, atomospheric pressures less escapes when the bottle is opened. are less. Thus the observed boiling point of Explanation: water in, say, Denver, will be less than the fa- ◦ miliar 100 C. Since the temperature at which Soda is carbonated by injecting carbon water boils is less under these conditions, it dioxide gas in under pressure. All gases will take longer to hard boil your egg. are more soluble at lower temperatures so a chilled soda will have less gas in the headspace 005 10.0 points above it and will have less gas evolve from so- The correct units for the equilibrium constant lution as the bottle is opened. The bottle does not appreciably change its volume on chilling. Kp for the reaction
N2 +3H2 ↽⇀ 2NH3 007 10.0 points are Calculate the standard free energy of the reaction 1. atm2. CCl3COOH(aq) + H2O(ℓ) ↽⇀ − + CCl3CO (aq)+ H3O (aq) , 2. atm. 2
− K = 0.30 at 298 K. 3. atm 1.
− 1. 19.1 kJ/mol 4. atm 2. − 2. 2.98 kJ/mol correct 5. None; Kp is dimensionless. correct Explanation: 3. 75.5 kJ/mol The unit of the components that define K is 4. 20.0 kJ/mol activity, which is the ratio of the measurement − to the standard state unit of the component, and so activity has no unit. 5. 68.2 kJ/mol
006 10.0 points 6. 2.00 kJ/mol
Why is it that the contents of a bottle of 7. 23.2 kJ/mol soda opened at room temperature tend to − expand all over the persons hand’s holding Explanation: the bottle, but the same bottle if opened when ◦ chilled just makes a slight hissing sound? ∆G = R T ln K r − 1. The nitrogen gas intentionally dissolved J = 8.314 (298 K) ln(0.30) in the soda is more soluble at lower tempera- − K mol · tures. = 2982.93 J/mol 2. The bottle allows more gas to expand through the pores of the plastic at lower tem- 008 10.0 points peratures. The equilibrium 3. The carbon dioxide gas intentionally dissolved in the soda is more soluble at lower N (g)+O (g) 2 NO(g) temperatures. CORRECT 2 2 ←→ 4. The bottle shrinks at low temperatures, exists in a closed container: If the partial creating less headspace for the gas above the pressures of N2, O2 and NO at equilibrium are Version137–Exam1TThONLY–sutcliffe–(52200) 3 1.5 atm, 2.5 atm and 0.75 atm, respectively, what is the equilbrium constant?
1. K = 6.67
2. K = 0.2
3. K = 0.15 correct
4. K=5
Explanation:
P 2 0.752 K = NO = =0.15 P P 1.5 2.5 N2 · O2 ·
009 10.0 points ◦ At 25 C the value of Kc for a particular re- ◦ action is 30. At 75 C the value of Kc for the same reaction is 40. Which one of the following statements about this reaction is necessarily true?
1. Increasing the pressure would decrease the value of Kc. 011 10.0 points 2. LeChatelier’s Principle does not apply to A solution of sucrose (C12H22O11 of MW 342 this reaction. grams/mole) in benzene is 0.40 molal. The so- lution’s density is 0.90 grams/mL. How many 3. The reaction is not at equilibrium. milliliters of the solution would contain 0.10 moles of sucrose? 4. The reaction is exothermic. 1. 225 mL 5. The reaction is endothermic. correct 2. 360 mL
Explanation: 3. 444 mL Endothermic reactions have a larger value of K at higher temperatures. 4. 316 mL correct
010 10.0 points 5. 250 mL Version137–Exam1TThONLY–sutcliffe–(52200) 4 Explanation: m =0.4 m MW = 342 g/mol 013 10.0 points nC12H22O11 =0.10mol density =0.9g/mL Both ammonia and phosphine (PH3) are sol- Thus the total mass of solution is uble in water. Which is least soluble and why? msolvent +msolute =1kg+(0.4 mol)(342 g/mol) 1. ammonia; it is too small to be hydrated by water molecules. = 1kg+136.8g =1.1368 kg 2. phosphine; it does not form hydrogen bonds with water molecules. correct Find the volume of this solution: m density = 3. ammonia; it does not form hydrogen V bonds with water molecules. m V = density 4. phosphine; the P H bonds are so strong 1.1368 kg − = that they cannot break to enable phosphine 0.9 kg/L to hydrogen-bond with water. =1.263L = 1263mL 5. ammonia; the N H bonds are so strong Since the solution is 0.4 molal, this value of that they cannot break− to enable the ammonia 1236 mL contains 0.4 moles. to hydrogen-bond with water. The volume containing 0.1 moles is 1/4 of this amount: Explanation: 1263 mL PH3 is least soluble because it cannot form = 315.8 mL 4 hydrogen bonds with water. 014 10.0 points 012 10.0 points Write the equilibrium expression for the reac- The equilibrium constant for the reaction tion 2NO(aq)+ 4H2O(ℓ) 4NH3(g)+5O2(g) 4NO(g)+6H2O(g) ←→+ − → 3H2(aq)+2H (aq)+2NO3 (aq) is K. Calculate the equilibrium constant for − [H+]2 [NO ]2 the reaction 1. K = · 3 [NO]2 5 − 2NO(g)+3H O(g) 2NH (g) + O (g) . [H ]3 [H+]2 [NO ]2 2 3 2 2. 2 3 → 2 K = · 2 · 4 [NO] [H2O] · − 1 [H ]3 [H+]2 [NO ]2 1. 3. K = 2 · · 3 correct K [NO]2 2 2. 2 K [NO] 4. K = − − [H+]2 [NO ]2 · 3 3. K [NO]2 − 5. K = 3 + 2 − 2 K [H2] [H ] [NO3 ] 4. · · 2 − Explanation: 1 5. correct Set up K, products in the numerator, reac- √K tants in the denominator, all raised to respec- Explanation: tive stoichiometric coefficients. Version137–Exam1TThONLY–sutcliffe–(52200) 5 freezing point of the resulting solution is ◦ 015 10.0 points 3.77 C. The freezing point of pure benzene is ◦ An unknown liquid has a vapor pressure of 88 5.48 C, and its freezing point depression con- ◦ ◦ ◦ mmHg at 45 C and 39 mmHg at 25 C. What stant is Kf = 5.12 C/molality. What is the is its heat of vaporization? molecular weight of the unknown compound?
1. 32000 kJ/mol 1. 160 grams/mole
2. 2000 J/mol 2. 100 grams/mole
3. 2000 kJ/mol 3. 120 grams/mole correct
4. 32 kJ/mol correct 4. 80.0 grams/mole Explanation: 5. 140 grams/mole Use the Clausius-Clapeyron Equation. Here, the only thing we don’t know is ∆H . Explanation: vap ◦ munknown =20g Tf =3.77 C 0 ◦ mbenzene = 500 g Tf =5.7 C P2 ∆Hvap 1 1 ln = ∆Tf = Km where ∆Tf is the freezing point P R T − T 1 1 2 (FP) depression, Kf is the FP depression con- P2 stant, and m is the molarity. R ln P1 ∆Hvap = = 32.0 kJ/mol mol solute 20.0(g/MW) 1 1 = kg solvent 0.500 kg T1 − T2 ∆Tf = Kf m 016 10.0 points ◦ ∆Tf Which combination of ∆G and K is possible m = Kf at standard conditions? 20.0(g/MW) 1.71◦C ◦ = 1. ∆G = 96 J, K =0.962 correct 0.500 kg 5.12◦C/(mol/kg) ◦ 2. ∆G = 96.6 kJ, K =1.04 20.0 g(5.12◦C kg/mol) MW = ◦ · ◦ − 1.71 C 0.0500 kg 3. ∆G = 98.3 J, K =1.66 10 17 × = 120 g/mol − × ◦ ◦ ∆Tf = 5.48 3.77 = 1.71 C 4. ∆G = 82 kJ, K =0.961 ◦ − − Kf = 5.12 C/molarity = moles/kg solute ◦ 5. ∆G = 98.5 J, K =1.76 1014 × 018 10.0 points Explanation: Assuming all of the following salts dissolve ◦ If ∆G is positive, K must be less than 1; completely in water, which one would be the ◦ if ∆G is negative, K must be greater than 1. best to use if you were trying to raise the ◦ Also, if ∆G is small (compared to RT), then boiling point of the solution? ◦ K will be close to 1, and if ∆G is large, K will be much smaller or greater than 1. 1. Na3PO4
017 10.0 points 2. KNO3 When 20.0 grams of an unknown compound are dissolved in 500 grams of benzene, the 3. Al2(SO4)3 correct Version137–Exam1TThONLY–sutcliffe–(52200) 6 If ammonia has a pure vapor pressure of 4. (NH4)2Cr2O7 1515 torr and propanol has a pure vapor pres- sure of 38 torr, a mixture of 2 moles of ammo- 5. NaCl nia and 6 moles of propanol would have what Explanation: total vapor pressure? Al2(SO4)3 is highly soluble and produces 5 ions upon dissolving, has the largest van’t 1. 776 torr hoff coefficient and is the best choice for at- tempting raise a solution’s boiling point. 2. 1164 torr
019 10.0 points 3. 1145 torr How many grams of Ca are contained in 113 4. 407 torr correct mL of 0.52 M Ca(OH)2 solution? 5. 1. 2.36 g correct 388 torr Explanation: 2. 8.71 g P = χ P + χ P + ... total a · a b · b 3. 1.18 g P =0.25 1515 torr + 0.75 38 torr total · · Ptotal = 407 torr 4. 4.71 g
5. 0.0588 g 022 10.0 points Explanation: The figure represents a reaction at 298 K.
VCa(OH)2 =113mL [Ca(OH)2] = 0.52 M bA B b ?gCa=0.113L E b G b 0.52 mol Ca(OH)2 b × L soln C D 1 mol Ca 40.08 g Ca × 1 mol Ca(OH)2 × 1 mol Ca =2.36 g Ca rxn progress Which statement is true? 020 10.0 points 1. At point B, the reaction will shift to the Water vapor condensing to liquid water is right to reach equilibrium. correct what type of process? 2. The reactants possess less free energy 1. endothermic than the products.
2. exothermic correct ◦ 3. The ∆G of reaction is zero at point C. 3. enthalpic 4. K is less than 1.
4. isotopic Explanation: Explanation: The reaction continues to shift to the right at point B because more products are needed 021 10.0 points to get to equilibrium point C. K is greater Version137–Exam1TThONLY–sutcliffe–(52200) 7 ◦ than 1 because ∆G is negative (point E is lower free energy than point A). Point C iden- 3. Kp = RTKc correct tifies where ∆ G is zero, whereas the differ- ◦ ence between points E and A identify ∆G . 4. Kc = RTKp Because point E is below point A, we know 2 the products have lower free energy than the 5. Kc =(R T ) Kp reactants. Explanation: 023 10.0 points 025 10.0 points What is the osmotic pressure of a solution − Which of the following would not result in that contains 4.56 10 3 moles of lactose in ◦ a phase transition? 100 mL of solution× at 25 C? 100 1. 848 torr correct 80 2. 113 torr b 60 3. 1053 torr 40 4. 71 torr Pressure, atm 20 5. 536 torr Explanation: 0 − n =4.56 10 3 mol V = 100 mL 0 200 400 600 800 1000 ◦ × T = 25 C+273.15 = 298.15 K Temperature, K 456 10−3 mol 1000 mL M = × 100 mL L − mol =4.56 10 2 1. at 275 K, increasing the pressure from 5 × L atm to 50 atm Osmotic pressure 2. at 10 atm, raising the temperature from π = M R T 0Kto275K − = 4.56 10 2 mol/L × 3. at 800 K, decreasing the pressure from 90 L torr to 20 atm correct 62.36 · (298.15 K) × K mol · = 847.8 torr 4. at 45 atm, lowering the temperature from 475 K to 50 K 024 10.0 points Explanation: Consider the reaction 026 10.0 points 2 HgO(s) 2Hg(ℓ) + O (g) . → 2 Suspensions of colloids can be discrimi- What is the relationship between K and p nated from solutions becuase the former Kc? 1. exist as gels 1. K = Kc 2. can scatter light CORRECT 3. are highly colored 2 2. Kp =(R T ) Kc 4. only exist at very high pressures Version137–Exam1TThONLY–sutcliffe–(52200) 8 Explanation: Colloidal suspensions scatter light (text looked at through the solution would appear blurred). This is called the Tyndall effect.
027 10.0 points The molar heat capacity of some molecule X(ℓ) is 10 cal/K mol, its heat of vaporization ◦ is 5000 cal/mol and· its boiling point is 75 C. ◦ For the conversion of one mol of X(g) at 75 C ◦ to one mol of X(ℓ) at 60 C,
1. 4850 cal of heat are released by X.
2. 5150 cal of heat are released by X. cor- rect
3. 150 cal of heat are released by X.
4. 5150 cal of heat are absorbed by X.
5. 4850 cal of heat are absorbed by X. Explanation: